Electrolysis (RP3) – GCSE Chemistry exam questions with mark scheme

1Grade 6–7Higher5 marks

The diagram shows apparatus for the electrolysis of molten lead bromide, $\text{PbBr}_2$.

(a) Name the products formed at electrode A (the anode) and electrode B (the cathode). [2]

(b) Explain why the lead bromide must be molten. [2]

$\text{Pb}^{2+} + \ldots \rightarrow \text{Pb}$

(a) A (anode): bromine allow $\text{Br}_2$; do not accept bromide / Br(1)
(a) B (cathode): lead allow Pb(1)
(b) (when molten) the ions are free to move allow ions can move to the electrodes / can carry the charge; do not accept electrons free to move(1)
(b) in the solid the ions are held in fixed positions (in a lattice) / cannot move(1)
(c) $2\text{e}^-$ allow 2e(1)
Answer: (a) bromine at the anode, lead at the cathode (b) ions free to move only when molten (c) Pb²⁺ + 2e⁻ → Pb
Examiner tip: Positive metal ions go to the negative cathode — the metal is always produced at the cathode for a molten binary compound.

2Grade 6–7Higher5 marks

Sodium chloride solution is electrolysed using inert electrodes.

(a) Name the product at the negative electrode and explain why sodium is not produced. [2]

(b) Write the half equation for the reaction at the positive electrode. [2]

(a) hydrogen allow $\text{H}_2$(1)
(a) sodium is more reactive than hydrogen (so hydrogen ions (from the water) are discharged instead) allow hydrogen is less reactive than sodium(1)
(b) correct species: $\text{Cl}^-$ forming $\text{Cl}_2$ with electrons lost (electrons on the right, or subtracted on the left)(1)
(b) $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-$ (balanced) allow $2\text{Cl}^- - 2\text{e}^- \rightarrow \text{Cl}_2$; ignore state symbols(1)
(c) oxidation, because chloride ions lose electrons (both needed) do not accept reasons in terms of oxygen(1)
Answer: (a) hydrogen; Na more reactive than H (b) 2Cl⁻ → Cl₂ + 2e⁻ (c) oxidation — electrons lost
Examiner tip: OIL RIG: Oxidation Is Loss of electrons — the reason must refer to electrons, not oxygen.

3Grade 1–3Foundation6 marks

The diagram shows the apparatus a teacher uses to electrolyse molten zinc chloride, $\text{ZnCl}_2$.

(a) Molten zinc chloride is an electrolyte. What is meant by an electrolyte? [1]

(b) Solid zinc chloride does not conduct electricity. Give the reason why. [1]

Which electrode, A or B, do the zinc ions move towards? Give a reason for your answer. [2]

(d) The teacher then electrolyses molten calcium bromide, $\text{CaBr}_2$, using the same apparatus.

Name the product formed at the positive electrode and the product formed at the negative electrode. [2]

a liquid or solution that contains ions / an ionic compound that is molten or dissolved (so that it conducts and is decomposed by electricity) allow 'a liquid that conducts electricity and is broken down by it'; ignore 'any liquid that conducts'(1)
the ions cannot move (they are held in fixed positions in the lattice) do not accept 'electrons cannot move' or 'no ions'(1)
B(1)
(because) zinc ions are positive and are attracted to the negative electrode / B is negative (opposite charges attract) dependent on B(1)
positive electrode: bromine do not accept bromide(1)
negative electrode: calcium(1)
Answer: (a) a molten or dissolved ionic compound (contains ions that can move); (b) ions cannot move in the solid; (c) B – Zn²⁺ is positive and B is the negative electrode; (d) bromine at positive, calcium at negative
Examiner tip: Writing 'bromide' for the product at the anode, or saying 'electrons cannot move' in the solid, loses the mark – it is ions that carry charge in an electrolyte.

4Grade 4–56 marks

Aluminium is extracted by the electrolysis of aluminium oxide. The diagram shows the cell used.

(a) Explain why aluminium cannot be extracted from aluminium oxide by heating it with carbon. [1]

(b) The aluminium oxide is dissolved in molten cryolite instead of being melted on its own.

