4.4 Chemical changes — coverage pack

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

4.4.1.1 · Metal oxides

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.

Tier 1 · Easy

1. Magnesium burns in oxygen to form $\mathrm{MgO}$ . Name the type of reaction and explain your choice in terms of oxygen.[2 marks]
Answer
- Oxidation; magnesium gains oxygen.
Method: 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.

Tier 2 · Standard

1. 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.[3 marks]
Answer
- $1.6\,\mathrm{g}$ of oxygen; the copper has been oxidised.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- Hydrogen is oxidised because it gains oxygen.
- Copper oxide is reduced because it loses oxygen.
Method: 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.

4.4.1.2 · The reactivity series

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.

Tier 1 · Easy

1. Place magnesium, copper and zinc in decreasing order of reactivity.[1 mark]
Answer
- Magnesium, zinc, copper.
Method: Use their positions in the reactivity series. Magnesium is above zinc, and zinc is above copper, so the decreasing order is magnesium, zinc, copper.

Tier 2 · Standard

1. 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.[3 marks]
Answer
- $\mathrm{P}>\mathrm{Q}>\mathrm{R}$ in reactivity.
Method: 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 $\mathrm{P}>\mathrm{Q}>\mathrm{R}$ .

Tier 3 · Hard

1. 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]
Answer
- $\mathrm{W}>\mathrm{Y}>\mathrm{X}>\mathrm{Z}$ ; Z is below hydrogen.
Method: 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 $\mathrm{W}>\mathrm{Y}>\mathrm{X}>\mathrm{Z}$ .

4.4.1.3 · Extraction of metals and reduction

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.

Tier 1 · Easy

1. Copper is less reactive than carbon. Explain why carbon can be used to obtain copper from copper oxide.[2 marks]
Answer
- Carbon is more reactive than copper and removes oxygen from copper oxide, reducing it to copper.
Method: 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.

Tier 2 · Standard

1. 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.[4 marks]
Answer
- Gold may be found as the metal itself; zinc can be extracted from its oxide using carbon; aluminium requires another method.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

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

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.
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.

Tier 1 · Easy

1. 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]
Answer
- Zinc is oxidised because it loses electrons.
Method: The electrons appear on the product side, so one zinc atom has lost two electrons. Loss of electrons is oxidation.

Tier 2 · Standard

1. Complete and balance the half equation $\mathrm{Cl_2}+\square\,\mathrm{e^-}\rightarrow\square\,\mathrm{Cl^-}$ and identify the process.[3 marks]
Answer
- $\mathrm{Cl_2}+2\mathrm{e^-}\rightarrow2\mathrm{Cl^-}$ ; reduction.
Method: 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.

Tier 3 · Hard

1. Magnesium is added to copper(II) ion solution. Write the oxidation half equation, the reduction half equation and the overall ionic equation. Identify the species oxidised and the species reduced.[5 marks]
Answer
- $\mathrm{Mg}\rightarrow\mathrm{Mg^{2+}}+2\mathrm{e^-}$
- $\mathrm{Cu^{2+}}+2\mathrm{e^-}\rightarrow\mathrm{Cu}$
- $\mathrm{Mg}+\mathrm{Cu^{2+}}\rightarrow\mathrm{Mg^{2+}}+\mathrm{Cu}$ ; magnesium is oxidised and $\mathrm{Cu^{2+}}$ is reduced.
Method: 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.

4.4.2.1 · Reactions of acids with metals

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. At Higher Tier, metal atoms lose electrons and are oxidised, while hydrogen ions gain electrons and are reduced.

Tier 1 · Easy

1. Name the two products when magnesium reacts with dilute hydrochloric acid.[2 marks]
Answer
- Magnesium chloride and hydrogen.
Method: 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.

Tier 2 · Standard

1. Write a balanced symbol equation for iron reacting with dilute sulfuric acid to form iron(II) sulfate and hydrogen.[3 marks]
Answer
- $\mathrm{Fe}+\mathrm{H_2SO_4}\rightarrow\mathrm{FeSO_4}+\mathrm{H_2}$
Method: 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 $1$ .

