4.4 Chemical changes — revision question pack

15 specification points · notes, questions, answers and worked methods

Checked against AQA 8462 section 4.4. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

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4.4.1.1 · Metal oxides

Explanation

A metal reacting with oxygen forms a metal oxide; the metal is oxidised because it gains oxygen.
Reduction is the loss of oxygen from a substance, while oxidation is the gain of oxygen.
For example, heating copper in oxygen forms copper oxide: $2\mathrm{Cu}+\mathrm{O_2}\rightarrow2\mathrm{CuO}$ .
Do not decide oxidation from the presence of oxygen alone: track whether oxygen is gained or lost by the named substance.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

A student heats $6.4\,\mathrm{g}$ of copper in oxygen and obtains $8.0\,\mathrm{g}$ of copper oxide. Calculate the mass of oxygen gained and state what has happened to the copper.

1.The oxygen gained accounts for the increase in mass: $8.0-6.4=1.6\,\mathrm{g}$ . Copper has gained that oxygen, so the copper has been oxidised.

Answer: $1.6\,\mathrm{g}$ of oxygen; the copper has been oxidised.

Common mistakes

Don't call gain of oxygen reduction instead of oxidation.
Don't describe oxygen as a catalyst even though it becomes part of the metal oxide.

Exam tip

For oxidation or reduction in terms of oxygen, state explicitly whether oxygen is gained or lost.

Tier 1 · Easy

Magnesium burns in oxygen to form $\mathrm{MgO}$ . Name the type of reaction and explain your choice in terms of oxygen.
[2 marks]
Total for this question: 2
A strip of calcium is heated in air and calcium oxide forms. State what happens to the calcium in terms of oxygen.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Copper powder gains mass when it is heated in air. Explain why its mass increases and state whether the copper is oxidised or reduced.
[2 marks]
Total for this question: 2
A $15.0\,\mathrm{g}$ sample of a metal oxide is reduced and leaves $12.0\,\mathrm{g}$ of metal. Calculate the mass of oxygen removed and explain why the change is reduction.
[3 marks]
Total for this question: 3
Write a balanced symbol equation for aluminium reacting with oxygen to form aluminium oxide, then explain why aluminium is oxidised.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Hydrogen is passed over hot copper oxide: $\mathrm{CuO}+\mathrm{H_2}\rightarrow\mathrm{Cu}+\mathrm{H_2O}$ . Identify what is oxidised and what is reduced, explaining each answer using oxygen transfer.
[4 marks]
Total for this question: 4
A student says both reactants are reduced in $\mathrm{Fe_2O_3}+3\mathrm{CO}\rightarrow2\mathrm{Fe}+3\mathrm{CO_2}$ . Evaluate the statement using oxygen transfer.
[4 marks]
Total for this question: 4
A metal oxide is heated in hydrogen and changes into the metal. Water also forms. The metal is then heated in air and changes back into the oxide. Explain the oxidation and reduction in both stages using oxygen transfer.
[4 marks]
Total for this question: 4
An $18.0\,\mathrm{g}$ sample of a metal oxide is completely reduced to $14.4\,\mathrm{g}$ of metal. The metal is then heated in air and forms $17.1\,\mathrm{g}$ of oxide. Calculate the mass of oxygen removed and the mass regained. Identify the change in each stage and evaluate whether the metal returned completely to its original oxide.
[5 marks]
Total for this question: 5
Magnesium removes oxygen from copper oxide: $\mathrm{CuO}+\mathrm{Mg}\rightarrow\mathrm{Cu}+\mathrm{MgO}$ . The reactants have a total mass of $20.7\,\mathrm{g}$ and $12.7\,\mathrm{g}$ of copper forms. Calculate the mass of magnesium oxide formed, then identify and explain the oxidation and reduction using oxygen transfer.
[5 marks]
Total for this question: 5

4.4.1.2 · The reactivity series

Explanation

A useful order is potassium, sodium, lithium, calcium, magnesium, carbon, zinc, iron, hydrogen and copper, from more reactive to less reactive. Metal reactivity is linked to how readily its atoms form positive ions; more reactive metals form positive ions more readily.
A more reactive metal displaces a less reactive metal from its compound, so displacement evidence can establish an order.
At room temperature, potassium, sodium and lithium react rapidly with cold water, calcium less vigorously and magnesium very slowly; zinc, iron and copper do not.
Metals above hydrogen react with dilute acids, increasingly slowly down to iron, while copper does not; reactions with steam are outside this point.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Metal P displaces Q from a solution of Q ions. Metal Q displaces R from a solution of R ions. Metal R does not displace P from a solution of P ions. Deduce the reactivity order of P, Q and R and justify it.

1.P displacing Q shows that P is more reactive than Q. Q displacing R shows that Q is more reactive than R. The final result is consistent because less-reactive R cannot displace P. Therefore $\mathrm{P}>\mathrm{Q}>\mathrm{R}$ .

Answer: $\mathrm{P}>\mathrm{Q}>\mathrm{R}$ in reactivity.

Common mistakes

Don't place hydrogen among the metals without recognising that it is a non-metal reference point.
Don't predict that a less reactive metal displaces a more reactive metal from its compound.

Exam tip

A displacement answer should compare both metals in the reactivity series and state which forms positive ions more readily.

Tier 1 · Easy

Place magnesium, copper and zinc in decreasing order of reactivity.
[1 mark]
Total for this question: 1
Metal J displaces metal K from a solution containing K ions. State which metal is more reactive.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Zinc is above iron in the reactivity series. Predict what happens when zinc is added to iron(II) sulfate solution, and explain the prediction in terms of forming positive ions.
[3 marks]
Total for this question: 3
Equal-sized pieces of metals P, Q and R are placed separately in the same dilute acid. P fizzes quickly, Q fizzes slowly and R does not react. Deduce the order of P, Q, R and hydrogen from most to least reactive.
[4 marks]
Total for this question: 4
Sodium and copper are compared. State which metal forms positive ions more readily, identify the more reactive metal, and predict whether copper displaces sodium from a sodium compound.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Four metals W, X, Y and Z are tested at room temperature. W reacts rapidly with cold water. X reacts slowly with dilute acid but not with cold water. Y does not react with cold water, reacts quickly with dilute acid and displaces X from an X salt solution. Z shows no reaction with dilute acid. Deduce their order from most to least reactive and explain the position of Z relative to hydrogen.
[5 marks]
Total for this question: 5
Metal L removes oxygen from the oxide of M. Metal M removes oxygen from the oxide of N. Metal P displaces L from a solution of L ions. Deduce the reactivity order of P, L, M and N and justify every comparison.
[5 marks]
Total for this question: 5
Three metals A, B and C are calcium, zinc and copper in an unknown order. A reacts with cold water. B does not react with cold water but reacts with dilute acid. C reacts with neither cold water nor dilute acid. Identify A, B and C and explain each choice.
[5 marks]
Total for this question: 5
Metal E displaces D and F from solutions of their ions, and F displaces D. A fourth result reports that D displaces E. Determine the reactivity order supported by the first three results, identify the anomalous result, and explain your decision in terms of forming positive ions.
[5 marks]
Total for this question: 5
Zinc powder is tested in dilute acid, while an iron strip is tested in acid that is three times more concentrated. The iron produces more hydrogen in the first minute, so a student concludes that iron is more reactive than zinc. Evaluate the conclusion, describe how to make the comparison fair, and state one displacement result that would establish the correct order.
[5 marks]
Total for this question: 5

4.4.1.3 · Extraction of metals and reduction

Explanation

Very unreactive metals can occur native, but most metals occur as compounds and must be extracted by chemical reactions.
A metal below carbon in the reactivity series can be extracted from its oxide by reduction with carbon.
In oxygen-transfer language, the metal oxide is reduced because it loses oxygen; carbon is oxidised because it gains oxygen.
Use supplied evidence to evaluate an extraction route; detailed industrial processes beyond reduction of oxides with carbon are not required here.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Aluminium is above carbon, zinc is below carbon and gold is very unreactive. For each metal, choose the most suitable description: found as the metal itself, extracted from its oxide using carbon, or requires a method other than carbon reduction. Explain each choice.

1.Gold's very low reactivity allows it to occur native. Zinc is below carbon, so its oxide can be reduced by carbon. Aluminium is above carbon, so carbon cannot remove oxygen from aluminium oxide and another extraction method is needed.

Answer: Gold may be found as the metal itself; zinc can be extracted from its oxide using carbon; aluminium requires another method.

Common mistakes

Don't choose carbon reduction for a metal above carbon in the reactivity series.
Don't call removal of oxygen oxidation instead of reduction.

Exam tip

For extraction, first locate the metal relative to carbon and then name the reduction method.

Tier 1 · Easy

Copper is less reactive than carbon. Explain why carbon can be used to obtain copper from copper oxide.
[2 marks]
Total for this question: 2
Lead is below carbon in the reactivity series, but potassium is above carbon. State which metal can be extracted from its oxide using carbon.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Iron is below carbon in the reactivity series, but magnesium is above carbon. Explain why carbon can extract iron from iron oxide but cannot extract magnesium from magnesium oxide.
[3 marks]
Total for this question: 3
In $2\mathrm{PbO}+\mathrm{C}\rightarrow2\mathrm{Pb}+\mathrm{CO_2}$ , identify the substance reduced and explain how carbon causes this change.
[3 marks]
Total for this question: 3
Gold can sometimes be found as the element, whereas zinc is usually found combined with other elements. Explain this difference using reactivity and state how zinc can be extracted from zinc oxide.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Consider $2\mathrm{CuO}+\mathrm{C}\rightarrow2\mathrm{Cu}+\mathrm{CO_2}$ . Identify the substance oxidised and the substance reduced, explaining each identification using oxygen transfer. A proposed alternative uses less fuel but produces only half as much copper per batch. State two pieces of additional information needed to judge which route is preferable.
[6 marks]
Total for this question: 6
Two extraction routes produce the same metal. Per kilogram of metal, route A uses $8.0\,\mathrm{MJ}$ and releases $3.0\,\mathrm{kg}$ of carbon dioxide directly; route B uses $12.0\,\mathrm{MJ}$ and releases $0.5\,\mathrm{kg}$ directly. A company prioritises the lower direct carbon dioxide release. Choose a route, justify the choice, and give one reason why the data do not establish which route has the lower total carbon footprint.
[4 marks]
Total for this question: 4
An ore-processing plant treats 200 kg of ore and obtains 60 kg of metal. Calculate the percentage of the ore mass obtained as metal. The metal oxide in the ore loses oxygen during the process. State the name of this change and explain why the calculation alone does not give the percentage yield of the extraction reaction.
[4 marks]
Total for this question: 4
Carbon does not remove oxygen from the oxide of metal J, but it does remove oxygen from the oxide of metal K. Metal L is found naturally as the uncombined element. State what can be concluded about J and K relative to carbon, explain what the observation about L suggests, and identify one ordering that the evidence does not establish.
[4 marks]
Total for this question: 4
Route P treats $500\,\mathrm{kg}$ of ore that is $40\%$ metal oxide and obtains metal equal to $70\%$ of that oxide mass. Route Q treats $320\,\mathrm{kg}$ of ore that is $75\%$ metal oxide and obtains metal equal to $65\%$ of that oxide mass. Calculate the metal output from each route, choose the route with the greater output, and state the reactivity condition needed if carbon is used for either extraction.
[6 marks]
Total for this question: 6

4.4.1.4 · Oxidation and reduction in terms of electrons (HT only)

Explanation

Higher tier: oxidation is loss of electrons and reduction is gain of electrons: OIL RIG. A half equation shows electrons explicitly and must balance both atoms and total charge.
Add state symbols when the question asks for them; they can be part of the mark for a complete ionic equation.
In a metal displacement, the more reactive metal loses electrons while the displaced metal ions gain electrons.
Do not identify redox from charge signs alone: compare the same species before and after the reaction and track electron transfer.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Complete and balance the half equation $\mathrm{Cl_2}+\square\,\mathrm{e^-}\rightarrow\square\,\mathrm{Cl^-}$ and identify the process.

1.Two chlorine atoms require $2\mathrm{Cl^-}$ on the right. Their total charge is $-2$ , so add two electrons on the left. Chlorine gains electrons, so it is reduced.

Answer: $\mathrm{Cl_2}+2\mathrm{e^-}\rightarrow2\mathrm{Cl^-}$ ; reduction.

Common mistakes

Don't use OIL RIG backwards and label electron loss as reduction.
Don't write an ionic equation that does not balance both atoms and charge.

Exam tip

Check both atom count and total charge after writing an ionic or half equation.

Tier 1 · Easy

The half equation is $\mathrm{Zn}\rightarrow\mathrm{Zn^{2+}}+2\mathrm{e^-}$ . State whether zinc is oxidised or reduced and explain why.
[2 marks]
Total for this question: 2
An oxide ion changes into an oxygen atom: $\mathrm{O^{2-}}\rightarrow\mathrm{O}+2\mathrm{e^-}$ . State whether the oxide ion is oxidised or reduced, and explain your answer.
[2 marks]
Total for this question: 2

Tier 2 · Standard

In the ionic equation $\mathrm{Mg}+\mathrm{Cu^{2+}}\rightarrow\mathrm{Mg^{2+}}+\mathrm{Cu}$ , identify the species oxidised and the species reduced. Explain both choices in terms of electrons.
[4 marks]
Total for this question: 4
Complete $\mathrm{O_2}+\square\,\mathrm{e^-}\rightarrow2\mathrm{O^{2-}}$ , then use the total charge to show that your equation is balanced and name the process.
[4 marks]
Total for this question: 4
An iron nail is placed in copper(II) sulfate solution and becomes coated with copper. Explain the observation in terms of electron transfer and identify the species oxidised and reduced.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Magnesium is added to copper(II) ion solution. Write the oxidation half equation, the reduction half equation and the overall ionic equation, including state symbols. Identify the species oxidised and the species reduced.
[5 marks]
Total for this question: 5
Aluminium displaces copper from copper(II) ions. Write balanced oxidation and reduction half equations, combine them to form the overall ionic equation, and identify the species oxidised and reduced.
[6 marks]
Total for this question: 6
The overall ionic equation is $2\mathrm{M}+3\mathrm{X^{2+}}\rightarrow2\mathrm{M^{3+}}+3\mathrm{X}$ . Determine the number of electrons lost by each M atom, the number gained by each $\mathrm{X^{2+}}$ ion and the total number transferred. Identify the species oxidised and reduced, then show that total charge balances.
[5 marks]
Total for this question: 5
Zinc reacts with copper(II) sulfate: $\mathrm{Zn}+\mathrm{CuSO_4}\rightarrow\mathrm{ZnSO_4}+\mathrm{Cu}$ . Write the oxidation and reduction half equations and the overall ionic equation. Identify the spectator ion and explain why it is omitted from the ionic equation.
[6 marks]
Total for this question: 6
The ionic equation is $2\mathrm{Fe^{3+}}+\mathrm{Sn^{2+}}\rightarrow2\mathrm{Fe^{2+}}+\mathrm{Sn^{4+}}$ . Write a half equation for each change, determine the total number of electrons transferred, identify the species oxidised and reduced, and verify the total charge of the overall equation.
[6 marks]
Total for this question: 6

4.4.2.1 · Reactions of acids with metals

Explanation

Acids react with magnesium, zinc and iron to produce a salt and hydrogen gas. Hydrochloric acid forms chloride salts, while sulfuric acid forms sulfate salts.
For example, $\mathrm{Zn}+2\mathrm{HCl}\rightarrow\mathrm{ZnCl_2}+\mathrm{H_2}$ ; the metal has replaced hydrogen from the acid.
Do not add water as a product: acid plus metal gives salt plus hydrogen.
Higher tier: metal atoms lose electrons and are oxidised, while hydrogen ions gain electrons and are reduced; explain these redox reactions in terms of electron transfer and identify both species.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Write a balanced symbol equation for iron reacting with dilute sulfuric acid to form iron(II) sulfate and hydrogen.

