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AQA GCSE Chemistry revision notes

Chemical changes

Section 4.4
15 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 8462 section 4.4

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

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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: 2Cu+O22CuO2\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 6.4g6.4\,\mathrm{g} of copper in oxygen and obtains 8.0g8.0\,\mathrm{g} of copper oxide. Calculate the mass of oxygen gained and state what has happened to the copper.

  1. 1.The oxygen gained accounts for the increase in mass: 8.06.4=1.6g8.0-6.4=1.6\,\mathrm{g}. Copper has gained that oxygen, so the copper has been oxidised.

Answer: 1.6g1.6\,\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

ORIGINAL

Magnesium burns in oxygen to form MgO\mathrm{MgO}. Name the type of reaction and explain your choice in terms of oxygen.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

Hydrogen is passed over hot copper oxide: CuO+H2Cu+H2O\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

Your progress and exam materials
4.4.1.2

The reactivity series

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 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 P>Q>R\mathrm{P}>\mathrm{Q}>\mathrm{R}.

Answer: P>Q>R\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

ORIGINAL

Place magnesium, copper and zinc in decreasing order of reactivity.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

4.4.1.3

Extraction of metals and reduction

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

Consider 2CuO+C2Cu+CO22\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

4.4.1.4

Oxidation and reduction in terms of electrons (HT only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 Cl2+eCl\mathrm{Cl_2}+\square\,\mathrm{e^-}\rightarrow\square\,\mathrm{Cl^-} and identify the process.

  1. 1.Two chlorine atoms require 2Cl2\mathrm{Cl^-} on the right. Their total charge is 2-2, so add two electrons on the left. Chlorine gains electrons, so it is reduced.

Answer: Cl2+2e2Cl\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

ORIGINAL

The half equation is ZnZn2++2e\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

Tier 2 · Standard

ORIGINAL

In the ionic equation Mg+Cu2+Mg2++Cu\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

Tier 3 · Hard

ORIGINAL

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

4.4.2.1

Reactions of acids with metals

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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, Zn+2HClZnCl2+H2\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. 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 11.

Answer: Fe+H2SO4FeSO4+H2\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

ORIGINAL

Name the two products when magnesium reacts with dilute hydrochloric acid.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

Higher Tier: magnesium reacts with hydrochloric acid according to Mg+2H+Mg2++H2\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

4.4.2.2

Neutralisation of acids and salt production

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 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

ORIGINAL

Potassium hydroxide is neutralised by nitric acid. Name the salt formed and the other product.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

Aluminium hydroxide reacts with sulfuric acid. Deduce the formula of aluminium sulfate and balance the symbol equation.

[5 marks]

Total for this question: 5

4.4.2.3

Soluble salts

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

4.4.2.4

The pH scale and neutralisation

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • The pH scale runs from 00 to 1414: acids have pH below 77, neutral solutions have pH 77, and alkalis have pH above 77. Universal or wide-range indicator gives an approximate pH from colour; a calibrated pH probe gives a numerical reading.
  • Acids in water produce H+\mathrm{H^+} ions, while aqueous alkalis contain OH\mathrm{OH^-} ions.
  • Neutralisation is H+(aq)+OH(aq)H2O(l)\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 55, 77 or 99. Describe how to determine its approximate pH and state one method that would give a more precise numerical value.

  1. 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

ORIGINAL

Classify solutions with pH 33, pH 77 and pH 1111 as acidic, neutral or alkaline.

[3 marks]

Total for this question: 3

Tier 2 · Standard

ORIGINAL

Hydrochloric acid is mixed with sodium hydroxide and the final solution has pH 1111. State which reactant is in excess and explain the result in terms of H+\mathrm{H^+} and OH\mathrm{OH^-} ions.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An alkali is added in small portions to an acidic solution until its pH changes from 22 to 77. 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 77.

[5 marks]

Total for this question: 5

4.4.2.5

Titrations (chemistry only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 dm3\mathrm{dm^3} to calculate chemical quantities and determine concentrations in moldm3\mathrm{mol\,dm^{-3}} and gdm3\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. 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

ORIGINAL

Name the apparatus used to transfer exactly 25.0cm325.0\,\mathrm{cm^3} of alkali and the apparatus used to add a measured, variable volume of acid.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A rough titre is 25.60cm325.60\,\mathrm{cm^3}. Three accurate titres are 24.80cm324.80\,\mathrm{cm^3}, 24.75cm324.75\,\mathrm{cm^3} and 24.70cm324.70\,\mathrm{cm^3}. Calculate the mean titre that should be used.

[2 marks]

Total for this question: 2

Tier 3 · Hard

ORIGINAL

Higher Tier: 25.0cm325.0\,\mathrm{cm^3} of sodium hydroxide is neutralised by 18.60cm318.60\,\mathrm{cm^3} of 0.150moldm30.150\,\mathrm{mol\,dm^{-3}} hydrochloric acid. The equation has a 1:11:1 ratio. Calculate the sodium hydroxide concentration in moldm3\mathrm{mol\,dm^{-3}} and gdm3\mathrm{g\,dm^{-3}}. Use Mr(NaOH)=40.0M_r(\mathrm{NaOH})=40.0.

[5 marks]

Total for this question: 5

4.4.2.6

Strong and weak acids (HT only)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 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 H+\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

ORIGINAL

State the difference between a strong acid and a weak acid in aqueous solution.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

Solution A has pH 22 and solution B has pH 55. Calculate how many times greater the hydrogen ion concentration is in A than in B. Explain why saying 'A is 33 times more acidic' is incorrect.

[4 marks]

Total for this question: 4

4.4.3.1

The process of electrolysis

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 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

ORIGINAL

State which electrode attracts positive ions during electrolysis and give the charge on that electrode.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

A molten ionic compound contains only M2+\mathrm{M^{2+}} and X\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

4.4.3.2

Electrolysis of molten ionic compounds

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 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

ORIGINAL

Predict the products at the cathode and anode when molten zinc chloride is electrolysed using inert electrodes.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

A binary compound has formula CaCl2\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

4.4.3.3

Using electrolysis to extract metals

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 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

ORIGINAL

Explain why aluminium is extracted using electrolysis rather than by reducing aluminium oxide with carbon.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

An aluminium plant considers two electrolytes. Mixture A melts at 950C950\,^{\circ}\mathrm{C}; mixture B melts at 1210C1210\,^{\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

4.4.3.4

Electrolysis of aqueous solutions

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An aqueous electrolyte contains ions from the dissolved compound as well as H+\mathrm{H^+} and OH\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. 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

ORIGINAL

Predict the products at inert electrodes during electrolysis of aqueous sodium chloride.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

4.4.3.5

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

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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. 1.Two bromide ions are needed to make one Br2\mathrm{Br_2} molecule. The left side has charge 2-2, so place two electrons on the right to balance charge. Electrons are lost, making the process oxidation.

Answer: 2BrBr2+2e2\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

ORIGINAL

Complete the cathode half equation Al3++eAl\mathrm{Al^{3+}}+\square\,\mathrm{e^-}\rightarrow\mathrm{Al} and name the process.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

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