Tier 3 · Hard

1. 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]
Answer
- $\mathrm{Mg}\rightarrow\mathrm{Mg^{2+}}+2\mathrm{e^-}$ ; magnesium is oxidised.
- $2\mathrm{H^+}+2\mathrm{e^-}\rightarrow\mathrm{H_2}$ ; hydrogen ions are reduced.
Method: 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.

4.4.2.2 · Neutralisation of acids and salt production

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.

Tier 1 · Easy

1. Potassium hydroxide is neutralised by nitric acid. Name the salt formed and the other product.[2 marks]
Answer
- Potassium nitrate and water.
Method: Nitric acid produces a nitrate, and potassium hydroxide supplies potassium ions. The salt is potassium nitrate; acid-alkali neutralisation also forms water.

Tier 2 · Standard

1. Complete the word equation and explain the salt name: copper oxide + sulfuric acid $\rightarrow$ ______ + ______.[3 marks]
Answer
- Copper sulfate and water.
Method: 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.

Tier 3 · Hard

1. Aluminium hydroxide reacts with sulfuric acid. Deduce the formula of aluminium sulfate and balance the symbol equation.[5 marks]
Answer
- $\mathrm{Al_2(SO_4)_3}$
- $2\mathrm{Al(OH)_3}+3\mathrm{H_2SO_4}\rightarrow\mathrm{Al_2(SO_4)_3}+6\mathrm{H_2O}$
Method: 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.

4.4.2.3 · Soluble salts

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.

Tier 1 · Easy

1. Why is an insoluble solid added to an acid until some remains unreacted when making a soluble salt?[2 marks]
Answer
- To ensure all the acid has reacted; the excess solid can then be removed by filtration.
Method: 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.

Tier 2 · Standard

1. 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.[4 marks]
Answer
- Warm the acid; add excess copper oxide; filter; evaporate some water; crystallise; dry the crystals.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.4.2.4 · The pH scale and neutralisation

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 $\mathrm{H^+}+\mathrm{OH^-}\rightarrow\mathrm{H_2O}$ ; equal volumes are not necessarily neutral if the reacting amounts differ.

Tier 1 · Easy

1. Classify solutions with pH $3$ , pH $7$ and pH $11$ as acidic, neutral or alkaline.[3 marks]
Answer
- pH $3$ : acidic; pH $7$ : neutral; pH $11$ : alkaline.
Method: Compare each value with $7$ . Values below $7$ are acidic, $7$ is neutral, and values above $7$ are alkaline.

Tier 2 · Standard

1. 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.[3 marks]
Answer
- Add universal or wide-range indicator and compare the colour with its pH chart; use a calibrated pH probe for greater precision.
Method: 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.

Tier 3 · Hard

1. 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, and explain why adding equal volumes of acid and alkali would not always give pH $7$ .[5 marks]
Answer
- $\mathrm{OH^-}$ ions react with $\mathrm{H^+}$ ions, decreasing the hydrogen ion concentration until the solution is neutral.
- $\mathrm{H^+}+\mathrm{OH^-}\rightarrow\mathrm{H_2O}$
- Equal volumes may contain different reacting amounts because the solutions may have different concentrations.
Method: 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.

4.4.2.5 · Titrations (chemistry only)

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.
At Higher Tier, use the balanced equation and volumes in $\mathrm{dm^3}$ to calculate concentrations in $\mathrm{mol\,dm^{-3}}$ and $\mathrm{g\,dm^{-3}}$ .

Tier 1 · Easy

1. 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]
Answer
- A volumetric pipette and a burette, respectively.
Method: 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.

Tier 2 · Standard

1. Describe how to obtain an accurate mean titre when titrating a strong alkali with a strong acid.[5 marks]
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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $0.112\,\mathrm{mol\,dm^{-3}}$ and $4.46\,\mathrm{g\,dm^{-3}}$ .
Method: 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 $4.46\,\mathrm{g\,dm^{-3}}$ .

4.4.2.6 · Strong and weak acids (HT only)

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.