1.Write the stated formulae, then count atoms. One iron atom, one sulfate group and two hydrogen atoms occur on each side, so all coefficients are $1$ .

Answer: $\mathrm{Fe}+\mathrm{H_2SO_4}\rightarrow\mathrm{FeSO_4}+\mathrm{H_2}$

Common mistakes

Don't predict hydrogen from copper and dilute acid even though copper is below hydrogen.
Don't identify the metal as reduced even though metal atoms lose electrons (Higher tier).

Exam tip

For acid–metal reactions, identify the salt from the acid and metal, then add hydrogen.

Tier 1 · Easy

Name the two products when magnesium reacts with dilute hydrochloric acid.
[2 marks]
Total for this question: 2
Copper is added to dilute sulfuric acid at room temperature. State the expected observation and explain it using the position of copper relative to hydrogen.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Zinc reacts with dilute hydrochloric acid. Give two observations you would see. How would you confirm the gaseous product, and which salt remains in solution?
[5 marks]
Total for this question: 5
Write a balanced symbol equation for iron reacting with dilute hydrochloric acid to form iron(II) chloride and hydrogen.
[3 marks]
Total for this question: 3
Equal-sized pieces of magnesium and iron are placed in equal volumes of the same dilute hydrochloric acid. Predict which metal produces hydrogen faster, explain why, and identify the salt formed by that metal.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Higher Tier: magnesium reacts with hydrochloric acid according to $\mathrm{Mg}+2\mathrm{H^+}\rightarrow\mathrm{Mg^{2+}}+\mathrm{H_2}$ . Write the oxidation and reduction half equations, then identify the species oxidised and the species reduced.
[4 marks]
Total for this question: 4
Higher Tier: zinc reacts with dilute sulfuric acid. Write the oxidation half equation, the reduction half equation and the overall ionic equation. Explain why sulfate ions do not appear in the ionic equation.
[6 marks]
Total for this question: 6
Describe an investigation that compares the reactions of magnesium, zinc, iron and copper with dilute hydrochloric acid. State the measurement, two control variables, and the expected order from fastest reaction to no reaction.
[5 marks]
Total for this question: 5
Equal-sized samples of metals A, B and C are placed in identical dilute acid, with each metal in excess. After $20\,\mathrm{s}$ , A has produced $34\,\mathrm{cm^3}$ of gas, B has produced $18\,\mathrm{cm^3}$ and C has produced none. A and B each eventually produce $50\,\mathrm{cm^3}$ , while C still produces none. Compare the reaction rates, explain what the equal final volumes do and do not show, and place C relative to hydrogen.
[5 marks]
Total for this question: 5
An unknown metal X is magnesium, aluminium or copper. It reacts with dilute hydrochloric acid, producing a gas, and the salt formed has formula $\mathrm{XCl_2}$ . Identify X, use both pieces of evidence to exclude the other metals, write the balanced reaction equation, and state the test for the gas.
[5 marks]
Total for this question: 5

4.4.2.2 · Neutralisation of acids and salt production

Explanation

An acid and an alkali or base form a salt and water; an acid and a metal carbonate form a salt, water and carbon dioxide.
Hydrochloric acid makes chlorides, nitric acid makes nitrates and sulfuric acid makes sulfates.
The positive ion in the alkali, base or carbonate supplies the metal part of the salt.
A common error is to choose the salt only from the acid: both the acid's negative ion and the other reactant's positive ion are needed.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Complete the word equation and explain the salt name: copper oxide + sulfuric acid $\rightarrow$ ______ + ______.

1.A metal oxide is a base, so it reacts with an acid to form salt and water. Copper oxide supplies copper ions and sulfuric acid supplies sulfate ions, giving copper sulfate.

Answer: Copper sulfate and water.

Common mistakes

Don't predict hydrogen when an acid reacts with a carbonate.
Don't choose the salt name from the base and ignore which acid supplies the negative ion.

Exam tip

Product prediction earns marks for the salt name and every additional product required by the reactant type.

Tier 1 · Easy

Potassium hydroxide is neutralised by nitric acid. Name the salt formed and the other product.
[2 marks]
Total for this question: 2
Magnesium oxide reacts with nitric acid. Name the salt and the other product.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Calcium carbonate reacts with hydrochloric acid. Name all three products (1 mark) and write a balanced symbol equation (2 marks).
[3 marks]
Total for this question: 3
Complete and balance the equation for calcium hydroxide reacting with hydrochloric acid: $\mathrm{Ca(OH)_2}+\mathrm{HCl}\rightarrow\mathrm{CaCl_2}+\mathrm{H_2O}$ .
[3 marks]
Total for this question: 3
Give the acid and one suitable base that could react to make copper nitrate solution. State the other product.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Aluminium hydroxide reacts with sulfuric acid. Deduce the formula of aluminium sulfate and balance the symbol equation.
[5 marks]
Total for this question: 5
Iron(III) oxide reacts with hydrochloric acid to form iron(III) chloride and water. Deduce the two iron-containing formulae and write the balanced symbol equation.
[5 marks]
Total for this question: 5
A student claims that every neutralisation reaction produces only a salt and water. Evaluate the claim using the reaction between sodium carbonate and sulfuric acid. State all products and write a balanced symbol equation.
[5 marks]
Total for this question: 5
A reaction produces calcium nitrate, water and carbon dioxide. Identify the acid and the calcium compound that reacted, explain how the products reveal both reactants, and write a balanced symbol equation.
[5 marks]
Total for this question: 5
A student proposes making aluminium nitrate by reacting aluminium oxide with sulfuric acid. Explain why this acid gives the wrong salt, choose the correct acid, state the other product, and write the balanced equation for the corrected reaction.
[5 marks]
Total for this question: 5

4.4.2.3 · Soluble salts

Explanation

Warm dilute acid gently, then add an insoluble metal oxide or carbonate in small portions until some solid remains unreacted.
Filter to remove the excess insoluble solid; the filtrate is the salt solution.
Use a water bath or electric heater to evaporate some water, allow the concentrated solution to cool and crystallise, then separate and dry the crystals.
Do not evaporate all the water with strong heating, because this can spit, lose product or leave an impure powder rather than good crystals.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Put these stages for making copper sulfate crystals from copper oxide and dilute sulfuric acid into a safe, logical order: crystallise, filter, warm the acid, add excess copper oxide, evaporate some water, dry the crystals.

1.Warm the acid to increase the reaction rate, then add copper oxide until it is in excess. Filter off unreacted oxide. Concentrate the filtrate using a water bath or electric heater, cool it to crystallise the salt, and dry the separated crystals.

Answer: Warm the acid; add excess copper oxide; filter; evaporate some water; crystallise; dry the crystals.

Common mistakes

Don't add only the exact reacting amount of insoluble solid, leaving acid that contaminates the product.
Don't evaporate the salt solution to dryness instead of concentrating and crystallising it.

Exam tip

A required-practical method must explain excess solid, filtration, concentration, crystallisation and drying in order.

Tier 1 · Easy

Why is an insoluble solid added to an acid until some remains unreacted when making a soluble salt?
[2 marks]
Total for this question: 2
Give one reason why a salt solution is concentrated gently instead of being heated strongly to dryness when preparing crystals.
[1 mark]
Total for this question: 1

Tier 2 · Standard

When making a soluble salt from an acid and an excess insoluble base, explain why the mixture is filtered before heating the filtrate and why the concentrated solution is then left to cool.
[3 marks]
Total for this question: 3
After zinc oxide has been added in excess to sulfuric acid, the mixture contains zinc sulfate solution and unreacted zinc oxide. Name the separation method and identify the residue and filtrate.
[3 marks]
Total for this question: 3
Choose copper oxide rather than sodium hydroxide for making pure copper sulfate crystals by the excess-solid method. Explain two reasons for the choice.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Describe how to prepare a pure, dry sample of zinc chloride crystals using zinc carbonate and dilute hydrochloric acid. Include the purpose of each separation or heating stage.
[6 marks]
Total for this question: 6
A student tries to make magnesium sulfate crystals from magnesium oxide and dilute sulfuric acid. The student adds one small portion of magnesium oxide, pours the mixture straight into an evaporating basin, and heats until completely dry. Describe three changes needed to obtain pure, dry crystals and explain the purpose of each change.
[6 marks]
Total for this question: 6
Explain why the excess-insoluble-solid method is unsuitable for preparing pure sodium chloride from sodium hydroxide and hydrochloric acid. Describe how titration results can instead be used to obtain sodium chloride crystals without indicator contaminating them.
[5 marks]
Total for this question: 5
A student obtains salt crystals, then rinses them with a large volume of warm distilled water before drying them. The final crystal mass is much lower than expected. Explain the low mass and describe two changes that would clean the crystals while limiting product loss.
[4 marks]
Total for this question: 4
An empty evaporating basin has a mass of $32.46\,\mathrm{g}$ . The basin and crystals have masses of $40.18\,\mathrm{g}$ , $39.72\,\mathrm{g}$ , $39.71\,\mathrm{g}$ and $39.71\,\mathrm{g}$ after successive periods of gentle drying. Determine when the crystals are dry, calculate their dry mass, and explain the purpose of repeated drying and weighing.
[4 marks]
Total for this question: 4

4.4.2.4 · The pH scale and neutralisation

Explanation

The pH scale runs from $0$ to $14$ : acids have pH below $7$ , neutral solutions have pH $7$ , and alkalis have pH above $7$ . Universal or wide-range indicator gives an approximate pH from colour; a calibrated pH probe gives a numerical reading.
Acids in water produce $\mathrm{H^+}$ ions, while aqueous alkalis contain $\mathrm{OH^-}$ ions.
Neutralisation is $\mathrm{H^+(aq)}+\mathrm{OH^-(aq)}\rightarrow\mathrm{H_2O(l)}$ ; include these state symbols when the stem requests them.
Equal volumes are not necessarily neutral if the reacting amounts differ.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

A colourless solution may have pH $5$ , $7$ or $9$ . Describe how to determine its approximate pH and state one method that would give a more precise numerical value.

1.A single indicator with a broad colour range distinguishes several pH values, so compare its colour with the supplied scale. A pH probe measures the numerical pH directly and is more precise than judging a colour.

Answer: Add universal or wide-range indicator and compare the colour with its pH chart; use a calibrated pH probe for greater precision.

Common mistakes

Don't call pH 7 acidic or alkaline rather than neutral.
Don't write the neutralisation ionic equation with unbalanced charge or missing water.

Exam tip

For neutralisation, use hydrogen ions plus hydroxide ions forming water and keep the charges balanced.

Tier 1 · Easy

Classify solutions with pH $3$ , pH $7$ and pH $11$ as acidic, neutral or alkaline.
[3 marks]
Total for this question: 3
State the pH of a neutral aqueous solution and name the negative ion present in an aqueous alkali.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Hydrochloric acid is mixed with sodium hydroxide and the final solution has pH $11$ . State which reactant is in excess and explain the result in terms of $\mathrm{H^+}$ and $\mathrm{OH^-}$ ions.
[3 marks]
Total for this question: 3
Compare universal indicator with a calibrated pH probe for measuring the pH of a colourless solution. Include the type of result from each method and which is more precise.
[3 marks]
Total for this question: 3
Universal indicator is added separately to solutions of pH 2, pH 7 and pH 12. State the expected colour for each solution.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An alkali is added in small portions to an acidic solution until its pH changes from $2$ to $7$ . Explain the pH change in terms of ions, write the ionic equation for neutralisation including state symbols, and explain why adding equal volumes of acid and alkali would not always give pH $7$ .
[5 marks]
Total for this question: 5
Dilute acid is added to an alkali. The mixture has pH $12$ before addition, pH $7$ at $18.0\,\mathrm{cm^3}$ of acid, and pH $2$ after $25.0\,\mathrm{cm^3}$ . Identify the ion in excess at each of the three stages and write the neutralisation ionic equation with state symbols.
[5 marks]
Total for this question: 5
Acid is added to an alkali. The pH readings at 12.4, 12.5, 12.6 and 12.7 cm³ of acid are 10.8, 8.1, 6.2 and 3.0 respectively. Determine the interval containing the neutral point, explain how to obtain a narrower interval, and explain why a calibrated pH probe is preferable to universal indicator for this investigation.
[4 marks]
Total for this question: 4
A particle diagram represents $160$ hydrogen ions in an acid. Each equal portion of alkali adds $40$ hydroxide ions, and every hydrogen ion reacts with one hydroxide ion. Determine the excess ion and whether the mixture is acidic, neutral or alkaline after three, four and five portions. Write the neutralisation ionic equation with state symbols.
[6 marks]
Total for this question: 6
A dark-coloured solution is neutralised. A student adds universal indicator but cannot distinguish its colour from the colour of the solution. Evaluate this method, name a better instrument, and describe how to use it to approach the neutral point accurately without contaminating the solutions.
[4 marks]
Total for this question: 4

4.4.2.5 · Titrations (chemistry only)

Explanation

Use a volumetric pipette to transfer a fixed volume to a conical flask and a burette to deliver the other solution accurately.
Add a suitable indicator, approach the end point dropwise while swirling, and record the burette difference as the titre.
Repeat until concordant titres are obtained, then calculate a mean from concordant results rather than including a rough value.
Higher tier: use the balanced equation and volumes in $\mathrm{dm^3}$ to calculate chemical quantities and determine concentrations in $\mathrm{mol\,dm^{-3}}$ and $\mathrm{g\,dm^{-3}}$ from titration data.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Titration apparatus with a burette delivering solution into an indicator-containing conical flask.

Worked example

Describe how to obtain an accurate mean titre when titrating a strong alkali with a strong acid.

1.Rinse and fill the burette with acid and record its initial reading. Pipette a fixed alkali volume into a conical flask and add a few indicator drops. Add acid while swirling, then use single drops near the colour change. Record the final reading and calculate the titre. Repeat and average only concordant titres, excluding the rough run.

Answer: Pipette alkali into a conical flask, add indicator, deliver acid from a burette while swirling, add acid dropwise near the end point, and repeat to obtain concordant titres before taking their mean.

Common mistakes

Don't rinse the burette with water immediately before filling it, diluting the solution.
Don't use the rough titre in the mean instead of selecting concordant accurate titres.

Exam tip

A titration method should name the apparatus, endpoint, dropwise addition and concordant repeats.