Explain why this is done. [2]

(d) The positive electrodes must be replaced regularly. Explain why. [2]

aluminium is more reactive than carbon (so carbon cannot remove the oxygen / reduce it) allow carbon is less reactive than aluminium(1)
(the mixture has a) lower melting point (than aluminium oxide) ignore 'dissolves more easily'(1)
so less energy is needed / reduces energy (and so cost) allow lower temperature needed so cheaper; ignore 'cheaper' alone(1)
aluminium ions are positive and are attracted to the negative electrode allow Al³⁺ attracted to the cathode(1)
oxygen is produced at the positive electrode(1)
(the oxygen) reacts with the carbon (electrodes) to form carbon dioxide (so the electrodes wear away) allow electrodes burn away; do not accept 'electrodes melt' or 'electrodes dissolve'(1)
Answer: (a) Al is more reactive than carbon; (b) the mixture has a lower melting point, so less energy is needed; (c) Al³⁺ ions are positive so attracted to the negative electrode; (d) oxygen forms at the positive electrodes and reacts with the carbon to make carbon dioxide
Examiner tip: In (d) say what is made at the anode (oxygen) and what it does (reacts with carbon to form CO₂); 'the electrodes wear away' alone scores nothing.

5Grade 4–56 marks

A student electrolyses four different solutions using inert electrodes.

Solution	Product at negative electrode	Product at positive electrode
silver nitrate	(i) ..........	oxygen
potassium iodide	(ii) ..........	(iii) ..........
copper(II) bromide	copper	(iv) ..........
magnesium sulfate	hydrogen	oxygen

(a) Complete the table by naming products (i) to (iv). [4]

(b) Magnesium sulfate solution contains $\text{Mg}^{2+}$ ions, but magnesium is not produced at the negative electrode.

Explain why hydrogen is produced instead. [2]

(i) silver(1)
(ii) hydrogen(1)
(iii) iodine do not accept iodide(1)
(iv) bromine do not accept bromide(1)
magnesium is more reactive than hydrogen allow hydrogen is less reactive than magnesium(1)
hydrogen ions (from water) are present / are discharged (instead of magnesium ions) allow water is ionised to H⁺ and OH⁻ ions(1)
Answer: (i) silver (ii) hydrogen (iii) iodine (iv) bromine; (b) Mg is more reactive than hydrogen, so H⁺ ions from the water are discharged instead
Examiner tip: Most lost marks here are for writing the ion name ('iodide', 'bromide') instead of the element formed at the positive electrode.

6Grade 6–7Higher6 marks

Copper(II) sulfate solution is blue. A student electrolyses copper(II) sulfate solution using inert carbon electrodes.

A pink-brown solid forms on the negative electrode and bubbles of gas form at the positive electrode.

(a) Write a half equation for the reaction at the negative electrode. [1]

(b) Oxygen is produced at the positive electrode. Balance the half equation for this reaction.

$$\ldots\,\text{OH}^- \rightarrow \text{O}_2 + \ldots\,\text{H}_2\text{O} + \ldots\,\text{e}^-$$ [1]

(d) Explain why oxygen, and not sulfur or a sulfur compound, is produced at the positive electrode. [2]

(e) After the current has passed for a long time, the solution becomes paler blue. Suggest why. [1]

$\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$ allow multiples; ignore state symbols(1)
$4\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^-$ (4, 2, 4) all three needed(1)
hydroxide ions lose electrons allow OH⁻ ions lose electrons / electrons are lost; do not accept 'oxygen is gained' alone(1)
there are no halide ions in the solution allow sulfate is not a halide(1)
so hydroxide ions (from water) are discharged (and oxygen is formed) allow OH⁻ ions are discharged in preference to sulfate ions(1)
copper(II) ions (which cause the blue colour) are removed from the solution / are turned into copper (at the negative electrode) allow concentration of Cu²⁺ ions decreases(1)
Answer: (a) Cu²⁺ + 2e⁻ → Cu; (b) 4, 2, 4; (c) OH⁻ ions lose electrons; (d) no halide ions present, so OH⁻ from water is discharged; (e) Cu²⁺ ions are removed as copper
Examiner tip: Oxidation at GCSE Higher must be explained as loss of electrons – 'gains oxygen' does not score when the species is an ion losing electrons.

