A Level Chemistry (A2) chapter 2: Electrochemistry MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

50 exam-style questions on Electrochemistry, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

50 questions · 2 topics · 10 per test

Test this chapter freeAll A Level Chemistry (A2) chapters

What chapter 2 covers

Ten Electrochemistry questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. A very dilute aqueous solution of sodium chloride is electrolysed using inert electrodes. What is the main product at the anode?

  1. A
    oxygen
  2. B
    chlorine
  3. C
    hydrogen
  4. D
    sodium
Show answer

Answer: A. E⦵ for O2/H2O (+1.23 V) is less positive than for Cl2/Cl− (+1.36 V), so water is oxidised more easily. Chlorine is the main product only when the chloride concentration is high. Hydrogen and sodium could only form at the cathode.

2. E⦵ values: Cl2 + 2e− ⇌ 2Cl− +1.36 V; O2 + 4H+ + 4e− ⇌ 2H2O +1.23 V. When concentrated aqueous sodium chloride is electrolysed with inert electrodes, chlorine, not oxygen, is the main anode product. Why?

  1. A
    chlorine is a stronger oxidising agent than oxygen
  2. B
    sodium ions remove the oxygen as sodium oxide
  3. C
    the E⦵ of Cl2/Cl− is more positive, so Cl− is always oxidised first
  4. D
    the high [Cl−] makes the electrode potential of Cl2/Cl− less positive, so Cl− is oxidised more readily
Show answer

Answer: D. Under standard conditions water would be oxidised first (+1.23 V is less positive). A high chloride concentration shifts Cl2 + 2e− ⇌ 2Cl− to the left, lowering E so chloride is discharged. The more positive E⦵ actually makes Cl− harder, not easier, to oxidise.

3. E⦵ values: Zn2+/Zn −0.76 V; Cr3+/Cr2+ −0.41 V; V3+/V2+ −0.26 V; Sn4+/Sn2+ +0.15 V. Which species is the strongest reducing agent?

  1. A
    Sn4+
  2. B
    Zn
  3. C
    Zn2+
  4. D
    Sn2+
Show answer

Answer: B. Reducing agents are on the right of the half-equations; the most negative E⦵ marks the one that gives up electrons most readily, Zn. Sn4+ is the strongest oxidising agent of the list, and Zn2+ is the weakest.

4. How is the standard electrode potential of the Fe3+/Fe2+ half-cell measured against a standard hydrogen electrode?

  1. A
    using a platinum electrode in a solution with Fe2+ and Fe3+ both at 1.00 mol dm−3
  2. B
    using an iron electrode in a solution with Fe2+ and Fe3+ both at 1.00 mol dm−3
  3. C
    using a platinum electrode in a solution of Fe3+ at 1.00 mol dm−3 with no Fe2+
  4. D
    using an iron electrode in a solution of Fe2+ at 1.00 mol dm−3 with no Fe3+
Show answer

Answer: A. For two ions of the same element, both ions are present at 1.00 mol dm−3 and an inert platinum electrode transfers the electrons. An iron electrode would react, and iron in Fe2+ sets up the Fe2+/Fe half-cell instead.

5. A cell is set up to measure E⦵ for the Ag+/Ag half-cell. Why is the salt bridge made with aqueous potassium nitrate rather than aqueous potassium chloride?

  1. A
    chloride ions would be oxidised to chlorine gas at the silver electrode
  2. B
    potassium nitrate carries electrons between the half-cells better than potassium chloride
  3. C
    nitrate ions are needed to oxidise the silver electrode
  4. D
    chloride ions would precipitate silver chloride, lowering [Ag+] and changing the potential
Show answer

Answer: D. The salt bridge completes the circuit by letting ions move between the half-cells, so its ions must not react with either solution. Ag+ + Cl− → AgCl(s) would lower [Ag+] below 1.00 mol dm−3. A salt bridge carries ions, never electrons.

6. A cell is made from a Zn2+/Zn half-cell (E⦵ = −0.76 V) and a Pb2+/Pb half-cell (E⦵ = −0.13 V). Which statement is correct?