Tier 1 · Easy

Name the apparatus used to transfer exactly $25.0\,\mathrm{cm^3}$ of alkali and the apparatus used to add a measured, variable volume of acid.
[2 marks]
Total for this question: 2
A burette reading changes from $1.45\,\mathrm{cm^3}$ to $23.80\,\mathrm{cm^3}$ . Calculate the titre.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A rough titre is $25.60\,\mathrm{cm^3}$ . Three accurate titres are $24.80\,\mathrm{cm^3}$ , $24.75\,\mathrm{cm^3}$ and $24.70\,\mathrm{cm^3}$ . Calculate the mean titre that should be used.
[2 marks]
Total for this question: 2
A student measures the fixed alkali volume with a measuring cylinder, adds acid rapidly near the end point, and leaves the flask still. Give one improvement for each action.
[3 marks]
Total for this question: 3
Give two actions that prevent reading errors when using a burette, and explain why the filling funnel should be removed before a titration begins.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Higher Tier: $25.0\,\mathrm{cm^3}$ of sodium hydroxide is neutralised by $18.60\,\mathrm{cm^3}$ of $0.150\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid. The equation has a $1:1$ ratio. Calculate the sodium hydroxide concentration in $\mathrm{mol\,dm^{-3}}$ and $\mathrm{g\,dm^{-3}}$ . Use $M_r(\mathrm{NaOH})=40.0$ .
[5 marks]
Total for this question: 5
Higher Tier: $25.0\,\mathrm{cm^3}$ of sulfuric acid is neutralised by $32.40\,\mathrm{cm^3}$ of $0.125\,\mathrm{mol\,dm^{-3}}$ potassium hydroxide. Use $\mathrm{H_2SO_4}+2\mathrm{KOH}\rightarrow\mathrm{K_2SO_4}+2\mathrm{H_2O}$ and $M_r(\mathrm{H_2SO_4})=98.0$ . Calculate the acid concentration in $\mathrm{mol\,dm^{-3}}$ and $\mathrm{g\,dm^{-3}}$ . Use unrounded values in your working.
[6 marks]
Total for this question: 6
A titration begins with an air bubble in the burette jet. The bubble disappears during the first run. Explain the effect on the recorded first titre, predict how later accurate titres compare, and state how the student should prevent and handle this error.
[5 marks]
Total for this question: 5
A titration gives these initial and final burette readings in $\mathrm{cm^3}$ : rough, $0.10$ and $25.60$ ; run 1, $0.35$ and $25.05$ ; run 2, $1.20$ and $25.95$ ; run 3, $0.80$ and $25.90$ ; run 4, $2.15$ and $26.90$ . Calculate the four accurate titres, identify the concordant set, and calculate its mean to two decimal places.
[4 marks]
Total for this question: 4
Student A rinses a conical flask with distilled water and leaves a few drops in it before pipetting in the alkali. Student B rinses a burette with distilled water and leaves drops in it before filling it with acid. Compare the effect of each action on the acid titre and state the correct rinsing procedure for both pieces of apparatus.
[5 marks]
Total for this question: 5

4.4.2.6 · Strong and weak acids (HT only)

Explanation

Higher tier: a strong acid ionises completely in water, whereas a weak acid ionises only partially.
Hydrochloric, nitric and sulfuric acids are strong; ethanoic, citric and carbonic acids are weak.
Strength describes degree of ionisation, while concentration describes the amount of solute per volume; either a strong or weak acid may be dilute.
For whole-number pH values, a decrease of one pH unit means a tenfold increase in hydrogen ion concentration, not an increase of one unit.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Equal-concentration solutions of hydrochloric acid and ethanoic acid are compared. Predict which has the lower pH and explain why. State whether either solution is more concentrated.

1.Hydrochloric acid is strong, so essentially all its acid particles ionise. Ethanoic acid is weak and only partially ionises. At equal starting concentration, hydrochloric acid therefore has the greater $\mathrm{H^+}$ concentration and lower pH. Strength does not change the given concentration comparison.

Answer: Hydrochloric acid has the lower pH because it ionises completely and produces a greater hydrogen ion concentration; neither is more concentrated because their concentrations are equal.

Common mistakes

Don't treat strong and concentrated as synonyms even though they describe different properties.
Don't say a one-unit pH fall doubles hydrogen-ion concentration instead of increasing it tenfold.

Exam tip

At equal concentration, link stronger acid → greater ionisation → more hydrogen ions → lower pH.

Tier 1 · Easy

State the difference between a strong acid and a weak acid in aqueous solution.
[2 marks]
Total for this question: 2
All the acid particles ionise in a dilute aqueous solution. State whether the acid is strong or weak and explain why the word dilute does not change that classification.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Equal-concentration solutions of hydrochloric acid and ethanoic acid are compared. Predict which has the lower pH and explain why. State whether the comparison tells you that either solution is more concentrated.
[4 marks]
Total for this question: 4
A bottle is labelled 'concentrated ethanoic acid'. Explain how the acid can be both concentrated and weak.
[3 marks]
Total for this question: 3
Equal-concentration hydrochloric acid and ethanoic acid react separately with equal pieces of magnesium. Predict which reaction is faster and explain the difference using ionisation and hydrogen ion concentration.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Solution A has pH $2$ and solution B has pH $5$ . Calculate how many times greater the hydrogen ion concentration is in A than in B. Explain why saying 'A is $3$ times more acidic' is incorrect.
[4 marks]
Total for this question: 4
A student claims that every acid of pH $3$ is weak and every acid of pH $1$ is strong. Evaluate the claim using acid strength, concentration and hydrogen ion concentration.
[4 marks]
Total for this question: 4
An acid initially has pH 1. It is diluted until its hydrogen ion concentration is 1000 times smaller. Determine the new pH and explain whether dilution has changed the strength of the acid.
[4 marks]
Total for this question: 4
Two hydrochloric acid solutions have pH $2$ and pH $4$ . Compare their hydrogen ion concentrations, determine which solution is more concentrated, and explain why both acids have the same strength despite the different pH values.
[4 marks]
Total for this question: 4
A hydrochloric acid solution and an ethanoic acid solution have the same pH. A student claims that they must therefore have the same acid strength and the same starting acid concentration. Evaluate both parts of the claim using hydrogen ion concentration and ionisation.
[5 marks]
Total for this question: 5

4.4.3.1 · The process of electrolysis

Explanation

An electrolyte is a molten ionic compound or ionic solution whose mobile ions allow it to conduct electricity. Positive ions move to the negative cathode, while negative ions move to the positive anode.
When ions are discharged at the electrodes, they form elements; the electric current drives this decomposition process.
In the electrolyte, charge is carried by moving ions rather than free electrons; a solid ionic compound does not conduct because its ions are fixed.
Higher tier: write, complete and balance half equations for electrode reactions.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Ion movement in an electrolytic cell connected to a direct-current supply.

Worked example

Solid sodium chloride does not conduct electricity, but molten sodium chloride does. Explain this difference in terms of ions.

1.Conduction requires charged particles that can move. The ionic lattice holds ions in fixed positions in the solid. Melting breaks down that fixed structure enough for the ions to move, so the liquid is an electrolyte.

Answer: In the solid, ions are fixed in the lattice and cannot carry charge; when molten, the ions are free to move and carry charge through the liquid.

Common mistakes

Don't send positive ions to the positive electrode.
Don't call a solid ionic compound an electrolyte even though its ions cannot move.

Exam tip

Label the cathode negative and anode positive before tracing ion movement.

Tier 1 · Easy

State which electrode attracts positive ions during electrolysis and give the charge on that electrode.
[2 marks]
Total for this question: 2
Define an electrolyte and state the type of charged particle that moves through it.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An electrolyte contains positive and negative ions. Explain how applying a direct current causes the electrolyte to decompose.
[3 marks]
Total for this question: 3
Solid potassium nitrate does not conduct electricity, but potassium nitrate solution does. Explain the difference in terms of the ions.
[3 marks]
Total for this question: 3
Compare how electric charge is carried through a metal wire and through an electrolyte during electrolysis.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A molten ionic compound contains only $\mathrm{M^{2+}}$ and $\mathrm{X^-}$ ions. Describe the movement of both ions when a direct current is applied, name the type of substance formed at each electrode, and state what carries charge through the melt.
[5 marks]
Total for this question: 5
A student states that anions move to the negative electrode, cations move to the positive electrode, and electrons carry charge through the electrolyte. Correct all three statements and explain how the moving particles allow electrolysis to occur.
[5 marks]
Total for this question: 5
A molten ionic compound conducts while connected to a direct-current supply, but the current stops when the compound solidifies. Explain the charge carriers in the wires and in the melt, the movement of both types of ion, and why solidification stops product formation.
[5 marks]
Total for this question: 5
An electrolytic cell contains $\mathrm{M^{2+}}$ and $\mathrm{X^-}$ ions. Initially the left electrode is negative and the right electrode is positive. The power-supply connections are then reversed. State the new name and charge of each electrode, describe the new direction of movement of both ions, and explain what happens to the physical locations of the products.
[5 marks]
Total for this question: 5
Four samples are connected separately to the same direct-current circuit: solid sodium chloride, molten sodium chloride, sodium chloride solution and sugar solution. Determine which samples allow sustained electrolysis and explain every choice in terms of the charged particles present and their mobility.
[5 marks]
Total for this question: 5

4.4.3.2 · Electrolysis of molten ionic compounds

Explanation

A molten binary ionic compound contains only its metal ions and non-metal ions, so its products are predictable from those ions. The metal forms at the negative cathode and the non-metal forms at the positive anode when inert electrodes are used.
For molten lead bromide, lead forms at the cathode and bromine forms at the anode.
Do not apply the aqueous-solution rules to a molten compound: there is no water supplying competing hydrogen or hydroxide ions.
Higher tier: write, complete and balance half equations for electrode reactions.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Molten lead bromide is electrolysed with inert electrodes. Name each product, identify its electrode, and explain why lead bromide must be molten.

1.Positive lead ions move to the cathode and form lead. Negative bromide ions move to the anode and form bromine. In solid lead bromide the ions are fixed, so the substance must be molten for its ions to move and conduct.

Answer: Lead forms at the negative cathode; bromine forms at the positive anode; melting frees the ions to move.

Common mistakes

Don't predict hydrogen or oxygen from a molten binary compound even though no water is present.
Don't reverse the electrodes and place the metal at the anode.

Exam tip

For a molten binary compound, predict the metal at the cathode and non-metal at the anode.

Tier 1 · Easy

Predict the products at the cathode and anode when molten zinc chloride is electrolysed using inert electrodes.
[2 marks]
Total for this question: 2
During electrolysis of molten lithium oxide, name the ion that moves to the anode and the element formed there.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A student predicts that hydrogen will form at the cathode during electrolysis of molten magnesium chloride. Explain why this prediction is wrong and state the actual cathode product.
[3 marks]
Total for this question: 3
A student says that chloride ions in molten calcium chloride move to the cathode and form chlorine atoms. Explain both errors and give the correct anode product.
[4 marks]
Total for this question: 4
A molten binary compound produces magnesium at one electrode and bromine at the other. Identify the two ions present, state the formula of the compound, and identify the electrode that produces magnesium.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A binary compound has formula $\mathrm{CaCl_2}$ . Predict both electrolysis products in the molten state and explain how the formula supports the relative numbers of calcium ions and chloride ions discharged.
[5 marks]
Total for this question: 5
Compare the products from molten lithium bromide with those from aqueous lithium bromide, using inert electrodes. Explain why the cathode products differ while the anode products are the same.
[5 marks]
Total for this question: 5
A sealed apparatus completely electrolyses 18.6 g of a molten binary ionic compound and collects both products. The metal collected at one electrode has a mass of 7.2 g. Calculate the mass of non-metal collected, identify the electrode for each product, and explain why the compound had to be molten.
[4 marks]
Total for this question: 4

4.4.3.3 · Using electrolysis to extract metals

Explanation

Electrolysis extracts metals that are too reactive for carbon reduction or that react with carbon. Extraction uses much energy because the ionic compound must be molten and an electric current must be maintained.
Aluminium is extracted from a molten mixture of aluminium oxide and cryolite; the mixture has a lower melting point than pure aluminium oxide.
Oxygen produced at the carbon anode reacts with carbon to form carbon dioxide, so the positive electrodes wear away and must be replaced.
Higher tier: write, complete and balance half equations for electrode reactions.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Cryolite is mixed with aluminium oxide before electrolysis. Explain how this reduces the energy cost of extraction.

1.Electrolysis needs mobile ions, so the electrolyte must be liquid. The mixture melts at a lower temperature than pure aluminium oxide, reducing the heating energy required and therefore the energy cost.

Answer: Cryolite lowers the melting point of the electrolyte, so less energy is needed to melt and keep it molten.

Common mistakes

Don't say cryolite increases the melting point of aluminium oxide.
Don't say the carbon anode is replaced because it melts rather than because it reacts with oxygen.

Exam tip

An aluminium-extraction explanation should cover both the lowered melting point and carbon-anode reaction.

Tier 1 · Easy

Explain why aluminium is extracted using electrolysis rather than by reducing aluminium oxide with carbon.
[2 marks]
Total for this question: 2
State the electrode at which aluminium forms during its extraction and explain why aluminium ions move there.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Give two environmental disadvantages linked directly to the electrolytic extraction of aluminium, and explain the source of each one.
[4 marks]
Total for this question: 4
Metal A is above carbon in the reactivity series, whereas metal B is below carbon. Choose electrolysis or carbon reduction for extracting each metal from its oxide, and explain both choices.
[4 marks]
Total for this question: 4
A carbon anode has a mass of 500 kg before use in an aluminium cell and 420 kg afterwards. Calculate the mass lost and explain the chemical reason for this loss.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An aluminium plant considers two electrolytes. Mixture A melts at $950\,^{\circ}\mathrm{C}$ ; mixture B melts at $1210\,^{\circ}\mathrm{C}$ . Both contain the same amount of aluminium oxide and give the same aluminium output. The cells use carbon anodes. Choose the likely lower-energy mixture and explain both your choice and why the anodes need regular replacement.
[5 marks]
Total for this question: 5
An aluminium plant lowers the melting point of its electrolyte and powers the cells using renewable electricity, yet it still reports direct carbon dioxide emissions from the cells. Explain how the lower melting point reduces energy use and why carbon dioxide is still produced inside the cells.
[5 marks]
Total for this question: 5
A company proposes replacing the carbon anodes in aluminium extraction with anodes that do not react with oxygen. Predict two possible advantages and give two pieces of evidence needed before deciding whether the change is worthwhile.
[4 marks]
Total for this question: 4
The total mass of carbon anodes in an aluminium cell falls from $640\,\mathrm{kg}$ to $520\,\mathrm{kg}$ . Calculate the carbon mass lost. Explain why it would be wrong to report this value as the mass of carbon dioxide made, and evaluate the claim that renewable electricity makes the cell carbon-free.
[4 marks]
Total for this question: 4
Two aluminium-cell designs each make $2000\,\mathrm{kg}$ of aluminium. A carbon-anode cell uses $14\,\mathrm{kWh}$ per kilogram of aluminium and consumes $0.45\,\mathrm{kg}$ of carbon per kilogram. An inert-anode cell uses $16\,\mathrm{kWh}$ per kilogram and consumes no carbon anode. Calculate the electricity use and carbon consumption for each design, then recommend a design if the priority is lower direct cell emissions and explain one trade-off.
[6 marks]
Total for this question: 6

4.4.3.4 · Electrolysis of aqueous solutions

Explanation

An aqueous electrolyte contains ions from the dissolved compound as well as $\mathrm{H^+}$ and $\mathrm{OH^-}$ originating from water. At the cathode, hydrogen forms if the metal is more reactive than hydrogen; otherwise the metal forms.
At the anode, a halide ion forms its halogen; if no halide is present, oxygen forms.
Apply both electrode rules separately and assume inert electrodes unless told otherwise; do not simply name the elements in the solute.
Higher tier: write, complete and balance half equations for electrode reactions.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Aqueous copper(II) sulfate is electrolysed using inert electrodes. Predict both products and justify each using the discharge rules.