7Grade 4–56 marks

A student electrolysed sodium sulfate solution using inert electrodes. The gas from each electrode was collected in an upturned measuring tube filled with the solution.

The student recorded the volume of gas collected every minute. The graph shows the results.

(a) Name the gas collected at the negative electrode (A) and describe a test to identify it. [2]

(b) Use the graph to give the volume of gas at the negative electrode after 4 minutes. [1]

(d) Use the graph to find the simplest whole-number ratio of

volume of gas at A : volume of gas at B after 6 minutes. [1]

(e) Give one hazard of this experiment and how it could be reduced. [1]

hydrogen(1)
(lighted splint) burns with a (squeaky) pop both the lighted splint and the pop needed; do not accept glowing splint(1)
12 (cm³) allow 11.8 – 12.2(1)
3 (minutes) / the positive electrode (B) result at 3 minutes do not accept a volume alone(1)
2 : 1 do not accept 18 : 9 (not simplest) or 1 : 2 (wrong order)(1)
hydrogen is flammable, so keep flames away / do not test the gas near a flame other than the splint allow electrical hazard – keep water away from the power supply / use low voltage; allow wear eye protection to protect from splashes of solution; hazard and precaution both needed(1)
Answer: (a) hydrogen – burns with a squeaky pop with a lighted splint; (b) 12 cm³; (c) 3 minutes (positive electrode, B); (d) 2 : 1; (e) e.g. hydrogen is flammable – keep naked flames away
Examiner tip: A 'test' needs the reagent and the result: 'squeaky pop' without 'lighted splint' scores only the name mark.

8Grade 6–76 marks

A student makes this hypothesis:

'When a solution is electrolysed using inert electrodes, chlorine is produced at the positive electrode only if the solution contains chloride ions.'

The student is given:

copper(II) chloride solution
copper(II) sulfate solution
a low-voltage d.c. power supply, carbon electrodes and normal laboratory apparatus

Plan an investigation to test the student's hypothesis.

Your plan should include how you would identify the product at the positive electrode. [6]

Pour a measured volume (e.g. 50 cm³) of copper(II) chloride solution into a beaker / electrolysis cell(1)
Place two (clean) carbon electrodes in the solution, not touching, and connect them to the d.c. supply(1)
Collect the gas at the positive electrode in a test tube filled with the solution, inverted over the electrode(1)
Switch on and observe both electrodes (copper forms on the negative electrode in both)(1)
Test the gas with damp (blue) litmus paper: chlorine bleaches it (white) allow turns red then white(1)
Test for oxygen with a glowing splint: oxygen relights it(1)
Repeat with the same volume and concentration of copper(II) sulfate solution(1)
Control variables: same voltage, same electrodes / distance apart, same time, same concentration and volume(1)
Copper(II) is the same positive ion in both, so only the negative ion is changed (fair test)(1)
Expected result supporting hypothesis: chlorine with copper(II) chloride, oxygen with copper(II) sulfate(1)
Safety: chlorine is toxic – use a fume cupboard / well-ventilated room / switch off soon after gas collected; wear eye protection(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, logically ordered method that could be followed to produce valid results. Uses two solutions that differ only in the anion (or justifies the choice), controls key variables, describes how the gas at the positive electrode is collected or observed and gives correct tests and results for both chlorine and oxygen, linked to the hypothesis.
Level 2	3–4	A method with most of the main steps, including setting up the electrolysis of at least two suitable solutions and at least one correct gas test with its result. Some control variables or links to the hypothesis may be missing.
Level 1	1–2	Simple statements about setting up electrolysis or testing a gas. The method lacks order or could not be followed to test the hypothesis.
No relevant content	0	Nothing worthy of credit.

Answer: Electrolyse equal volumes of equal-concentration CuCl₂(aq) and CuSO₄(aq) with carbon electrodes at the same voltage; collect the gas at the positive electrode; test with damp litmus (chlorine bleaches) and a glowing splint (oxygen relights); chlorine only from CuCl₂ supports the hypothesis; use a fume cupboard.
Examiner tip: A top-band plan names BOTH gas tests with results and keeps the positive ion the same (copper) so only the negative ion changes.