Tier 1 · Easy

1. State the difference between a strong acid and a weak acid in aqueous solution.[2 marks]
Answer
- A strong acid ionises completely; a weak acid ionises only partially.
Method: Use degree of ionisation, not concentration. Complete ionisation defines a strong acid, while only a fraction of acid particles ionising defines a weak acid.

Tier 2 · Standard

1. 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]
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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- A has $1000$ times the hydrogen ion concentration of B.
Method: 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.

4.4.3.1 · The process of electrolysis

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.

Tier 1 · Easy

1. State which electrode attracts positive ions during electrolysis and give the charge on that electrode.[2 marks]
Answer
- The cathode; it is negative.
Method: Opposite charges attract. Positive ions therefore move towards the negative electrode, which is called the cathode.

Tier 2 · Standard

1. Solid sodium chloride does not conduct electricity, but molten sodium chloride does. Explain this difference in terms of ions.[3 marks]
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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $\mathrm{M^{2+}}$ 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.
Method: 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.

4.4.3.2 · Electrolysis of molten ionic compounds

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.

Tier 1 · Easy

1. Predict the products at the cathode and anode when molten zinc chloride is electrolysed using inert electrodes.[2 marks]
Answer
- Zinc at the cathode; chlorine at the anode.
Method: 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.

Tier 2 · Standard

1. Molten lead bromide is electrolysed with inert electrodes. Name each product, identify its electrode, and explain why lead bromide must be molten.[4 marks]
Answer
- Lead forms at the negative cathode; bromine forms at the positive anode; melting frees the ions to move.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.4.3.3 · Using electrolysis to extract metals

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.

Tier 1 · Easy

1. Explain why aluminium is extracted using electrolysis rather than by reducing aluminium oxide with carbon.[2 marks]
Answer
- Aluminium is more reactive than carbon, so carbon cannot reduce aluminium oxide.
Method: Use the reactivity series. A metal above carbon cannot be displaced from its oxide by carbon, so aluminium requires electrolysis.

Tier 2 · Standard

1. Cryolite is mixed with aluminium oxide before electrolysis. Explain how this reduces the energy cost of extraction.[3 marks]
Answer
- Cryolite lowers the melting point of the electrolyte, so less energy is needed to melt and keep it molten.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- Mixture A; its lower melting point requires less heating. Oxygen formed at the anode reacts with carbon to make carbon dioxide, consuming the anode.
Method: 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.

4.4.3.4 · Electrolysis of aqueous solutions

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.

Tier 1 · Easy

1. Predict the products at inert electrodes during electrolysis of aqueous sodium chloride.[2 marks]
Answer
- Hydrogen at the cathode; chlorine at the anode.
Method: 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.

Tier 2 · Standard

1. Aqueous copper(II) sulfate is electrolysed using inert electrodes. Predict both products and justify each using the discharge rules.[4 marks]
Answer
- Copper at the cathode; oxygen at the anode.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- Magnesium chloride: hydrogen at the cathode and chlorine at the anode.
- Silver nitrate: silver at the cathode and oxygen at the anode.
Method: 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.

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

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.

Tier 1 · Easy

1. Complete the cathode half equation $\mathrm{Al^{3+}}+\square\,\mathrm{e^-}\rightarrow\mathrm{Al}$ and name the process.[2 marks]
Answer
- $\mathrm{Al^{3+}}+3\mathrm{e^-}\rightarrow\mathrm{Al}$ ; reduction.
Method: Three electrons give total charge $-3$ , balancing the ion's $+3$ charge to form neutral aluminium. The ion gains electrons, so this is reduction.

Tier 2 · Standard

1. Write the balanced half equation for bromide ions forming bromine at the anode and explain why it is oxidation.[3 marks]
Answer
- $2\mathrm{Br^-}\rightarrow\mathrm{Br_2}+2\mathrm{e^-}$ ; bromide ions lose electrons.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- $2\mathrm{H^+}+2\mathrm{e^-}\rightarrow\mathrm{H_2}$ : reduction.
- $4\mathrm{OH^-}\rightarrow\mathrm{O_2}+2\mathrm{H_2O}+4\mathrm{e^-}$ : oxidation.
Method: 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.