1.Copper is less reactive than hydrogen, so copper ions are discharged as copper at the cathode. Sulfate is not a halide, so hydroxide ions from water are discharged and oxygen forms at the anode.

Answer: Copper at the cathode; oxygen at the anode.

Common mistakes

Don't always predict the dissolved metal at the cathode and ignore its position relative to hydrogen.
Don't predict oxygen at the anode when a halide ion is present.

Exam tip

For aqueous electrolysis, apply the cathode and anode selection rules separately.

Tier 1 · Easy

Predict the products at inert electrodes during electrolysis of aqueous sodium chloride.
[2 marks]
Total for this question: 2
Predict the cathode product during electrolysis of aqueous zinc sulfate with inert electrodes, and justify the prediction using the reactivity series.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An aqueous solution produces copper at the cathode and chlorine at the anode when electrolysed with inert electrodes. Choose the solute from sodium chloride, copper(II) sulfate and copper(II) chloride, and justify both products.
[3 marks]
Total for this question: 3
Electrolysis of one aqueous solution gives a gas that pops with a burning splint at the cathode and a gas that relights a glowing splint at the anode. Choose the solute from sodium sulfate, sodium chloride and copper(II) sulfate, then justify the choice.
[4 marks]
Total for this question: 4
Aqueous sodium bromide is electrolysed with inert electrodes. Describe the cathode gas and its test, then state the observation at the anode.
[3 marks]
Total for this question: 3

Tier 3 · Hard

Predict and compare the electrode products for aqueous magnesium chloride and aqueous silver nitrate, both with inert electrodes. Explain every product using relative reactivity or the halide rule.
[6 marks]
Total for this question: 6
Describe how to investigate the products formed when aqueous sodium chloride is electrolysed with inert electrodes. Include the predicted product and test at each electrode, and one safety precaution specific to the products.
[6 marks]
Total for this question: 6
Compare the products from aqueous sodium sulfate and aqueous copper(II) sulfate using inert electrodes. Identify the product that changes when sodium ions are replaced by copper(II) ions, the product that stays the same, and explain both results.
[5 marks]
Total for this question: 5
Aqueous calcium bromide and aqueous calcium nitrate are electrolysed using inert electrodes. Compare the products from the two solutions and explain why the cathode products are the same but the anode products differ.
[6 marks]
Total for this question: 6
Aqueous copper(II) sulfate is electrolysed for a long time using inert electrodes. At first a solid forms at the cathode and the blue solution becomes paler. Later, bubbles begin at the cathode after the blue colour has disappeared; bubbles occur at the anode throughout. Identify all three electrode products and explain the sequence using the ions available and the discharge rules.
[5 marks]
Total for this question: 5

4.4.3.5 · Representation of reactions at electrodes as half equations (HT only)

Explanation

Higher tier: at the cathode, positive ions gain electrons, so cathode reactions are reductions.
At the anode, negative ions lose electrons, so anode reactions are oxidations.
Balance a half equation by conserving atoms and total charge; electrons go on the side needed to balance charge.
Check electron direction: electrons are reactants in a reduction half equation and products in an oxidation half equation.
Exam answers should link the observed product or property to the relevant particles, electron transfer or position in the reactivity series.

Worked example

Write the balanced half equation for bromide ions forming bromine at the anode and explain why it is oxidation.

1.Two bromide ions are needed to make one $\mathrm{Br_2}$ molecule. The left side has charge $-2$ , so place two electrons on the right to balance charge. Electrons are lost, making the process oxidation.

Answer: $2\mathrm{Br^-}\rightarrow\mathrm{Br_2}+2\mathrm{e^-}$ ; bromide ions lose electrons.

Common mistakes

Don't place electrons on the wrong side of a half equation for oxidation or reduction.
Don't balance atoms but leave unequal total charge on the two sides.

Exam tip

A half equation must balance atoms, charge and electrons, with reduction at the cathode and oxidation at the anode.

Tier 1 · Easy

Complete the cathode half equation $\mathrm{Al^{3+}}+\square\,\mathrm{e^-}\rightarrow\mathrm{Al}$ and name the process.
[2 marks]
Total for this question: 2
The anode equation is $2\mathrm{I^-}\rightarrow\mathrm{I_2}+2\mathrm{e^-}$ . Name the process and explain the answer using electrons.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Copper(II) sulfate solution is electrolysed using inert electrodes. Write the balanced half equation for the cathode reaction, name the process, and state what is discharged at the cathode once the copper(II) ions have been used up.
[3 marks]
Total for this question: 3
Complete the oxygen-forming anode equation $\square\,\mathrm{OH^-}\rightarrow\mathrm{O_2}+\square\,\mathrm{H_2O}+\square\,\mathrm{e^-}$ . Show that both atoms and charge balance.
[4 marks]
Total for this question: 4
The cathode half equation is $\mathrm{M^{n+}}+3\mathrm{e^-}\rightarrow\mathrm{M}$ . Determine $n$ , show how the total charge balances, and identify the process.
[3 marks]
Total for this question: 3

Tier 3 · Hard

During electrolysis of aqueous sodium sulfate with inert electrodes, hydrogen forms at the cathode and oxygen forms at the anode. Write a balanced half equation for each electrode and label oxidation and reduction.
[6 marks]
Total for this question: 6
Molten sodium oxide contains $\mathrm{Na^+}$ and $\mathrm{O^{2-}}$ ions. Write a balanced half equation for each electrode, label oxidation and reduction, and deduce the ratio of sodium atoms to oxygen molecules formed.
[6 marks]
Total for this question: 6
Four proposed electrode half equations are shown. A: $\mathrm{Cl^-}\rightarrow\mathrm{Cl_2}+\mathrm{e^-}$ . B: $2\mathrm{Cl^-}\rightarrow\mathrm{Cl_2}+2\mathrm{e^-}$ . C: $\mathrm{Cu^{2+}}\rightarrow\mathrm{Cu}+2\mathrm{e^-}$ . D: $\mathrm{Cu^{2+}}+2\mathrm{e^-}\rightarrow\mathrm{Cu}$ . Choose the two equations that balance atoms and charge, identify the electrode and process for each chosen equation, and explain one error in each rejected equation.
[6 marks]
Total for this question: 6
Electrolysis of a molten metal chloride produces two M atoms for every three $\mathrm{Cl_2}$ molecules. Determine the formula of the chloride and the charge on the metal ion. Write both balanced electrode half equations and label each process as oxidation or reduction.
[6 marks]
Total for this question: 6
During electrolysis of an aqueous electrolyte, the cathode equation is $2\mathrm{H^+}+2\mathrm{e^-}\rightarrow\mathrm{H_2}$ and the anode equation is $4\mathrm{OH^-}\rightarrow\mathrm{O_2}+2\mathrm{H_2O}+4\mathrm{e^-}$ . Use electron balance to determine the expected hydrogen-to-oxygen volume ratio at the same temperature and pressure, then predict the oxygen volume when $48\,\mathrm{cm^3}$ of hydrogen is collected.
[3 marks]
Total for this question: 3

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.4.1.1 · Metal oxides

Tier 1 · Easy

Mark scheme for 4.4.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Oxidation; magnesium gains oxygen.	Identify the substance being followed. Magnesium starts as the element and ends combined with oxygen in $\mathrm{MgO}$ , so it has gained oxygen and has been oxidised.	2
Total Question 1			2
02.1	Calcium gains oxygen, so the calcium is oxidised.	Compare calcium before and after heating. It begins as the element and ends combined with oxygen, so it has gained oxygen; gain of oxygen is oxidation.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Oxygen from the air combines with the copper, so copper oxide forms; the copper is oxidised.	The product contains the original copper plus oxygen atoms from the air, which accounts for the mass increase. Gain of oxygen is oxidation.	2
Total Question 1			2
02.1	$3.0\,\mathrm{g}$ ; the metal oxide loses oxygen, and loss of oxygen is reduction.	The removed oxygen accounts for the mass decrease, so its mass is $15.0-12.0=3.0\,\mathrm{g}$ . The oxide becomes the metal by losing oxygen, which is reduction.	3
Total Question 2			3
03.1	4Al + 3O₂ → 2Al₂O₃; aluminium gains oxygen, so it is oxidised.	Balance aluminium by placing 2 before Al₂O₃ and 4 before Al. The product then contains six oxygen atoms, requiring 3O₂. Aluminium changes from the element to an oxide, so it has gained oxygen.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Hydrogen is oxidised because it gains oxygen. Copper oxide is reduced because it loses oxygen.	Compare each reactant with its product. $\mathrm{H_2}$ becomes $\mathrm{H_2O}$ , so hydrogen gains oxygen and is oxidised. $\mathrm{CuO}$ becomes $\mathrm{Cu}$ , so copper oxide loses oxygen and is reduced.	4
Total Question 1			4
02.1	The statement is incorrect. Iron(III) oxide is reduced because it loses oxygen, whereas carbon monoxide is oxidised because it gains oxygen.	$\mathrm{Fe_2O_3}$ becomes iron, so oxygen has been removed from it. Carbon monoxide becomes carbon dioxide, so it has gained oxygen. The two changes are reduction and oxidation respectively, not two reductions.	4
Total Question 2			4
03.1	In the first stage the metal oxide is reduced because it loses oxygen, while hydrogen is oxidised because it gains oxygen to form water. In the second stage the metal is oxidised because it gains oxygen from the air.	Track the oxygen rather than relying on the direction of the arrows. Oxygen leaves the oxide and joins hydrogen in the first reaction, so reduction and oxidation occur respectively. Heating the recovered metal in air transfers oxygen back to it, which oxidises the metal.	4
Total Question 3			4
04.1	$3.6\,\mathrm{g}$ of oxygen is removed in the first stage. $2.7\,\mathrm{g}$ of oxygen is regained in the second stage. The oxide is reduced first, and the metal is then oxidised. The final sample is $0.9\,\mathrm{g}$ lighter than the original oxide, so the metal has not returned completely to its original oxide.	The first mass change is $18.0-14.4=3.6\,\mathrm{g}$ , which is oxygen lost by the oxide, so this is reduction. Reheating increases the mass by $17.1-14.4=2.7\,\mathrm{g}$ as the metal gains oxygen, so this is oxidation. The metal would need to regain all $3.6\,\mathrm{g}$ to reproduce the original oxide, but it is short by $3.6-2.7=0.9\,\mathrm{g}$ .	5
Total Question 4			5
05.1	$8.0\,\mathrm{g}$ of magnesium oxide forms. Copper oxide is reduced because it loses oxygen. Magnesium is oxidised because it gains that oxygen.	Mass is conserved, so the magnesium oxide mass is $20.7-12.7=8.0\,\mathrm{g}$ . Copper oxide changes into copper after oxygen is removed, which is reduction. Magnesium accepts the transferred oxygen and becomes magnesium oxide, which is oxidation.	5
Total Question 5			5

4.4.1.2 · The reactivity series

Tier 1 · Easy

Mark scheme for 4.4.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Magnesium, zinc, copper.	Use their positions in the reactivity series. Magnesium is above zinc, and zinc is above copper, so the decreasing order is magnesium, zinc, copper.	1
Total Question 1			1
02.1	Metal J is more reactive than metal K.	Only a more reactive metal displaces a less reactive metal from its compound. J displaces K, so J is above K in the reactivity series.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.4.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Zinc displaces iron from the solution because zinc atoms form positive ions more readily than iron atoms. Iron is deposited as a grey or dark solid and the pale green solution fades.	A metal higher in the reactivity series forms positive ions more readily. Zinc therefore becomes zinc ions and forces the less-reactive iron ions to form iron atoms, seen as a grey or dark deposit while the pale green iron(II) sulfate colour fades.	3
Total Question 1			3
02.1	$\mathrm{P}>\mathrm{Q}>\mathrm{H}>\mathrm{R}$ .	Under the same conditions, the faster acid reaction places P above Q. Both P and Q displace hydrogen from the acid, so both are above hydrogen. R does not react, placing it below hydrogen.	4
Total Question 2			4
03.1	Sodium forms positive ions more readily and is more reactive. Copper does not displace sodium because copper is less reactive.	Higher reactivity means a metal forms positive ions more readily. Sodium is far above copper in the reactivity series, so sodium is the more reactive metal and the less-reactive copper cannot displace it from a compound.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$\mathrm{W}>\mathrm{Y}>\mathrm{X}>\mathrm{Z}$ ; Z is below hydrogen.	Rapid reaction with cold water places W highest. Y displaces X and reacts faster with acid, so Y is above X. X still reacts with acid, so it is above hydrogen. Z does not react with dilute acid, so it is below hydrogen and below X. Hence $\mathrm{W}>\mathrm{Y}>\mathrm{X}>\mathrm{Z}$ .	5
Total Question 1			5
02.1	$\mathrm{P}>\mathrm{L}>\mathrm{M}>\mathrm{N}$ .	Removing oxygen from another metal's oxide shows greater reactivity, so L is above M and M is above N. P displacing L shows that P is above L. Combining the evidence gives $\mathrm{P}>\mathrm{L}>\mathrm{M}>\mathrm{N}$ .	5
Total Question 2			5
03.1	A is calcium because calcium reacts with cold water. B is zinc because zinc does not react with cold water but is above hydrogen and reacts with dilute acid. C is copper because copper is below hydrogen and does not react with dilute acid.	Match each observation to the known reactivity pattern. Calcium reacts with cold water. Zinc is less reactive than calcium but remains above hydrogen, so it reacts with dilute acid. Copper is below hydrogen, so it does not displace hydrogen from dilute acid.	5
Total Question 3			5
04.1	The supported order is $\mathrm{E}>\mathrm{F}>\mathrm{D}$ . E displacing D and F places E above both metals. F displacing D places F above D. The report that D displaces E is anomalous because it contradicts the other results. E forms positive ions more readily than D, so the less-reactive D should not displace E ions.	Translate each displacement into a comparison: the metal that displaces another is more reactive and forms positive ions more readily. The consistent comparisons are E above D, E above F and F above D, giving $\mathrm{E}>\mathrm{F}>\mathrm{D}$ . D displacing E would require the reverse comparison, so that observation is the anomaly.	5
Total Question 4			5
05.1	The conclusion is not valid because acid concentration and metal surface area were not controlled. Use acid of the same concentration and volume, the same temperature, and metal samples with matched surface area or form. Zinc is more reactive than iron. Zinc displaces iron from a solution containing iron ions, whereas iron does not displace zinc from zinc ions.	A faster observed reaction can be caused by conditions as well as by the metal's position in the reactivity series. Here both acid concentration and the powder-versus-strip surface area differ, so the rate evidence cannot isolate reactivity. With all other variables matched, zinc should react faster because it is above iron; its ability to displace iron provides independent evidence.	5
Total Question 5			5

