AQA A-Level Chemistry: Electrode potentials and electrochemical cells (7405) – 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 section 3.1.11, Electrode potentials and electrochemical cells, of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. The whole section is marked “A-level only”, so none of it is AS content. It is listed under A-level Paper 1, and Paper 3 can assess any content, including required practical 8. Use only the E⦵ data given in each question.

Before you start, you may want the study guide or the revision notes. The AQA A-Level Chemistry hub and printable checklist cover the rest of the course, and the free diagnostics show where to focus.

Questions

1. State the three conditions that apply when a standard electrode potential, E⦵, is measured. [3]

2. A student writes the half-equation for a nickel electrode as Ni → Ni²⁺ + 2e⁻. Rewrite it using the IUPAC convention, and state what a negative E⦵ value tells you about nickel compared with hydrogen. [2]

3. The standard hydrogen electrode is the reference for electrode potentials.

(a) Describe the standard hydrogen electrode. [3] (b) Explain why platinum is used as the electrode. [1]

4. A non-rechargeable cell is made from two half-cells. Data: Mg²⁺ + 2e⁻ ⇌ Mg, E⦵ = −2.37 V; Ni²⁺ + 2e⁻ ⇌ Ni, E⦵ = −0.25 V.

(a) Write the conventional representation of the cell made from these two half-cells. [2] (b) Calculate the EMF of the cell. [2] (c) Write the overall equation for the cell reaction. [1]

5. Data: Sn⁴⁺ + 2e⁻ ⇌ Sn²⁺, E⦵ = +0.15 V; Ag⁺ + e⁻ ⇌ Ag, E⦵ = +0.80 V.

(a) Explain why the tin half-cell needs a platinum electrode. [1] (b) Write the conventional representation of the cell. [2] (c) Calculate the EMF. [1] (d) Write the overall cell equation. [1]

6. Data: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ ⇌ 2Cr³⁺ + 7H₂O, E⦵ = +1.33 V; Cl₂ + 2e⁻ ⇌ 2Cl⁻, E⦵ = +1.36 V; Br₂ + 2e⁻ ⇌ 2Br⁻, E⦵ = +1.07 V; Zn²⁺ + 2e⁻ ⇌ Zn, E⦵ = −0.76 V.

(a) Deduce whether acidified dichromate(VI) ions can oxidise chloride ions under standard conditions. Explain. [2] (b) Deduce whether acidified dichromate(VI) ions can oxidise bromide ions. Calculate the EMF. [2] (c) Identify the strongest reducing agent in the data. [1] (d) Give one reason why a reaction predicted to be feasible may not be observed. [1]

7. A standard copper half-cell (E⦵ = +0.34 V) is used as a reference.

(a) Metal X is connected to it. X is the negative electrode and the EMF is 1.04 V. Calculate E⦵ for X²⁺/X. [2] (b) Metal Y is the positive electrode and the EMF is 0.46 V. Calculate E⦵ for Y⁺/Y. [1]

8. The cell Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s) has an EMF of 1.10 V under standard conditions. The copper(II) sulfate solution is replaced by one of concentration 0.10 mol dm⁻³. Explain the effect on the EMF. The Nernst equation is not required. [3]

9. This question is about required practical 8.

(a) Explain why the metal strips are rubbed with emery paper before use. [1] (b) State the function of the salt bridge. [2] (c) Explain why potassium chloride is not used in a salt bridge for a silver half-cell. [1] (d) Explain why a high-resistance voltmeter is used. [1]

10. The simplified electrode reactions in a lithium cell are:

Positive electrode: Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻ Negative electrode: Li → Li⁺ + e⁻

(a) Give the oxidation state of cobalt in CoO₂ and in Li⁺[CoO₂]⁻. [2] (b) Write the overall equation for the reaction during discharge. [1] (c) Describe what happens at each electrode when the cell is recharged. [2] (d) State one risk associated with lithium cells. [1]

11. A lead–acid rechargeable cell uses these half-equations:

PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ ⇌ PbSO₄ + 2H₂O, E⦵ = +1.69 V PbSO₄ + 2e⁻ ⇌ Pb + SO₄²⁻, E⦵ = −0.36 V

(a) Deduce the reaction at the negative electrode during discharge. [1] (b) Write the overall equation for discharge. [2] (c) Calculate the EMF of the cell. [1] (d) Give the oxidation state of lead in PbO₂, PbSO₄ and Pb, and use them to show that the overall reaction is a redox reaction. [2] (e) Explain how the cell is recharged. [1] (f) Explain how the electrode reactions generate an electric current. [2]

12. Data: Zn²⁺ + 2e⁻ ⇌ Zn, −0.76 V; Fe²⁺ + 2e⁻ ⇌ Fe, −0.44 V; Fe³⁺ + e⁻ ⇌ Fe²⁺, +0.77 V; Ag⁺ + e⁻ ⇌ Ag, +0.80 V.

(a) Choose the two half-cells that give the cell with the largest EMF. Calculate the EMF and write the conventional representation. [3] (b) Iron filings are added to a solution of iron(III) ions. Predict whether a reaction occurs. Write an equation and calculate the EMF. [3] (c) Ag⁺ ions are mixed with Fe²⁺ ions under standard conditions. Calculate the EMF and suggest why the reaction may not go as predicted if conditions change. [2]

Answers

1. Temperature 298 K [1]; pressure 100 kPa [1]; 1.00 mol dm⁻³ concentration of ions [1] [3] Examiner insight: Give the specification’s values; “room temperature” or “1 atm” does not match them.

