Pearson Edexcel International GCSE Chemistry 4CH1: Electrolysis – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sub-topic 1(i) Electrolysis of the Pearson Edexcel International GCSE Chemistry (4CH1) specification, Issue 3 (September 2024): statements 1.55C–1.60C. The qualification is untiered, and every statement in this sub-topic has a C reference, so all of it is Paper 2 only. A calculator may be used in both 4CH1 papers. Course links: 4CH1 Chemistry hub, printable checklist, the electrolysis study guide and the electrolysis revision notes. The topic-level Principles of chemistry practice has further questions on concentrated sodium chloride solution.

Where needed: molar volume of a gas = 24 000 cm³ at rtp.

Questions

1. State what is meant by:

(a) a cation [1] (b) an anion [1]

2. Tetrachloromethane, CCl₄, is a covalent liquid. Explain why it does not conduct electricity. [2]

3. A student tests four substances, W, X, Y and Z.

Substance	Solid conducts?	Molten conducts?	Dissolves in water?	Solution conducts?
W	No	Yes	Yes	Yes
X	Yes	Yes	No	–
Y	No	No	Yes	No
Z	No	Yes	No	–

(a) Identify the substance that is a metal. [1] (b) Identify the two ionic compounds and explain your choice. [3]

4. Solid lead(II) bromide is electrolysed using graphite electrodes. A bulb is connected in series. The solid is heated until it melts.

(a) Explain why the bulb lights only after the lead(II) bromide has melted. [2] (b) State what is seen at the anode. [1] (c) Write the half-equation for the reaction at the cathode and explain why it is a reduction. [2] (d) Explain why the experiment is carried out in a fume cupboard. [1]

5. Dilute sulfuric acid is electrolysed with inert electrodes. The gases are collected separately.

(a) Name the gas formed at each electrode. [2] (b) Write the half-equation for the reaction at the cathode. [1] (c) 30 cm³ of oxygen is collected. Calculate the volume of hydrogen formed at the same time and the amount, in moles, of hydrogen at rtp. [2] (d) Explain why the sulfuric acid becomes more concentrated. [1]

6. Copper(II) sulfate solution is electrolysed with inert electrodes. (A_r Cu = 63.5)

(a) Describe two observations, other than bubbles at the anode. [2] (b) Write the half-equations for the reactions at the cathode and at the anode. [2] (c) 0.508 g of copper is deposited. Calculate the volume of oxygen formed at rtp. [3]

7. Predict the product at the cathode and at the anode when each solution is electrolysed with inert electrodes.

(a) copper(II) nitrate solution [2] (b) potassium sulfate solution [2] (c) concentrated magnesium bromide solution [2]

8. A student electrolyses sodium chloride solution with graphite electrodes. Universal indicator is added.

(a) The solution near the cathode turns purple. Explain why. [2] (b) Describe the test for the gas formed at the anode, and give the result. [2] (c) With a very dilute solution, a second gas is also formed at the anode. Name it and write the half-equation for its formation. [2]

9. Describe an experiment to find the products at each electrode when concentrated potassium chloride solution and copper(II) sulfate solution are electrolysed. Include how you would identify each product. [6]

10. Molten sodium chloride is electrolysed. 2.30 g of sodium forms at the cathode. (A_r Na = 23, Cl = 35.5)

(a) Calculate the amount, in moles, of electrons transferred. [2] (b) Calculate the volume of chlorine formed at rtp, in dm³. [2] (c) Calculate the mass of sodium chloride decomposed. [1] (d) Explain, in terms of electrons, why the reaction at the anode is oxidation. [1]

11. Concentrated copper(II) chloride solution is electrolysed with inert electrodes. (A_r Cu = 63.5)

(a) Predict the product at each electrode. Explain each prediction. [4] (b) Write the two half-equations. [2] (c) 0.381 g of copper forms. Calculate the volume of chlorine expected at rtp. [3] (d) The volume of chlorine measured is less than expected. Suggest one reason. [1]

Answers

1. (a) A positive ion [1]. (b) A negative ion [1]. Examiner insight: “Positive particle” or “positive atom” is not enough; the word ion is needed for each mark.

2. It is made of molecules, so there are no ions [1]. Its electrons are held in covalent bonds, so there are no charged particles free to move [1]. Examiner insight: Two separate points are needed for both marks: no ions, and no mobile electrons. Saying only “it is covalent” restates the question and scores nothing.

3. (a) X [1]. (b) W and Z [1]. Neither conducts as a solid [1], but both conduct when molten, because the ions become free to move [1]. Examiner insight: Z is insoluble, so you can only use its molten result. Y conducts in no state, so it is covalent; don’t pick it as ionic just because it dissolves.

4. (a) In the solid, the ions are held in fixed positions in the lattice [1]. When molten, the ions are free to move and carry the charge [1]. (b) Red-brown fumes (of bromine) [1]. (c) Pb²⁺ + 2e⁻ → Pb [1]. Lead ions gain electrons [1]. (d) Bromine vapour is toxic [1]. Examiner insight: In (a), “electrons move when molten” is a contradiction and loses the mark; the mobile particles in a molten salt are ions.