4.4.1.3 · Extraction of metals and reduction

Tier 1 · Easy

Mark scheme for 4.4.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Carbon is more reactive than copper and removes oxygen from copper oxide, reducing it to copper.	Compare copper with carbon in the reactivity series. Because copper is below carbon, carbon can take oxygen from copper oxide. Loss of oxygen reduces the oxide to copper.	2
Total Question 1			2
02.1	Lead.	Carbon can reduce the oxide of a metal below it in the reactivity series. Lead is below carbon, whereas potassium is above it.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.4.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Carbon is more reactive than iron, so it can remove oxygen from iron oxide. Carbon is less reactive than magnesium, so it cannot remove oxygen from magnesium oxide.	Compare each metal with carbon. An oxide can be reduced by carbon only when carbon is more reactive than the metal and can take its oxygen.	3
Total Question 1			3
02.1	Lead oxide is reduced because it loses oxygen. Carbon removes that oxygen and is oxidised to carbon dioxide.	Track oxygen between reactants and products. Lead oxide becomes lead after oxygen is removed, so it is reduced. Carbon accepts the oxygen and forms $\mathrm{CO_2}$ .	3
Total Question 2			3
03.1	Gold is very unreactive, so it may remain uncombined. Zinc is more reactive and forms compounds, but zinc is below carbon, so carbon can reduce zinc oxide to zinc.	Very unreactive metals are least likely to have reacted with substances in the environment and can occur native. Zinc commonly occurs in compounds. Its position below carbon means carbon can remove oxygen from zinc oxide during extraction.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Carbon is oxidised because it gains oxygen; copper oxide is reduced because it loses oxygen. Any two relevant factors, such as total energy use, carbon dioxide released per mass of copper, raw-material cost, yield or waste produced.	Carbon gains oxygen to become $\mathrm{CO_2}$ , so it is oxidised. $\mathrm{CuO}$ loses oxygen to become copper, so it is reduced. The stated fuel and batch-output facts are not enough for a fair comparison; compare routes on a common basis such as energy, emissions, cost, yield or waste per mass of copper.	6
Total Question 1			6
02.1	Choose route B because its direct carbon dioxide release is lower. The total footprint also depends on information such as how the energy for each route is generated.	The stated priority makes B the keyed choice: $0.5\,\mathrm{kg}$ is less than $3.0\,\mathrm{kg}$ per kilogram of metal. However, B uses more energy, and the table omits emissions from supplying that energy, so direct emissions alone cannot determine the full carbon footprint.	4
Total Question 2			4
03.1	30%; the metal oxide is reduced because it loses oxygen. The ore contains substances other than the metal oxide, and no theoretical mass of metal from the reacting oxide is given, so the percentage yield cannot be calculated.	The fraction of the original ore obtained as metal is 60 ÷ 200 = 0.30, so the percentage is 30%. Loss of oxygen is reduction. Percentage yield compares actual product with the theoretical product from the reacting material, but the prompt gives only the total ore mass.	4
Total Question 3			4
04.1	J is above carbon because carbon cannot reduce its oxide. K is below carbon because carbon can reduce its oxide. L is very unreactive, which allows it to occur native. The evidence does not by itself establish the exact order of K and L.	Carbon can remove oxygen only from the oxide of a metal below carbon, so the two oxide tests place J above carbon and K below it. Finding L uncombined is evidence of very low reactivity, but it supplies no direct displacement comparison between K and L, so their exact relative positions are not fixed by the stated results.	4
Total Question 4			4
05.1	P contains $200\,\mathrm{kg}$ of metal oxide. P produces $140\,\mathrm{kg}$ of metal. Q contains $240\,\mathrm{kg}$ of metal oxide. Q produces $156\,\mathrm{kg}$ of metal. Choose Q because it produces $16\,\mathrm{kg}$ more metal. Carbon reduction is suitable only if the metal is below carbon in the reactivity series.	For P, $0.40\times500=200\,\mathrm{kg}$ of oxide and $0.70\times200=140\,\mathrm{kg}$ of metal. For Q, $0.75\times320=240\,\mathrm{kg}$ of oxide and $0.65\times240=156\,\mathrm{kg}$ of metal. Q therefore gives $156-140=16\,\mathrm{kg}$ more. These output calculations do not override the chemistry: carbon can reduce the oxide only when carbon is more reactive than the metal.	6
Total Question 5			6

4.4.1.4 · Oxidation and reduction in terms of electrons (HT only)

Tier 1 · Easy

Mark scheme for 4.4.1.4 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Zinc is oxidised because it loses electrons.	The electrons appear on the product side, so one zinc atom has lost two electrons. Loss of electrons is oxidation.	2
Total Question 1			2
02.1	It is oxidised because it loses two electrons.	The electrons are on the product side, so the oxide ion loses them. Loss of electrons is oxidation.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.1.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Magnesium is oxidised because each magnesium atom loses two electrons. Copper(II) ions are reduced because each $\mathrm{Cu^{2+}}$ ion gains two electrons.	Track each species across the equation. Magnesium changes from charge $0$ to $+2$ , so it loses electrons. Copper changes from $+2$ to charge $0$ , so the ion gains electrons.	4
Total Question 1			4
02.1	$\mathrm{O_2}+4\mathrm{e^-}\rightarrow2\mathrm{O^{2-}}$ ; both sides have total charge $-4$ , and oxygen is reduced because it gains electrons.	Two oxide ions have a combined charge of $-4$ , so four electrons are required on the left. There are two oxygen atoms on each side and charge $-4$ on each side. Electron gain makes the change reduction.	4
Total Question 2			4
03.1	Iron atoms lose electrons to form $\mathrm{Fe^{2+}}$ ions, so iron is oxidised. $\mathrm{Cu^{2+}}$ ions gain those electrons to form copper atoms, so the copper ions are reduced and the copper forms the coating.	The deposited solid must come from $\mathrm{Cu^{2+}}$ ions becoming neutral copper atoms by electron gain. Iron supplies the electrons as it changes from neutral atoms to $\mathrm{Fe^{2+}}$ ions. Electron loss is oxidation and electron gain is reduction.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.4.1.4 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$\mathrm{Mg(s)}\rightarrow\mathrm{Mg^{2+}(aq)}+2\mathrm{e^-}$ $\mathrm{Cu^{2+}(aq)}+2\mathrm{e^-}\rightarrow\mathrm{Cu(s)}$ $\mathrm{Mg(s)}+\mathrm{Cu^{2+}(aq)}\rightarrow\mathrm{Mg^{2+}(aq)}+\mathrm{Cu(s)}$ ; magnesium is oxidised and $\mathrm{Cu^{2+}}$ is reduced.	Magnesium loses two electrons, so it is oxidised. Each copper(II) ion gains those two electrons, so it is reduced. Adding the half equations cancels $2\mathrm{e^-}$ and gives the ionic equation.	5
Total Question 1			5
02.1	$2\mathrm{Al}\rightarrow2\mathrm{Al^{3+}}+6\mathrm{e^-}$ $3\mathrm{Cu^{2+}}+6\mathrm{e^-}\rightarrow3\mathrm{Cu}$ $2\mathrm{Al}+3\mathrm{Cu^{2+}}\rightarrow2\mathrm{Al^{3+}}+3\mathrm{Cu}$ ; aluminium is oxidised and copper(II) ions are reduced.	Each aluminium atom loses three electrons and each copper(II) ion gains two. Multiply the aluminium half equation by two and the copper half equation by three so that six electrons cancel. Atom counts and the total charge of $+6$ then balance in the overall equation.	6
Total Question 2			6
03.1	Each M atom loses three electrons and is oxidised. Each $\mathrm{X^{2+}}$ ion gains two electrons and is reduced. Six electrons are transferred in total. The left side has charge +6 and the right side also has charge +6.	Changing M from charge 0 to +3 requires a loss of three electrons per atom, so two M atoms lose six. Changing X from +2 to 0 requires two electrons per ion, so three ions gain six. The electron totals match, as do the equation charges: 3 × +2 on the left and 2 × +3 on the right.	5
Total Question 3			5
04.1	$\mathrm{Zn}\rightarrow\mathrm{Zn^{2+}}+2\mathrm{e^-}$ ; zinc is oxidised. $\mathrm{Cu^{2+}}+2\mathrm{e^-}\rightarrow\mathrm{Cu}$ ; copper(II) ions are reduced. $\mathrm{Zn}+\mathrm{Cu^{2+}}\rightarrow\mathrm{Zn^{2+}}+\mathrm{Cu}$ . Sulfate ions are spectators because they remain unchanged on both sides.	Zinc atoms lose two electrons and copper(II) ions gain the same two electrons. Adding the half equations cancels the electrons and gives the ionic equation. Sulfate ions accompany the metal ions in the full equation but undergo no change, so identical sulfate ions cancel from both sides.	6
Total Question 4			6
05.1	$2\mathrm{Fe^{3+}}+2\mathrm{e^-}\rightarrow2\mathrm{Fe^{2+}}$ ; iron(III) ions are reduced. $\mathrm{Sn^{2+}}\rightarrow\mathrm{Sn^{4+}}+2\mathrm{e^-}$ ; tin(II) ions are oxidised. Two electrons are transferred. The total charge is $+8$ on each side.	Each $\mathrm{Fe^{3+}}$ ion gains one electron, so two ions gain two electrons in total. The $\mathrm{Sn^{2+}}$ ion loses two electrons as its charge rises to $+4$ . Electron gain is reduction and electron loss is oxidation. The reactant charge is $2(+3)+(+2)=+8$ , and the product charge is $2(+2)+(+4)=+8$ .	6
Total Question 5			6

4.4.2.1 · Reactions of acids with metals

Tier 1 · Easy

Mark scheme for 4.4.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Magnesium chloride and hydrogen.	A metal plus an acid gives a salt plus hydrogen. Hydrochloric acid supplies chloride, so the salt is magnesium chloride and the gas is hydrogen.	2
Total Question 1			2
02.1	There is no reaction; copper is below hydrogen in the reactivity series and cannot displace hydrogen from the acid.	A metal must be above hydrogen to react with a dilute acid and release hydrogen gas. Copper is below hydrogen, so no fizzing or other reaction is expected.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	For the two observation marks, accept two distinct observations from: bubbles, effervescence or gas given off (one observation only); the zinc gets smaller or dissolves; the mixture gets warm or hot, or its temperature rises. A burning splint at the mouth of the tube gives a pop. The salt is zinc chloride.	A metal-acid reaction produces hydrogen gas, causing effervescence as the metal is used up; it is also exothermic, so a temperature rise is an accepted alternative observation. Hydrogen gives a pop with a burning splint. Hydrochloric acid forms chloride salts, so zinc forms zinc chloride.	5
Total Question 1			5
02.1	$\mathrm{Fe}+2\mathrm{HCl}\rightarrow\mathrm{FeCl_2}+\mathrm{H_2}$	Use $\mathrm{FeCl_2}$ for iron(II) chloride and $\mathrm{H_2}$ for diatomic hydrogen. A coefficient of two before $\mathrm{HCl}$ supplies two chlorine atoms and two hydrogen atoms, balancing every element.	3
Total Question 2			3
03.1	Magnesium produces hydrogen faster because it is more reactive than iron. Magnesium chloride forms.	The acid conditions and metal sizes are matched, so reactivity determines the rate comparison. Magnesium is above iron in the reactivity series and reacts faster. Hydrochloric acid forms a chloride salt, giving magnesium chloride.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.2.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$\mathrm{Mg}\rightarrow\mathrm{Mg^{2+}}+2\mathrm{e^-}$ ; magnesium is oxidised. $2\mathrm{H^+}+2\mathrm{e^-}\rightarrow\mathrm{H_2}$ ; hydrogen ions are reduced.	Magnesium loses two electrons to form $\mathrm{Mg^{2+}}$ , so it is oxidised. Two hydrogen ions gain the same two electrons and form $\mathrm{H_2}$ , so the hydrogen ions are reduced. The electrons cancel when the half equations are added.	4
Total Question 1			4
02.1	$\mathrm{Zn}\rightarrow\mathrm{Zn^{2+}}+2\mathrm{e^-}$ $2\mathrm{H^+}+2\mathrm{e^-}\rightarrow\mathrm{H_2}$ $\mathrm{Zn}+2\mathrm{H^+}\rightarrow\mathrm{Zn^{2+}}+\mathrm{H_2}$ ; sulfate ions are spectator ions because they remain unchanged.	Zinc loses two electrons and is oxidised. Two hydrogen ions gain those electrons and form hydrogen, so they are reduced. Adding the half equations cancels the electrons. Sulfate ions are present before and after the reaction without changing, so they cancel from the ionic equation.	6
Total Question 2			6
03.1	Measure the volume of hydrogen produced in a fixed time, or the time taken to collect a fixed volume. Keep the acid volume and concentration the same and use equal-sized or equal-mass metal samples. The expected order is magnesium, zinc, iron, then copper with no reaction.	Change only the metal. A gas syringe can measure hydrogen volume against time. Matching acid concentration, acid volume, temperature and metal surface area makes the comparison fair. The reactivity series predicts progressively slower reactions from magnesium to iron, while copper is below hydrogen and does not react with the dilute acid.	5
Total Question 3			5
04.1	A reacts faster than B because it produces more gas in the same first $20\,\mathrm{s}$ . A and B produce the same total amount of hydrogen, but this does not mean that their rates are equal. C does not react and is below hydrogen in the reactivity series.	Rate is compared using gas produced per unit time, so $34\,\mathrm{cm^3}$ against $18\,\mathrm{cm^3}$ in the same interval makes A faster. The shared final volume describes the total gas yield after completion, not how quickly it formed. A metal that gives no hydrogen with dilute acid is below hydrogen.	5
Total Question 4			5
05.1	X is magnesium. Copper is excluded because it is below hydrogen and does not react with dilute hydrochloric acid; aluminium is excluded because it forms $\mathrm{AlCl_3}$ , not $\mathrm{XCl_2}$ . $\mathrm{Mg}+2\mathrm{HCl}\rightarrow\mathrm{MgCl_2}+\mathrm{H_2}$ . The hydrogen gives a pop with a burning splint.	The reaction shows that X is above hydrogen, eliminating copper. Two chloride ions in $\mathrm{XCl_2}$ show that X forms $2+$ ions, which matches magnesium rather than aluminium's $3+$ ions. Two hydrochloric acid units balance chlorine and hydrogen in the equation, and the gaseous product is confirmed by the burning-splint pop test.	5
Total Question 5			5

