S5 Chemistry Practice Paper: Redox Reactions And Electrolysis (12 Questions, 100 Marks, 75 Minutes, With Answers)

S5 Chemistry Practice Paper: Redox Reactions and Electrolysis (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-09

HKDSE Chemistry: Redox Reactions and Electrolysis

12 questions, 100 marks, 75 minutes on redox and electron transfer, oxidation numbers, electrolysis calculations, electrode products, half-equations and corrosion protection.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show all working. Answers without working earn no method marks. Write down a formula before substituting numbers.

Give numerical answers to three significant figures unless stated otherwise, and state the correct units.

Spend about 35 minutes on Section A, then use the remaining time for the charge calculations in Section B and the half-equation balancing in Section C.

Section A: Oxidation Numbers and Redox Concepts

Definitions of oxidation and reduction, oxidation numbers, oxidising and reducing agents

#	Question	Answer and marking notes
1	Define oxidation and reduction in terms of electron transfer, and state the role of an oxidising agent.	(a) Oxidation is the loss of electrons and an increase in oxidation number (1 mark). (b) Reduction is the gain of electrons and a decrease in oxidation number (1 mark). (c) An oxidising agent is itself reduced by accepting electrons, causing the other species to be oxidised (2 marks). (d) The oxidation number of an element in the oxidising agent decreases during the reaction (2 marks).
2	Find the oxidation number of manganese in KMnO₄. Given that MnO₄⁻ is reduced to Mn²⁺ in acidic solution, find the change in oxidation number and state how many electrons each manganese atom gains.	(a) Let the oxidation number of manganese be x, then (+1) + x + 4(-2) = 0 (2 marks). (b) x = +7 (3 marks). (c) Manganese is +7 in MnO₄⁻ and +2 in Mn²⁺, a change of 7 – 2 = 5 (3 marks). (d) The oxidation number falls by 5, so each manganese atom gains 5 electrons (2 marks).
3	Find the oxidation number of chromium in K₂Cr₂O₇, and state whether the change of chromium from +6 to +3 is oxidation or reduction.	(a) Let the oxidation number of chromium be y, then 2(+1) + 2y + 7(-2) = 0 (2 marks). (b) 2y = 12, so y = +6 (3 marks). (c) A change from +6 to +3 is a decrease in oxidation number, so it is reduction (3 marks).
4	A current of 2.0 A is passed through copper(II) sulphate solution for 965 seconds. Given Cu = 63.5, find (a) the charge passed; (b) the number of moles of electrons; (c) the mass of copper deposited at the cathode.	(a) Charge Q = It = 2.0 × 965 = 1930 C (3 marks). (b) Moles of electrons = 1930 ÷ 96500 = 0.02 mol (3 marks). (c) From Cu²⁺ + 2e⁻ → Cu, moles of copper = 0.01 mol (2 marks). (d) Mass = 0.01 × 63.5 = 0.635 g (2 marks).

Section B: Electrolysis and Quantitative Electrolysis

Faraday’s laws, electrode products and charge calculations

#	Question	Answer and marking notes
5	A current of 4.0 A is passed through copper(II) sulphate solution for 1930 seconds. Find the mass of copper deposited at the cathode. Given Cu = 63.5 and the Faraday constant = 96500 C mol⁻¹.	(a) Q = It = 4.0 × 1930 = 7720 C (2 marks). (b) Moles of electrons = 7720 ÷ 96500 = 0.08 mol (2 marks). (c) Moles of copper = 0.08 ÷ 2 = 0.04 mol (2 marks). (d) Mass = 0.04 × 63.5 = 2.54 g (2 marks).
6	During the electrolysis of dilute sulphuric acid, 0.02 mol of electrons is transferred. Find the number of moles of hydrogen produced at the cathode and its volume at room temperature and pressure, taking the molar volume as 24 dm³ mol⁻¹.	(a) The cathode reaction is 2H⁺ + 2e⁻ → H₂, so moles of hydrogen = 0.02 ÷ 2 = 0.01 mol (4 marks). (b) Volume = 0.01 × 24 = 0.24 dm³ (4 marks). (c) That is 240 cm³ (2 marks).
7	A current of 5.0 A is passed through copper(II) sulphate solution until 3.2 g of copper is deposited at the cathode. Find the charge passed and the time taken. Given Cu = 64.	(a) Moles of copper = 3.2 ÷ 64 = 0.05 mol (2 marks). (b) Moles of electrons = 0.05 × 2 = 0.10 mol (2 marks). (c) Q = 0.10 × 96500 = 9650 C (2 marks). (d) t = Q ÷ I = 9650 ÷ 5.0 = 1930 s (2 marks).
8	Compare electrolytic cells and galvanic cells in terms of energy conversion and the spontaneity of the reaction.	(a) An electrolytic cell converts electrical energy into chemical energy (2 marks). (b) A galvanic cell converts chemical energy into electrical energy (2 marks). (c) The reaction in an electrolytic cell is non-spontaneous and needs an applied voltage (2 marks). (d) The reaction in a galvanic cell is spontaneous and can supply current (2 marks).

Section C: Redox Equations and Applications

Half-equation balancing, corrosion and sacrificial protection

#	Question	Answer and marking notes
9	In acidic solution, MnO₄⁻ reacts with Fe²⁺ to form Mn²⁺ and Fe³⁺. Write the two half-equations and the balanced overall ionic equation, and identify the oxidising agent.	(a) Reduction half-equation: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (3 marks). (b) Oxidation half-equation: Fe²⁺ → Fe³⁺ + e⁻ (2 marks). (c) Multiplying the oxidation half-equation by 5 and adding: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺ (3 marks). (d) MnO₄⁻ is the oxidising agent and is itself reduced (2 marks).
10	Explain why a small amount of dilute sulphuric acid is added when electrolysing water, and state the gas produced at the anode together with its electrode reaction.	(a) Pure water conducts very poorly, so a little dilute sulphuric acid provides ions that allow current to flow (3 marks). (b) The acid acts only as an electrolyte, and it is the water that is electrolysed (2 marks). (c) Oxygen is produced at the anode (1 mark). (d) The anode reaction is 2H₂O → O₂ + 4H⁺ + 4e⁻ (2 marks).
11	State whether each of the following is a redox reaction, with a reason: (a) Zn with CuSO₄ solution; (b) NaOH with HCl; (c) 2H₂O₂ decomposing into 2H₂O and O₂.	(a) It is a redox reaction: the oxidation number of zinc rises from 0 to +2 and that of copper falls from +2 to 0 (3 marks). (b) It is not a redox reaction: all oxidation numbers are unchanged, and it is an acid-base neutralisation (2 marks). (c) It is a redox reaction: oxygen changes from -1 to -2 and to 0, which is a disproportionation (3 marks).
12	Explain how sacrificial protection prevents iron from rusting, and name two metals commonly used.	(a) Zinc or magnesium is more reactive than iron and loses electrons first, so it is oxidised preferentially (2 marks). (b) The sacrificial metal acts as the anode while the iron becomes the cathode and is protected (2 marks). (c) Zinc and magnesium are commonly used (2 marks).

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