5. (a) Cathode: hydrogen [1]. Anode: oxygen [1]. (b) 2H⁺ + 2e⁻ → H₂ [1]. (c) Volume of H₂ = 2 × 30 = 60 cm³ [1]. Moles = 60 ÷ 24 000 = 0.0025 mol [1]. (d) Hydrogen and oxygen come from water, so water is removed while the acid stays in the solution [1]. Examiner insight: In (c) the second mark can be awarded as a follow-through from a wrong volume, provided you divide by 24 000 for cm³.

6. (a) A pink-brown solid forms on the cathode [1]. The blue colour of the solution fades [1]. (b) Cathode: Cu²⁺ + 2e⁻ → Cu [1]. Anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻ [1]. (c) n(Cu) = 0.508 ÷ 63.5 = 0.00800 mol [1]. 4e⁻ give 2Cu and 1O₂, so n(O₂) = 0.00800 ÷ 2 = 0.00400 mol [1]. Volume = 0.00400 × 24 000 = 96 cm³ [1]. Examiner insight: The middle mark is for the 2 : 1 ratio from balanced electrons. Using a 1 : 1 ratio gives 192 cm³ and loses that mark and the final one.

7. (a) Cathode: copper [1]. Anode: oxygen [1]. (b) Cathode: hydrogen [1]. Anode: oxygen [1]. (c) Cathode: hydrogen [1]. Anode: bromine [1]. Examiner insight: Each mark is for one named product at the correct electrode. Naming a product without its electrode, or giving “potassium” or “magnesium” for an aqueous solution, scores zero for that mark.

8. (a) Hydrogen ions are discharged at the cathode [1], leaving hydroxide ions (sodium hydroxide) in solution, which is alkaline [1]. (b) Hold damp blue litmus paper in the gas [1]. It turns red, then white (bleached) [1]. (c) Oxygen [1]. 4OH⁻ → O₂ + 2H₂O + 4e⁻ [1]. Examiner insight: “Turns red” on its own is the test for an acid, not for chlorine; the bleaching is what earns the result mark.

9. One mark per point:

• Graphite (inert) electrodes in the solution, connected to a low-voltage d.c. supply [1].
• Small test tubes filled with the solution and inverted over each electrode to collect gas [1].
• Watch the cathode with copper(II) sulfate: a pink-brown coating shows copper [1].
• Test a cathode gas with a lighted splint: squeaky pop shows hydrogen (from potassium chloride solution) [1].
• Test the anode gas from copper(II) sulfate with a glowing splint: it relights, showing oxygen [1].
• Test the anode gas from potassium chloride with damp blue litmus paper: it is bleached, showing chlorine [1]. Examiner insight: Each test must be paired with its result and the product it shows. Rinsing the electrodes between solutions is good practice but would not replace a missing test.

10. (a) n(Na) = 2.30 ÷ 23 = 0.100 mol [1]. Na⁺ + e⁻ → Na, so n(e⁻) = 0.100 mol [1]. (b) 2Cl⁻ → Cl₂ + 2e⁻, so n(Cl₂) = 0.100 ÷ 2 = 0.0500 mol [1]. Volume = 0.0500 × 24 = 1.20 dm³ [1]. (c) 0.100 × 58.5 = 5.85 g [1]. (d) Chloride ions lose electrons [1]. Examiner insight: Check the units asked for: an answer of 1200 with no unit, when dm³ was requested, does not score the final mark in (b).

11. (a) Cathode: copper [1], because copper is less reactive than hydrogen [1]. Anode: chlorine [1], because chloride ions are discharged in preference to hydroxide ions in a concentrated solution [1]. (b) Cu²⁺ + 2e⁻ → Cu [1]. 2Cl⁻ → Cl₂ + 2e⁻ [1]. (c) n(Cu) = 0.381 ÷ 63.5 = 0.00600 mol [1]. Both half-equations use 2e⁻, so n(Cl₂) = 0.00600 mol [1]. Volume = 0.00600 × 24 000 = 144 cm³ [1]. (d) Some chlorine dissolves in the water of the solution [1]. Examiner insight: In (a), each explanation mark depends on the correct product; a wrong product with a correct-sounding reason does not score.

Where marks are usually lost

• Saying “electrons move” in a molten or dissolved electrolyte instead of “ions move”.
• Writing that a solid ionic compound “has no ions”.
• Forgetting H⁺ and OH⁻ from water, so giving the metal for a reactive-metal solution.
• Writing Cl or Br for the anode product instead of Cl₂ or Br₂.
• Putting electrons on the wrong side of an anode half-equation, or leaving charges unbalanced.
• Using a 1 : 1 ratio for copper and oxygen, or hydrogen and oxygen, instead of balancing electrons.
• Dividing a cm³ volume by 24 instead of 24 000.
• Giving a gas test without its result, or “litmus turns red” for chlorine.

Next steps

• Recap the rules in the electrolysis revision notes.
• Rework anything you missed with the electrolysis study guide.
• Practise mole and gas-volume methods with the formulae, equations and calculations practice.
• Course hub: 4CH1 Chemistry and the printable checklist.
• Take the free all-topics diagnostic.
• Book a free trial class.

Official syllabus

Pearson Edexcel International GCSE in Chemistry (4CH1), Specification, Issue 3, September 2024, published by Pearson Education Limited. Topic 1: Principles of chemistry, sub-topic (i) Electrolysis, statements 1.55C–1.60C.