4.4.2.2 · Neutralisation of acids and salt production

Tier 1 · Easy

Mark scheme for 4.4.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Potassium nitrate and water.	Nitric acid produces a nitrate, and potassium hydroxide supplies potassium ions. The salt is potassium nitrate; acid-alkali neutralisation also forms water.	2
Total Question 1			2
02.1	Magnesium nitrate and water.	Nitric acid forms nitrate salts, and magnesium oxide supplies magnesium ions. An acid reacting with a metal oxide also produces water.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Products (1 mark): calcium chloride, water and carbon dioxide. Correct formulae (1 mark): $\mathrm{CaCO_3}$ , $\mathrm{HCl}$ , $\mathrm{CaCl_2}$ , $\mathrm{H_2O}$ and $\mathrm{CO_2}$ in the correct equation positions. Balancing (1 mark): place 2 before $\mathrm{HCl}$ , giving $\mathrm{CaCO_3}+2\mathrm{HCl}\rightarrow\mathrm{CaCl_2}+\mathrm{H_2O}+\mathrm{CO_2}$ .	Award one mark only when all three products are named. For the equation, award one mark for the correct formulae and one for balancing them; two $\mathrm{HCl}$ supply the two chlorine atoms and two hydrogen atoms.	3
Total Question 1			3
02.1	$\mathrm{Ca(OH)_2}+2\mathrm{HCl}\rightarrow\mathrm{CaCl_2}+2\mathrm{H_2O}$	Two hydrochloric acid units provide the two chloride ions needed in $\mathrm{CaCl_2}$ . Their two hydrogen ions combine with the two hydroxide groups to make two water molecules. Calcium, oxygen, hydrogen and chlorine are then balanced.	3
Total Question 2			3
03.1	Use nitric acid and copper oxide (or copper hydroxide). The other product is water.	A nitrate salt requires nitric acid, and the copper ion must come from a copper base. Copper oxide or copper hydroxide supplies copper ions and neutralises the acid, forming copper nitrate and water.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.2.2 Tier 3 · Hard
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01.1	$\mathrm{Al_2(SO_4)_3}$ $2\mathrm{Al(OH)_3}+3\mathrm{H_2SO_4}\rightarrow\mathrm{Al_2(SO_4)_3}+6\mathrm{H_2O}$	Aluminium ions are $\mathrm{Al^{3+}}$ and sulfate ions are $\mathrm{SO_4^{2-}}$ , so the smallest neutral ratio is $2:3$ , giving $\mathrm{Al_2(SO_4)_3}$ . Use two aluminium hydroxide units and three sulfuric acid units; the six hydroxide groups and six acid hydrogens then form six waters.	5
Total Question 1			5
02.1	$\mathrm{Fe_2O_3}$ and $\mathrm{FeCl_3}$ $\mathrm{Fe_2O_3}+6\mathrm{HCl}\rightarrow2\mathrm{FeCl_3}+3\mathrm{H_2O}$	Iron(III) ions are $\mathrm{Fe^{3+}}$ . Balancing them with oxide ions gives $\mathrm{Fe_2O_3}$ , while balancing them with chloride ions gives $\mathrm{FeCl_3}$ . Two iron atoms require two $\mathrm{FeCl_3}$ ; six $\mathrm{HCl}$ then supply six chlorines and six hydrogens, forming three waters with the three oxygens.	5
Total Question 2			5
03.1	The claim is incorrect. The products are sodium sulfate, water and carbon dioxide. $\mathrm{Na_2CO_3}+\mathrm{H_2SO_4}\rightarrow\mathrm{Na_2SO_4}+\mathrm{H_2O}+\mathrm{CO_2}$ .	A carbonate reacting with an acid forms three products, not two. Sodium and sulfate form sodium sulfate, while the carbonate also gives water and carbon dioxide. The written equation has two sodium, one sulfur, one carbon, two hydrogen and seven oxygen atoms on each side.	5
Total Question 3			5
04.1	The reactants are nitric acid and calcium carbonate. Nitrate in the salt identifies nitric acid, while carbon dioxide and water together identify a carbonate reacting with an acid. $\mathrm{CaCO_3}+2\mathrm{HNO_3}\rightarrow\mathrm{Ca(NO_3)_2}+\mathrm{H_2O}+\mathrm{CO_2}$ .	Work backwards from both parts of the salt and from the extra products. Nitric acid supplies nitrate ions, and calcium carbonate supplies calcium ions while producing water and carbon dioxide. Two nitric acid units are required to provide the two nitrate groups and two hydrogen atoms, balancing the equation.	5
Total Question 4			5
05.1	Sulfuric acid would form aluminium sulfate, not aluminium nitrate. Use nitric acid; the other product is water. $\mathrm{Al_2O_3}+6\mathrm{HNO_3}\rightarrow2\mathrm{Al(NO_3)_3}+3\mathrm{H_2O}$ .	The acid supplies the negative ion in the salt, so nitrate requires nitric acid. Aluminium oxide is a base, so the second product is water. Two aluminium atoms give two aluminium nitrate units; these contain six nitrate groups, requiring six nitric acid units, and the six hydrogens and three oxide oxygens form three waters.	5
Total Question 5			5

4.4.2.3 · Soluble salts

Tier 1 · Easy

Mark scheme for 4.4.2.3 Tier 1 · Easy
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01.1	To ensure all the acid has reacted; the excess solid can then be removed by filtration.	Solid remaining shows that the acid is no longer able to react with more solid, so the acid has been used up. Because the added solid is insoluble, any excess can be filtered off.	2
Total Question 1			2
02.1	Gentle concentration avoids losing product by spitting (or allows crystals to form on cooling instead of leaving a dry powder).	Only some water needs to be removed. Strong heating to dryness can make the solution spit and lose salt, while leaving a concentrated solution allows crystals to grow as it cools.	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.4.2.3 Tier 2 · Standard
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01.1	Filtering removes the unreacted excess base. Heating evaporates some water to concentrate the salt solution, and cooling then allows salt crystals to form.	The excess reactant is an insoluble solid, so filtration separates it from the salt solution. Partial evaporation makes a concentrated solution; cooling then allows the dissolved salt to crystallise.	3
Total Question 1			3
02.1	Filter the mixture; zinc oxide is the residue and zinc sulfate solution is the filtrate.	The excess zinc oxide is insoluble, so filtration traps it on the filter paper as the residue. The soluble zinc sulfate passes through in solution as the filtrate.	3
Total Question 2			3
03.1	Copper oxide supplies the copper needed for copper sulfate and is insoluble, so excess copper oxide can be filtered off. Sodium hydroxide would form sodium sulfate and any excess is soluble, so it could not be removed by filtration.	The positive ion in the solid becomes part of the salt, so a copper compound is required. The excess-solid method also needs an insoluble reactant: unreacted copper oxide remains as a residue, whereas dissolved sodium hydroxide would pass through the filter and contaminate the solution.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.2.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Warm the acid, add zinc carbonate until it is in excess, filter, partly evaporate the filtrate, cool to crystallise, then separate and dry the crystals.	Warm the dilute hydrochloric acid gently and add zinc carbonate in portions until fizzing stops and solid remains. This uses up the acid. Filter to remove excess insoluble zinc carbonate. Use a water bath or electric heater to evaporate some water from the zinc chloride filtrate, then cool so crystals form. Filter or decant the crystals and dry them between filter papers or in a warm place.	6
Total Question 1			6
02.1	Add magnesium oxide until some remains, ensuring that all acid is neutralised. Filter to remove the unreacted solid. Evaporate only some water and cool the concentrated solution so crystals form, then separate and dry the crystals.	Add the insoluble oxide in portions until it is visibly in excess, ensuring no acid remains. Filter so unreacted magnesium oxide does not contaminate the product. Concentrate the filtrate gently rather than drying it out, then cool it so magnesium sulfate crystallises. Finally separate the crystals and dry their surfaces.	6
Total Question 2			6
03.1	Sodium hydroxide is soluble, so any excess cannot be removed by filtration. Use a titration with indicator to find the exact reacting volumes, then repeat using those volumes without indicator. Gently concentrate the neutral sodium chloride solution, cool it to crystallise, and separate and dry the crystals.	The excess-solid method relies on an insoluble base leaving a filterable residue. Sodium hydroxide stays dissolved, so an excess would remain in the product. A preliminary titration identifies the neutral volumes; repeating them without indicator gives an uncontaminated salt solution that can be concentrated and crystallised.	5
Total Question 3			5
04.1	The soluble salt dissolves in the large volume of warm water, so some product passes away in the washings. Rinse with only a small volume of cold distilled water. Remove surface water by pressing the crystals between dry filter papers or drying them in a warm place.	The desired crystals are soluble, and both a larger solvent volume and a higher temperature allow more salt to dissolve. A minimal cold rinse removes soluble impurities while dissolving less product. Drying afterwards removes water from the crystal surfaces without washing away further salt.	4
Total Question 4			4
05.1	The crystals are dry when the last two readings are both $39.71\,\mathrm{g}$ . Their dry mass is $39.71-32.46=7.25\,\mathrm{g}$ . Repeated drying and weighing checks for constant mass, showing that no more water is being removed.	The falling readings show water leaving the sample. The repeated value of $39.71\,\mathrm{g}$ is constant mass, so further gentle drying causes no measurable loss. Subtracting the empty-basin mass gives $7.25\,\mathrm{g}$ of dry crystals.	4
Total Question 5			4

4.4.2.4 · The pH scale and neutralisation

Tier 1 · Easy

Mark scheme for 4.4.2.4 Tier 1 · Easy
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01.1	pH $3$ : acidic; pH $7$ : neutral; pH $11$ : alkaline.	Compare each value with $7$ . Values below $7$ are acidic, $7$ is neutral, and values above $7$ are alkaline.	3
Total Question 1			3
02.1	pH $7$ ; hydroxide ions, $\mathrm{OH^-}$ .	Neutral solutions have pH $7$ . Alkalis dissolve in water and produce hydroxide ions, written $\mathrm{OH^-}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.2.4 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Sodium hydroxide is in excess. Hydrogen ions and hydroxide ions react to form water, but hydroxide ions remain after all the hydrogen ions have reacted, so the solution is alkaline.	Neutralisation removes $\mathrm{H^+}$ and $\mathrm{OH^-}$ in reacting amounts. A final pH above $7$ shows that unreacted $\mathrm{OH^-}$ remains, so the alkali was in excess.	3
Total Question 1			3
02.1	Both methods estimate the pH of the solution. Universal indicator gives an approximate pH by matching a colour to a chart, whereas a calibrated pH probe gives a numerical reading and is more precise.	Indicator colours cover ranges and depend on a visual comparison, so the result is approximate. A calibrated probe measures pH directly as a number and resolves smaller differences.	3
Total Question 2			3
03.1	pH 2: red; pH 7: green; pH 12: purple.	Use the universal-indicator scale. Strongly acidic solutions are red, neutral solution is green, and strongly alkaline solutions are purple.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.2.4 Tier 3 · Hard
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01.1	$\mathrm{OH^-}$ ions react with $\mathrm{H^+}$ ions, decreasing the hydrogen ion concentration until the solution is neutral. $\mathrm{H^+(aq)}+\mathrm{OH^-(aq)}\rightarrow\mathrm{H_2O(l)}$ Equal volumes may contain different reacting amounts because the solutions may have different concentrations.	The alkali supplies $\mathrm{OH^-}$ ions, which remove $\mathrm{H^+}$ ions by forming water. As the hydrogen ion concentration falls, pH rises; at pH $7$ the mixture is neutral. Neutrality depends on equal reacting amounts of $\mathrm{H^+}$ and $\mathrm{OH^-}$ , not equal solution volumes, because concentration also affects how many ions are present.	5
Total Question 1			5
02.1	At pH $12$ , $\mathrm{OH^-}$ is in excess; at pH $7$ , neither $\mathrm{H^+}$ nor $\mathrm{OH^-}$ is in excess; at pH $2$ , $\mathrm{H^+}$ is in excess. $\mathrm{H^+(aq)}+\mathrm{OH^-(aq)}\rightarrow\mathrm{H_2O(l)}$	The initially alkaline mixture contains excess hydroxide ions. At neutral pH the reacting amounts of hydrogen and hydroxide ions have removed each other as water. Adding acid beyond that point leaves excess hydrogen ions, making the mixture acidic. The ionic equation balances atoms and total charge.	5
Total Question 2			5
03.1	The neutral point lies between 12.5 and 12.6 cm³. Add acid in smaller volume steps within that interval and record more readings. A calibrated pH probe gives numerical values and is more precise than judging indicator colours.	Neutral pH 7 lies between the measured values 8.1 and 6.2, so the corresponding volume is bracketed by 12.5 and 12.6 cm³. Smaller additions provide closer bracketing readings. A probe resolves the rapid pH change without relying on subjective colour matching.	4
Total Question 3			4
04.1	After three portions, $40$ $\mathrm{H^+}$ ions remain, so the mixture is acidic. After four portions, neither ion is in excess, so the mixture is neutral. After five portions, $40$ $\mathrm{OH^-}$ ions remain, so the mixture is alkaline. $\mathrm{H^+(aq)}+\mathrm{OH^-(aq)}\rightarrow\mathrm{H_2O(l)}$ .	Three portions supply $3\times40=120$ hydroxide ions, leaving $160-120=40$ hydrogen ions. Four portions supply exactly $160$ , so all represented hydrogen and hydroxide ions react. A fifth portion adds $40$ hydroxide ions after neutralisation, leaving hydroxide in excess. The ionic equation shows the one-to-one removal of the ions as water.	6
Total Question 4			6
05.1	Universal indicator is unsuitable because the solution colour masks the indicator colour and makes the pH estimate unreliable. Use a calibrated pH probe. Rinse the probe with distilled water between readings, add the neutralising solution in small portions or dropwise near pH $7$ , and record the pH after mixing.	Indicator requires a reliable visual colour match, which the dark sample prevents. A calibrated probe gives a numerical pH independent of the sample colour. Rinsing prevents carry-over between samples, while smaller additions near pH $7$ reduce the risk of passing the neutral point.	4
Total Question 5			4

4.4.2.5 · Titrations (chemistry only)