  1. A
    electrons flow in the external circuit from the lead electrode to the zinc electrode
  2. B
    the lead electrode is the negative electrode of the cell
  3. C
    electrons flow in the external circuit from the zinc electrode to the lead electrode
  4. D
    electrons flow through the salt bridge from the zinc half-cell to the lead half-cell
Show answer

Answer: C. Zinc has the more negative E⦵, so it is oxidised and is the negative electrode. Electrons leave it and travel through the wire to the lead electrode, where Pb2+ is reduced. Only ions move through the salt bridge.

7. Half-equations: MnO4− + 8H+ + 5e− → Mn2+ + 4H2O; H2O2 → O2 + 2H+ + 2e−. In the balanced ionic equation for the reaction of 2 mol of MnO4− with hydrogen peroxide, how many moles of H2O2 react and how many moles of H+ are used up?

  1. A
    2 mol H2O2; 16 mol H+
  2. B
    5 mol H2O2; 6 mol H+
  3. C
    5 mol H2O2; 10 mol H+
  4. D
    5 mol H2O2; 16 mol H+
Show answer

Answer: B. 2MnO4− take 10e−, supplied by 5H2O2. These release 10H+, which cancel against the 16H+ used: 2MnO4− + 6H+ + 5H2O2 → 2Mn2+ + 8H2O + 5O2. 16 mol forgets to cancel the H+ on both sides.

8. Iron(III) ions dissolve copper: 2Fe3+(aq) + Cu(s) → 2Fe2+(aq) + Cu2+(aq). E⦵(Fe3+/Fe2+) = +0.77 V and E⦵(Cu2+/Cu) = +0.34 V. What is ΔG⦵ for the reaction as written? (F = 96 500 C mol−1)

  1. A
    −41.5 kJ mol−1
  2. B
    −232 kJ mol−1
  3. C
    −83.0 kJ mol−1
  4. D
    −214 kJ mol−1
Show answer

Answer: C. E⦵cell = 0.77 − 0.34 = +0.43 V. Two electrons are transferred (Cu → Cu2+ + 2e−), so ΔG⦵ = −2 × 0.43 × 96 500 = −83 000 J mol−1 = −83.0 kJ mol−1. Doubling the Fe3+ potential to 1.54 V gives −232.

9. A cell is made from a Cu2+/Cu half-cell (E⦵ = +0.34 V) and an Ag+/Ag half-cell (E⦵ = +0.80 V). Aqueous sodium chloride is added to the silver half-cell and a white precipitate forms. How do the silver electrode potential and the cell e.m.f. change?

  1. A
    the silver electrode potential becomes less positive and the e.m.f. decreases
  2. B
    the silver electrode potential becomes more positive and the e.m.f. increases
  3. C
    the silver electrode potential becomes less positive and the e.m.f. increases
  4. D
    neither changes, because E⦵ values are constants
Show answer

Answer: A. Precipitating AgCl lowers [Ag+], moving Ag+ + e− ⇌ Ag to the left, so E(Ag) becomes less positive and closer to the copper value: the e.m.f. falls. E⦵ is constant, but E changes with concentration.

10. Chlorine oxidises bromide ions: Cl2 + 2Br− → 2Cl− + Br2. E⦵(Cl2/Cl−) = +1.36 V and E⦵(Br2/Br−) = +1.07 V. What is ΔG⦵ for the reaction as written? (F = 96 500 C mol−1)

  1. A
    −56 kJ mol−1
  2. B
    −28 kJ mol−1
  3. C
    +56 kJ mol−1
  4. D
    −5.6 × 104 kJ mol−1
Show answer

Answer: A. E⦵cell = 1.36 − 1.07 = +0.29 V and two electrons are transferred, so ΔG⦵ = −nE⦵cellF = −2 × 0.29 × 96 500 = −56 000 J mol−1 = −56 kJ mol−1. A positive Ecell always gives a negative ΔG.

Start the chapter testAbout the A Level Chemistry (A2) test

Read the notes for this chapter

Ways to practise this chapter

Electrochemistry: questions students ask

How many Electrochemistry MCQs are there?

50, each with the answer and a short explanation, across 2 topics: Electrolysis, Standard electrode potentials E , standard cell potentials Ecell and the Nernst equation. Every chapter test draws ten of them, so you can take it many times.

Is the A Level Chemistry (A2) chapter 2 test free?

Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.

How is the chapter test marked?

The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

Keep going

← Chapter 1: Chemical energeticsChapter 3: Equilibria →All chaptersStuck? Ask a question