9Grade 8–9Higher7 marks

Lithium is extracted by the electrolysis of molten lithium chloride, $\text{LiCl}$.

(a) Write half equations for the reactions at each electrode. [2]

(b) Which ion is oxidised? Give a reason for your answer. [1]

$$2\text{LiCl} \rightarrow 2\text{Li} + \text{Cl}_2$$

Calculate the maximum mass of lithium that can be produced from 2.55 kg of lithium chloride.

Relative atomic masses ($A_r$): Li = 7 Cl = 35.5 [3]

(d) Calculate the maximum mass of chlorine produced from the same 2.55 kg of lithium chloride. Give your answer in kg. [1]

negative electrode: $\text{Li}^+ + \text{e}^- \rightarrow \text{Li}$ allow multiples; ignore state symbols(1)
positive electrode: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-$ allow $2\text{Cl}^- - 2\text{e}^- \rightarrow \text{Cl}_2$; allow correct equations given in either order if electrodes are identified(1)
chloride ions, because they lose electrons do not accept chlorine loses electrons(1)
$M_r$ LiCl = 42.5 or moles LiCl = 2550 ÷ 42.5 = 60 (mol)(1)
moles Li = 60 (mol) (1 : 1 ratio)(1)
mass Li = 60 × 7 = 420 g / 0.42 kg allow ecf from incorrect moles; correct answer with no working scores 3; allow 2550 × 7/42.5 = 420 g for 3 marks(1)
mass Cl₂ = 2.13 kg (30 mol × 71 = 2130 g) allow 2.55 − 0.42 = 2.13 kg; allow ecf from (c); answer must be in kg(1)
Answer: (a) Li⁺ + e⁻ → Li; 2Cl⁻ → Cl₂ + 2e⁻; (b) chloride ions – they lose electrons; (c) 420 g (0.42 kg); (d) 2.13 kg
Examiner tip: Forgetting to convert 2.55 kg to 2550 g, or using M_r = 71 for chlorine in the 'Cl' of LiCl, are the usual ways to lose the calculation marks.

10Grade 8–9Higher8 marks

A teacher has two solutions, X and Y. Each is a solution of one of these compounds:

copper(II) chloride
copper(II) sulfate
sodium iodide
sodium sulfate

Each solution is electrolysed using inert electrodes. The table shows the observations.

Solution	Negative electrode	Positive electrode
X	pink-brown solid forms	gas forms that relights a glowing splint
Y	gas forms that burns with a squeaky pop	colourless solution turns brown near the electrode; no gas

(a) Identify solution X and solution Y. Explain your answers using the observations. [4]

(b) Write a balanced half equation for the reaction at the positive electrode in solution Y. [2]

(c) After electrolysis, the solution near the negative electrode in Y turns universal indicator purple. Explain why. [2]

X is copper(II) sulfate(1)
copper forms (so copper ions present) and oxygen forms at the positive electrode so there are no halide ions / chloride is ruled out both parts of reasoning needed(1)
Y is sodium iodide(1)
hydrogen forms (so the metal is more reactive than hydrogen – sodium) and iodine forms (brown) so iodide ions present / not oxygen so a halide is present both parts of reasoning needed(1)
$\text{I}^-$ on the left and $\text{I}_2$ and $\text{e}^-$ on the right(1)
$2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-$ fully balanced allow $2\text{I}^- - 2\text{e}^- \rightarrow \text{I}_2$; allow multiples(1)
hydrogen ions are removed / discharged (as hydrogen gas) at the negative electrode(1)
so hydroxide ions are left (in excess) / sodium hydroxide forms, making the solution alkaline allow OH⁻ concentration greater than H⁺; ignore 'pH increases' alone(1)
Answer: (a) X = copper(II) sulfate (Cu at cathode; O₂ so no halide); Y = sodium iodide (H₂ so metal more reactive than H; I₂ so iodide present); (b) 2I⁻ → I₂ + 2e⁻; (c) H⁺ ions are discharged as H₂, leaving OH⁻ ions (sodium hydroxide), so the solution is alkaline
Examiner tip: Identifying X and Y scores only half the marks – each identification must be backed by evidence from BOTH electrodes.

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