Tier 1 · Easy

Mark scheme for 4.4.2.5 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	A volumetric pipette and a burette, respectively.	A volumetric pipette delivers one fixed accurate volume. A burette has a graduated scale and tap, so it measures the variable volume added to the flask.	2
Total Question 1			2
02.1	$22.35\,\mathrm{cm^3}$ .	The titre is the final burette reading minus the initial reading: $23.80-1.45=22.35\,\mathrm{cm^3}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.2.5 Tier 2 · Standard
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01.1	Mean titre $=24.75\,\mathrm{cm^3}$ , from $\dfrac{24.80+24.75+24.70}{3}$ . The unit may be omitted, but the rough titre must be excluded — including it gives $24.96\,\mathrm{cm^3}$ , which is wrong.	Use the concordant accurate titres and exclude the rough titre. Their total is $74.25\,\mathrm{cm^3}$ , so dividing by three gives $24.75\,\mathrm{cm^3}$ .	2
Total Question 1			2
02.1	Use a volumetric pipette for the fixed volume; add acid dropwise near the end point; swirl the flask so the solutions mix completely.	A volumetric pipette transfers the fixed volume more accurately than a measuring cylinder. Single drops reduce the chance of passing the end point. Swirling brings added acid into contact with all the alkali, so the indicator change represents the whole mixture.	3
Total Question 2			3
03.1	Read the bottom of the meniscus at eye level and record readings to two decimal places. Remove the funnel so extra drops cannot enter the burette after the initial reading and change the measured titre.	Eye-level viewing avoids parallax and the meniscus gives a consistent reference point. A burette scale supports readings to the nearest 0.05 cm³, written with two decimal places. A funnel left in place can drip between readings, adding unmeasured solution.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.2.5 Tier 3 · Hard
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01.1	$0.112\,\mathrm{mol\,dm^{-3}}$ and $4.46\,\mathrm{g\,dm^{-3}}$ .	Convert the acid volume: $18.60\,\mathrm{cm^3}=0.01860\,\mathrm{dm^3}$ . Acid amount $=0.150\times0.01860=0.002790\,\mathrm{mol}$ . The $1:1$ ratio gives the same amount of $\mathrm{NaOH}$ . Its volume is $0.0250\,\mathrm{dm^3}$ , so concentration $=0.002790/0.0250=0.1116\,\mathrm{mol\,dm^{-3}}$ , or $0.112\,\mathrm{mol\,dm^{-3}}$ . Multiplying by $40.0$ gives $4.464\,\mathrm{g\,dm^{-3}}$ , or $4.46\,\mathrm{g\,dm^{-3}}$ .	5
Total Question 1			5
02.1	$0.0810\,\mathrm{mol\,dm^{-3}}$ and $7.94\,\mathrm{g\,dm^{-3}}$ (accept $7.93$ to $7.95\,\mathrm{g\,dm^{-3}}$ ).	Potassium hydroxide amount $=0.125\times0.03240=0.004050\,\mathrm{mol}$ . The $2:1$ ratio gives $0.002025\,\mathrm{mol}$ of $\mathrm{H_2SO_4}$ . Divide by $0.0250\,\mathrm{dm^3}$ to obtain $0.0810\,\mathrm{mol\,dm^{-3}}$ . Multiplying the unrounded concentration by $98.0$ gives $7.938\,\mathrm{g\,dm^{-3}}$ , which is $7.94\,\mathrm{g\,dm^{-3}}$ to three significant figures.	6
Total Question 2			6
03.1	The first recorded titre is too large because part of the measured volume fills the jet instead of entering the flask. Later titres are smaller because the jet is already full. Run solution through the jet before taking the initial reading, and treat the affected first run as rough or exclude it from the mean.	The burette reading records all solution leaving the barrel. During the faulty run, some of that solution removes the air bubble and fills the jet, so the barrel change exceeds the volume delivered to the flask. Filling the jet before the initial reading removes the systematic error; the anomalous run must not contribute to the concordant mean.	5
Total Question 3			5
04.1	The accurate titres are $24.70$ , $24.75$ , $25.10$ and $24.75\,\mathrm{cm^3}$ . Runs 1, 2 and 4 are concordant. Mean $=(24.70+24.75+24.75)/3=24.733\ldots\,\mathrm{cm^3}=24.73\,\mathrm{cm^3}$ to two decimal places.	Subtract each initial reading from its final reading: $25.05-0.35=24.70$ , $25.95-1.20=24.75$ , $25.90-0.80=25.10$ and $26.90-2.15=24.75\,\mathrm{cm^3}$ . The first, second and fourth accurate results lie within $0.10\,\mathrm{cm^3}$ of one another. Their total is $74.20\,\mathrm{cm^3}$ , giving $74.20/3=24.733\ldots\,\mathrm{cm^3}$ .	4
Total Question 4			4
05.1	The water in A's flask does not change the amount of alkali delivered by the pipette, so it does not change the titre. The water in B's burette dilutes the acid, so a larger volume is needed and the titre is too large. Rinse the conical flask with distilled water, but rinse the burette with the acid solution before filling it.	Extra distilled water in the flask increases total volume but does not add or remove alkali particles, so the reacting amount is unchanged. Water left in the burette lowers the acid concentration; more of that diluted acid is then required to neutralise the fixed alkali amount. The flask may therefore be water-rinsed, whereas the burette must be conditioned with its own solution.	5
Total Question 5			5

4.4.2.6 · Strong and weak acids (HT only)

Tier 1 · Easy

Mark scheme for 4.4.2.6 Tier 1 · Easy
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01.1	A strong acid ionises completely; a weak acid ionises only partially.	Use degree of ionisation, not concentration. Complete ionisation defines a strong acid, while only a fraction of acid particles ionising defines a weak acid.	2
Total Question 1			2
02.1	The acid is strong because it ionises completely. Dilute describes the amount of acid per volume, not its degree of ionisation.	Strength is defined by the fraction of acid particles that ionise. Complete ionisation means strong even when the solution contains only a small amount of acid per unit volume.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.2.6 Tier 2 · Standard
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01.1	Hydrochloric acid has the lower pH because it ionises completely and produces a greater hydrogen ion concentration; neither is more concentrated because their concentrations are equal.	Hydrochloric acid is strong, so essentially all its acid particles ionise. Ethanoic acid is weak and only partially ionises. At equal starting concentration, hydrochloric acid therefore has the greater $\mathrm{H^+}$ concentration and lower pH. Strength does not change the given concentration comparison.	4
Total Question 1			4
02.1	Concentrated means there is a large amount of ethanoic acid per unit volume, whereas weak means that only a small proportion of its particles ionise in water.	The two terms describe different properties. Concentration concerns how much solute is present in a volume; acid strength concerns the degree of ionisation. A solution can therefore satisfy both descriptions.	3
Total Question 2			3
03.1	Hydrochloric acid reacts faster because it ionises completely and therefore has a greater concentration of H⁺ ions. Ethanoic acid only partially ionises.	The starting acid concentrations and magnesium pieces are matched, so acid strength controls the available hydrogen ion concentration. Complete ionisation of hydrochloric acid supplies more H⁺ ions at once than partial ionisation of ethanoic acid, increasing the reaction rate.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.2.6 Tier 3 · Hard
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01.1	A has $1000$ times the hydrogen ion concentration of B.	The pH difference is $5-2=3$ units. Each unit represents a factor of $10$ , so the ratio is $10^3=1000$ . The pH scale is logarithmic, so a three-unit difference is not a factor of three.	4
Total Question 1			4
02.1	The claim is incorrect. pH indicates hydrogen ion concentration, while strength is the degree of ionisation. A dilute strong acid can have pH $3$ , and a sufficiently concentrated weak acid can have a lower pH than a more dilute strong acid.	A pH value describes the resulting $\mathrm{H^+}$ concentration, which depends on both the starting acid concentration and how far it ionises. Strength alone therefore cannot be inferred from one pH reading without concentration information.	4
Total Question 2			4
03.1	The new pH is 4. A 1000-fold decrease is three factors of 10, so pH rises by three units. Dilution lowers concentration but does not change whether the acid ionises completely or partially, so its strength is unchanged.	Write 1000 as 10³. Each tenfold fall in H⁺ concentration increases pH by one, giving 1 + 3 = 4. Strength is a property of degree of ionisation, whereas dilution changes the amount of acid per volume.	4
Total Question 3			4
04.1	The pH $2$ solution has $10^2=100$ times the hydrogen ion concentration of the pH $4$ solution. The pH $2$ hydrochloric acid is more concentrated. Both are strong acids because hydrochloric acid ionises completely; concentration and strength are different properties.	The two-unit pH difference represents two factors of ten, so the hydrogen ion concentration ratio is $100:1$ . Because both samples contain the same strong, completely ionising acid, the greater hydrogen ion concentration comes from a greater acid concentration. Their degree of ionisation remains complete in both solutions, so their strength classification is unchanged.	4
Total Question 4			4
05.1	The same pH means the solutions have the same hydrogen ion concentration. They do not have the same strength: hydrochloric acid is strong and ionises completely, while ethanoic acid is weak and ionises only partially. Their starting acid concentrations need not be equal; more weak-acid particles may be needed to produce the same hydrogen ion concentration, and the exact concentration cannot be found from pH alone.	pH describes the resulting $\mathrm{H^+}$ concentration, not how completely the acid ionised or how many acid particles were initially dissolved. Hydrochloric acid supplies hydrogen ions by complete ionisation, whereas only a fraction of ethanoic acid particles ionise. Equal pH therefore does not imply equal strength or equal starting concentration.	5
Total Question 5			5

4.4.3.1 · The process of electrolysis

Tier 1 · Easy

Mark scheme for 4.4.3.1 Tier 1 · Easy
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01.1	The cathode; it is negative.	Opposite charges attract. Positive ions therefore move towards the negative electrode, which is called the cathode.	2
Total Question 1			2
02.1	An electrolyte is a molten ionic compound or ionic solution that conducts electricity; moving ions carry the charge.	An ionic substance acts as an electrolyte only when its ions can move. The mobile positive and negative ions transport charge through the liquid or solution.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.3.1 Tier 2 · Standard
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01.1	The mobile positive ions move to the negative cathode and the mobile negative ions move to the positive anode. The ions are discharged at the electrodes to form products, decomposing the electrolyte.	The electric field drives oppositely charged ions towards opposite electrodes. Electron transfer at the electrodes discharges the ions, so the ionic compound is split into products.	3
Total Question 1			3
02.1	In the solid, the ions are fixed in a lattice and cannot move. In solution, the ions are free to move and carry charge.	Both samples contain ions, but conduction needs mobile charged particles. Dissolving separates the ions from fixed lattice positions, allowing them to move through the solution.	3
Total Question 2			3
03.1	Electrons carry charge through the metal wire, whereas mobile positive and negative ions carry charge through the electrolyte.	Metals contain delocalised electrons that can move through the external circuit. In the liquid electrolyte, charged ions move instead: cations travel towards the cathode and anions towards the anode.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.3.1 Tier 3 · Hard
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01.1	$\mathrm{M^{2+}}$ moves to the negative cathode and forms metal M; $\mathrm{X^-}$ moves to the positive anode and forms non-metal X; moving ions carry charge through the melt.	Cations move towards the negative cathode and are discharged as the metal. Anions move towards the positive anode and are discharged as the non-metal. Within the molten electrolyte, both types of mobile ion transport charge.	5
Total Question 1			5
02.1	Anions move to the positive anode; cations move to the negative cathode; ions, not electrons, carry charge through the electrolyte. Mobile ions reach the electrodes and are discharged to form products.	Opposite charges attract, so negative ions travel to the positive electrode and positive ions to the negative electrode. The electrolyte conducts because these ions move. Their discharge at the electrodes decomposes the ionic substance into products.	5
Total Question 2			5
03.1	Electrons carry charge through the wires, while mobile ions carry charge through the melt. Cations move to the negative cathode and anions move to the positive anode. On solidifying, the ions become fixed in a lattice, so charge cannot pass through the compound and ions cannot reach the electrodes to form products.	Separate the external circuit from the electrolyte. Electron movement provides current in the metal conductors, but ion movement completes the circuit through the molten compound. Freezing removes that ion mobility, so discharge and decomposition cease even though ions are still present.	5
Total Question 3			5
04.1	The left electrode becomes the positive anode and the right electrode becomes the negative cathode. $\mathrm{X^-}$ ions now move left to the anode, while $\mathrm{M^{2+}}$ ions move right to the cathode. The products form at the opposite physical electrodes because cathode and anode names follow polarity, not a fixed left or right position.	Reversing the supply reverses each electrode's charge. Cations always move to the negative cathode, so $\mathrm{M^{2+}}$ changes direction and moves right. Anions always move to the positive anode, so $\mathrm{X^-}$ changes direction and moves left. Each discharge product therefore switches sides with its electrode process.	5
Total Question 4			5
05.1	Molten sodium chloride and sodium chloride solution allow sustained electrolysis because they contain mobile ions. Solid sodium chloride contains ions, but they are fixed in a lattice and cannot carry charge to the electrodes. Sugar solution contains neutral sugar molecules rather than mobile ions, so it does not act as an electrolyte.	Electrolysis requires mobile charged particles in the sample. Melting sodium chloride frees its ions, and dissolving it separates the ions so they can move through water. The solid still has ions but no ion mobility. Sugar dissolves as neutral molecules, so its solution lacks the ions needed to complete the circuit and undergo electrolysis.	5
Total Question 5			5

4.4.3.2 · Electrolysis of molten ionic compounds

Tier 1 · Easy

Mark scheme for 4.4.3.2 Tier 1 · Easy
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01.1	Zinc at the cathode; chlorine at the anode.	The molten compound contains zinc ions and chloride ions only. Metal ions form the metal at the cathode, while chloride ions form chlorine at the anode.	2
Total Question 1			2
02.1	Oxide ions, $\mathrm{O^{2-}}$ ; oxygen, $\mathrm{O_2}$ .	Negative oxide ions are attracted to the positive anode. When discharged, oxide ions form the element oxygen, which exists as diatomic $\mathrm{O_2}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.3.2 Tier 2 · Standard
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01.1	Molten magnesium chloride contains no water or hydrogen ions. Its positive ions are $\mathrm{Mg^{2+}}$ , so magnesium forms at the cathode.	Do not apply the competing-ion rules for aqueous solutions to a molten compound. The only cations in molten $\mathrm{MgCl_2}$ are magnesium ions, which gain electrons at the cathode.	3
Total Question 1			3
02.1	Chloride ions are negative, so they move to the positive anode, not the cathode. They form diatomic chlorine, $\mathrm{Cl_2}$ , at the anode.	The positive anode attracts $\mathrm{Cl^-}$ ions. Chlorine is diatomic, so discharged chlorine atoms pair to make $\mathrm{Cl_2}$ rather than remaining as separate atoms.	4
Total Question 2			4
03.1	The ions are $\mathrm{Mg^{2+}}$ and $\mathrm{Br^-}$ , so the compound is $\mathrm{MgBr_2}$ . Magnesium forms at the cathode.	The products identify magnesium ions and bromide ions. One $\mathrm{Mg^{2+}}$ requires two $\mathrm{Br^-}$ ions for a neutral formula, giving $\mathrm{MgBr_2}$ . Positive magnesium ions move to the negative cathode and form magnesium.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.3.2 Tier 3 · Hard
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01.1	Calcium forms at the cathode and chlorine forms at the anode; there are two chloride ions for each calcium ion, and two chloride ions combine to form one $\mathrm{Cl_2}$ molecule.	Molten $\mathrm{CaCl_2}$ contains $\mathrm{Ca^{2+}}$ and $\mathrm{Cl^-}$ ions. Calcium ions form calcium metal at the cathode. Chloride ions form $\mathrm{Cl_2}$ at the anode. Charge balance in the formula requires two chloride ions for each calcium ion, matching one calcium atom and one chlorine molecule formed.	5
Total Question 1			5
02.1	Molten lithium bromide produces lithium at the cathode and bromine at the anode. Aqueous lithium bromide produces hydrogen at the cathode and bromine at the anode. The molten compound contains lithium ions as its only cations. In water, lithium is more reactive than hydrogen, so hydrogen is discharged instead; bromide is a halide and forms bromine in both cases.	Apply the molten rule first: the ions of the binary compound form its two elements. In the aqueous sample, water supplies competing hydrogen ions, and lithium's position above hydrogen selects hydrogen at the cathode. Bromide ions still give bromine at the anode under the aqueous halide rule.	5
Total Question 2			5
03.1	11.4 g of non-metal is collected. The metal forms at the cathode and the non-metal at the anode. The compound must be molten so its ions are mobile and can carry charge to the electrodes.	Because both products are captured in the sealed apparatus, mass is conserved: 18.6 − 7.2 = 11.4 g. Metal cations move to the cathode, while non-metal anions move to the anode. In a solid lattice the ions would be fixed and electrolysis could not continue.	4
Total Question 3			4

4.4.3.3 · Using electrolysis to extract metals

Tier 1 · Easy

Mark scheme for 4.4.3.3 Tier 1 · Easy
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01.1	Aluminium is more reactive than carbon, so carbon cannot reduce aluminium oxide.	Use the reactivity series. A metal above carbon cannot be displaced from its oxide by carbon, so aluminium requires electrolysis.	2
Total Question 1			2
02.1	Aluminium forms at the negative cathode because positive aluminium ions are attracted to the negative electrode.	The electrolyte contains positively charged aluminium ions. Opposite charges attract, so these ions travel to the negative cathode, where aluminium is produced.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.3.3 Tier 2 · Standard
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01.1	Large amounts of electricity are needed to keep the electrolyte molten and drive electrolysis, so generating that electricity may release greenhouse gases. Oxygen formed at the anode reacts with the carbon anodes to produce carbon dioxide.	Link each effect to the process. Heating and maintaining the current create a large electricity demand, while consumption of the carbon anodes forms carbon dioxide.	4
Total Question 1			4
02.1	Use electrolysis for A because carbon cannot reduce the oxide of a more reactive metal. Use carbon reduction for B because carbon is more reactive than B and can remove oxygen from its oxide.	Compare each metal with carbon. An oxide of a metal above carbon cannot be reduced by carbon and needs electrolysis. Carbon can take oxygen from the oxide of a metal below it.	4
Total Question 2			4
03.1	80 kg is lost. Oxygen formed at the anode reacts with the carbon to form carbon dioxide, consuming the anode.	Subtract the final mass from the initial mass: 500 − 420 = 80 kg. The electrode is not merely worn mechanically; oxygen produced from oxide ions reacts with carbon, so carbon leaves the cell in carbon dioxide.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.3.3 Tier 3 · Hard
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01.1	Mixture A; its lower melting point requires less heating. Oxygen formed at the anode reacts with carbon to make carbon dioxide, consuming the anode.	Because both mixtures give the same output, the lower operating temperature makes A the likely lower-energy option. During electrolysis, oxygen is produced at the positive electrode. It reacts with the carbon electrode to form $\mathrm{CO_2}$ , so carbon is steadily lost and the anode must be replaced.	5
Total Question 1			5
02.1	A lower melting point means less heating is needed to melt and maintain the electrolyte, reducing energy use. Oxide ions form oxygen at the positive carbon anodes; the oxygen reacts with carbon to make carbon dioxide, so direct emissions remain even when the electricity is renewable.	Electrolysis requires mobile ions in a liquid electrolyte, so lowering its melting point lowers the required operating temperature. The electricity source does not change the electrode chemistry: oxygen made at the anode consumes carbon as $\mathrm{CO_2}$ .	5
Total Question 2			5
03.1	Possible advantages are lower direct carbon dioxide emissions and less frequent anode replacement. Relevant evidence includes the new anodes' lifetime, cost, energy use, effect on aluminium purity, or whether they remain unreactive at the operating temperature.	Current carbon anodes are consumed when oxygen forms carbon dioxide, so an unreactive material could remove both that emission source and replacement demand. A decision still needs operational evidence: the material must survive the cell conditions without increasing energy use, contaminating the product or costing more overall.	4
Total Question 3			4
04.1	$120\,\mathrm{kg}$ of carbon is lost. The carbon dioxide mass is greater than the carbon mass lost because oxygen from the electrolyte also contributes to the carbon dioxide mass. The cell is not carbon-free: renewable electricity can remove electricity-generation emissions, but carbon anodes still react with oxygen and produce carbon dioxide directly.	The anode loss is $640-520=120\,\mathrm{kg}$ . That is the mass of carbon consumed, not the mass of the compound formed; carbon dioxide also contains oxygen supplied by oxide ions. Changing the electricity source does not change this electrode reaction, so direct cell emissions remain.	4
Total Question 4			4
05.1	The carbon-anode cell uses $28\,000\,\mathrm{kWh}$ . It consumes $900\,\mathrm{kg}$ of carbon. The inert-anode cell uses $32\,000\,\mathrm{kWh}$ . It consumes no carbon anode. Choose the inert-anode cell for lower direct cell emissions. It uses $4000\,\mathrm{kWh}$ more electricity, so cost and emissions from electricity generation must also be considered.	Multiply each per-kilogram value by $2000$ . The carbon design uses $14\times2000=28\,000\,\mathrm{kWh}$ and $0.45\times2000=900\,\mathrm{kg}$ of carbon. The inert design uses $16\times2000=32\,000\,\mathrm{kWh}$ and no carbon. It meets the stated direct-emissions priority but requires $32\,000-28\,000=4000\,\mathrm{kWh}$ more electricity.	6
Total Question 5			6

4.4.3.4 · Electrolysis of aqueous solutions

Tier 1 · Easy

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01.1	Hydrogen at the cathode; chlorine at the anode.	Sodium is more reactive than hydrogen, so hydrogen is produced at the cathode. Chloride is a halide ion, so chlorine is produced at the anode.	2
Total Question 1			2
02.1	Hydrogen; zinc is more reactive than hydrogen, so hydrogen is formed instead of zinc at the cathode.	For an aqueous electrolyte, compare the dissolved metal with hydrogen. Zinc is above hydrogen in the reactivity series, so the water-derived hydrogen ions are discharged.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.3.4 Tier 2 · Standard
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01.1	The solute is copper(II) chloride. Copper forms because it is less reactive than hydrogen, and chlorine forms because chloride is a halide ion.	The cathode product requires copper ions, eliminating sodium chloride. The anode product requires halide ions, eliminating copper(II) sulfate. Copper(II) chloride supplies both $\mathrm{Cu^{2+}}$ and $\mathrm{Cl^-}$ .	3
Total Question 1			3
02.1	Sodium sulfate. Sodium is more reactive than hydrogen, so hydrogen forms at the cathode. Sulfate is not a halide, so oxygen forms at the anode.	The pop identifies hydrogen and the relighting test identifies oxygen. Sodium sulfate gives that pair: sodium is not deposited from water and there is no halide to form a halogen. Sodium chloride would give chlorine, while copper(II) sulfate would deposit copper.	4
Total Question 2			4
03.1	Hydrogen bubbles at the cathode and gives a pop with a burning splint. Bromine forms at the anode, producing an orange-brown colour.	Sodium is more reactive than hydrogen, so hydrogen forms instead of sodium at the cathode and is confirmed by the pop test. Bromide is a halide ion, so it forms bromine at the anode; bromine gives the orange-brown observation.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.4.3.4 Tier 3 · Hard
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01.1	Magnesium chloride: hydrogen at the cathode and chlorine at the anode. Silver nitrate: silver at the cathode and oxygen at the anode.	Magnesium is above hydrogen, so water-derived hydrogen is discharged at the magnesium chloride cathode; chloride gives chlorine at its anode. Silver is below hydrogen, so silver forms at the silver nitrate cathode. Nitrate is not a halide, so oxygen forms at that solution's anode.	6
Total Question 1			6
02.1	Connect inert electrodes in aqueous sodium chloride to a direct-current supply. Hydrogen forms at the cathode and gives a pop with a burning splint. Chlorine forms at the anode and bleaches damp litmus paper. Use small quantities in a fume cupboard or well-ventilated area and avoid inhaling chlorine.	Place graphite or other inert electrodes in the solution without letting them touch, then pass a direct current. Test the cathode gas with a burning splint for hydrogen. Test the anode gas with damp litmus paper for chlorine bleaching. Chlorine is toxic, so minimise exposure, use ventilation and wear eye protection.	6
Total Question 2			6
03.1	Sodium sulfate gives hydrogen at the cathode and oxygen at the anode. Copper(II) sulfate gives copper at the cathode and oxygen at the anode. The cathode product changes because sodium is above hydrogen but copper is below hydrogen. The anode product stays oxygen because sulfate is not a halide in either solution.	Apply the cathode rule separately to each metal ion: sodium is too reactive to be deposited from water, whereas copper is deposited. At the anode, neither solution contains a halide, so hydroxide ions from water form oxygen in both cases.	5
Total Question 3			5
04.1	Aqueous calcium bromide produces hydrogen at the cathode. It produces bromine at the anode. Aqueous calcium nitrate also produces hydrogen at the cathode. It produces oxygen at the anode. Both cathodes produce hydrogen because calcium is more reactive than hydrogen. The bromide ions form bromine, whereas nitrate is not a halide, so oxygen forms from the water in calcium nitrate solution.	Apply each discharge rule independently. Calcium is above hydrogen in the reactivity series, so water-derived hydrogen is discharged at both cathodes. Bromide is a halide and forms bromine at its anode. Nitrate is not a halide, so hydroxide ions from water form oxygen at the other anode.	6
Total Question 4			6
05.1	Copper forms at the cathode first because copper is less reactive than hydrogen. Removing $\mathrm{Cu^{2+}}$ ions makes the blue colour fade. After the copper(II) ions are used up, hydrogen forms at the cathode from water-derived hydrogen ions. Oxygen forms at the anode throughout because sulfate is not a halide.	Initially, copper(II) ions are selected at the cathode because copper is below hydrogen, so copper deposits and its ions are removed from the blue solution. Once no copper(II) ions remain, the competing water-derived hydrogen ions are discharged and hydrogen bubbles appear. The anode rule is unchanged throughout: sulfate is not a halide, so oxygen forms at the inert anode.	5
Total Question 5			5

4.4.3.5 · Representation of reactions at electrodes as half equations (HT only)

Tier 1 · Easy

Mark scheme for 4.4.3.5 Tier 1 · Easy
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01.1	$\mathrm{Al^{3+}}+3\mathrm{e^-}\rightarrow\mathrm{Al}$ ; reduction.	Three electrons give total charge $-3$ , balancing the ion's $+3$ charge to form neutral aluminium. The ion gains electrons, so this is reduction.	2
Total Question 1			2
02.1	Oxidation; iodide ions lose electrons.	Electrons appear on the product side, showing that the iodide ions release them. Loss of electrons is oxidation.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.4.3.5 Tier 2 · Standard
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01.1	$\mathrm{Cu^{2+}}+2\mathrm{e^-}\rightarrow\mathrm{Cu}$ Reduction, because the copper(II) ions gain electrons. Hydrogen is then produced instead, because hydrogen ions from the water are discharged once no copper(II) ions remain.	A copper(II) ion gains two electrons to become a neutral copper atom, so the cathode reaction is reduction. With inert electrodes nothing replenishes the copper(II) ions, so when they run out the next-least-reactive cation present is discharged — the hydrogen ions from the water — and hydrogen is given off.	3
Total Question 1			3
02.1	$4\mathrm{OH^-}\rightarrow\mathrm{O_2}+2\mathrm{H_2O}+4\mathrm{e^-}$ ; each side has four oxygen atoms, four hydrogen atoms and total charge $-4$ .	Four hydroxide ions supply four oxygen atoms and four hydrogen atoms. One $\mathrm{O_2}$ and two $\mathrm{H_2O}$ balance those atoms. Four electrons on the right make its total charge $-4$ , matching the left.	4
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03.1	$n=3$ , so the ion is $\mathrm{M^{3+}}$ . The charges +3 and −3 sum to zero on the left, matching neutral M on the right. The process is reduction.	Three electrons contribute total charge −3. The ion must therefore have charge +3 for the reactant side to be neutral. The ion gains electrons at the cathode, so the change is reduction.	3
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Tier 3 · Hard

Mark scheme for 4.4.3.5 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$2\mathrm{H^+}+2\mathrm{e^-}\rightarrow\mathrm{H_2}$ : reduction. $4\mathrm{OH^-}\rightarrow\mathrm{O_2}+2\mathrm{H_2O}+4\mathrm{e^-}$ : oxidation.	Two hydrogen ions gain two electrons to form one hydrogen molecule, so the cathode reaction is reduction. At the anode, four hydroxide ions form one oxygen molecule and two water molecules; four electrons are released to balance charge, so this reaction is oxidation. Both atoms and total charge balance in each equation.	6
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02.1	$\mathrm{Na^+}+\mathrm{e^-}\rightarrow\mathrm{Na}$ : reduction. $2\mathrm{O^{2-}}\rightarrow\mathrm{O_2}+4\mathrm{e^-}$ : oxidation. The product ratio is $4$ sodium atoms to $1$ oxygen molecule.	Each sodium ion gains one electron at the cathode. Two oxide ions lose four electrons in total at the anode to form one $\mathrm{O_2}$ molecule. Multiply the sodium half equation by four so that four electrons cancel, giving the ratio $4\mathrm{Na}:1\mathrm{O_2}$ .	6
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03.1	B is balanced and occurs at the anode; chloride ions lose electrons, so it is oxidation. D is balanced and occurs at the cathode; $\mathrm{Cu^{2+}}$ ions gain electrons, so it is reduction. A has one chlorine atom on the left but two on the right. C puts electrons on the wrong side and does not balance charge.	Check atom count and total charge independently. Equation B has two chlorine atoms and charge −2 on each side; electron loss identifies anode oxidation. Equation D has total charge 0 on both sides after +2 and −2 combine; electron gain identifies cathode reduction. A fails atom balance, while C gives +2 on the left and −2 on the right.	6
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04.1	The compound is $\mathrm{MCl_3}$ and the metal ion is $\mathrm{M^{3+}}$ . $\mathrm{M^{3+}}+3\mathrm{e^-}\rightarrow\mathrm{M}$ at the cathode; reduction. $2\mathrm{Cl^-}\rightarrow\mathrm{Cl_2}+2\mathrm{e^-}$ at the anode; oxidation.	Three chlorine molecules contain six chlorine atoms. With two M atoms, this gives three chlorine atoms per metal atom and therefore formula $\mathrm{MCl_3}$ . Three chloride ions require a $3+$ metal ion for a neutral compound. The metal ion gains three electrons at the cathode, while pairs of chloride ions lose two electrons at the anode.	6
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05.1	Double the cathode equation so that four electrons are transferred in each equation. Two $\mathrm{H_2}$ molecules form for each $\mathrm{O_2}$ molecule, so the hydrogen-to-oxygen volume ratio is $2:1$ . $24\,\mathrm{cm^3}$ of oxygen is expected.	The cathode reaction uses two electrons per hydrogen molecule, while the anode reaction releases four electrons per oxygen molecule. Multiply the cathode equation by two so four electrons cancel; this gives two hydrogen molecules for every oxygen molecule. Gas volumes at the same temperature and pressure follow the same ratio, so $48/2=24\,\mathrm{cm^3}$ of oxygen.	3
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