2. Ni²⁺(aq) + 2e⁻ ⇌ Ni(s) [1]; negative E⦵ means nickel loses electrons more readily than hydrogen, so Ni is a better reducing agent than H₂ [1] [2] Examiner insight: The half-equation must be a reduction with electrons on the left; reversing only the arrow is not enough.

3. (a) Hydrogen gas at 100 kPa [1]; bubbled over platinum in 1.00 mol dm⁻³ H⁺(aq) [1]; at 298 K [1] (b) Platinum is inert and conducts electrons to and from the H₂/H⁺ equilibrium [1] Examiner insight: A labelled-diagram description still needs the three conditions written; a sketch alone shows no concentration or pressure.

4. (a) Mg is more negative so goes on the left [1]; Mg(s) | Mg²⁺(aq) || Ni²⁺(aq) | Ni(s) [1] (b) EMF = −0.25 − (−2.37) [1] = +2.12 V [1] (c) Mg + Ni²⁺ → Mg²⁺ + Ni [1] Examiner insight: A correctly written cell always has a positive EMF, so a negative answer in (b) signals the electrodes are the wrong way round.

5. (a) Both Sn²⁺ and Sn⁴⁺ are ions in solution, so an inert conductor is needed to transfer electrons [1] (b) Tin half-cell on the left, as the negative electrode [1]; Pt(s) | Sn²⁺(aq), Sn⁴⁺(aq) || Ag⁺(aq) | Ag(s) [1] (c) 0.80 − 0.15 = +0.65 V [1] (d) 2Ag⁺ + Sn²⁺ → 2Ag + Sn⁴⁺ [1] Examiner insight: In (d) the silver half-equation is doubled but its E⦵ is not; doubling 0.80 V in (c) gives a wrong EMF.

6. (a) Cl₂/Cl⁻ (+1.36 V) is more positive than Cr₂O₇²⁻/Cr³⁺ (+1.33 V) [1]; so dichromate cannot oxidise Cl⁻ (EMF would be −0.03 V) [1] (b) +1.33 V is more positive than +1.07 V, so dichromate can oxidise Br⁻ [1]; EMF = 1.33 − 1.07 = +0.26 V [1] (c) Zn [1] (d) The activation energy may be high so the reaction is too slow; or conditions are non-standard [1] Examiner insight: In (c) the answer is the species Zn, not Zn²⁺; naming the ion confuses reducing agent with oxidising agent.

7. (a) EMF = E(Cu) − E(X), so 1.04 = 0.34 − E(X) [1]; E(X) = −0.70 V [1] (b) E(Y) = 0.34 + 0.46 = +0.80 V [1] Examiner insight: Decide from the polarity which electrode is subtracted before substituting; a sign slip gives +1.38 V.

8. Lowering [Cu²⁺] shifts Cu²⁺ + 2e⁻ ⇌ Cu to the left [1]; the copper electrode potential becomes less positive [1]; so the EMF decreases below 1.10 V [1] [3] Examiner insight: A bare “decreases” is not an explanation; link the equilibrium shift to the copper electrode potential.

9. (a) To remove the oxide layer, which would affect the measured potential [1] (b) Completes the circuit [1]; lets ions move to balance charge without the solutions mixing [1] (c) Chloride ions would precipitate silver chloride, changing [Ag⁺] [1] (d) So that almost no current flows and the reading is the EMF [1] Examiner insight: In (b) “lets electrons flow” is wrong: only ions move through the salt bridge.

10. (a) CoO₂: +4 [1]; Li⁺[CoO₂]⁻: +3 [1] (b) Li + CoO₂ → Li⁺[CoO₂]⁻ [1] (c) An external supply drives the reactions in reverse [1]; Li⁺ is reduced back to Li at one electrode while Li⁺[CoO₂]⁻ is oxidised back to Li⁺ + CoO₂ at the other [1] (d) Lithium is reactive; a damaged cell may catch fire [1] Examiner insight: In (a) the charge on [CoO₂]⁻ must be used; ignoring it gives +4 twice.

11. (a) The more negative half-equation runs in reverse: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ [1] (b) Add the positive-electrode reduction to (a), 2e⁻ each [1]; Pb + PbO₂ + 4H⁺ + 2SO₄²⁻ → 2PbSO₄ + 2H₂O [1] (c) 1.69 − (−0.36) = +2.05 V [1] (d) PbO₂ +4, PbSO₄ +2, Pb 0 [1]; Pb is oxidised (0 → +2) and PbO₂ is reduced (+4 → +2) [1] (e) An external supply forces current backwards, reversing both electrode reactions to regenerate Pb and PbO₂ [1] (f) Oxidation of Pb at the negative electrode releases electrons [1]; they flow through the external circuit to the positive electrode, where PbO₂ is reduced [1] Examiner insight: In (b) an equation with uncancelled electrons, or SO₄²⁻ on both sides, is not a finished overall equation.

12. (a) Largest gap is Zn and Ag [1]; EMF = 0.80 − (−0.76) = +1.56 V [1]; Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s) [1] (b) Fe³⁺/Fe²⁺ (+0.77) is more positive than Fe²⁺/Fe (−0.44), so reaction occurs [1]; Fe + 2Fe³⁺ → 3Fe²⁺ [1]; EMF = 0.77 − (−0.44) = +1.21 V [1] (c) EMF = 0.80 − 0.77 = +0.03 V [1]; the EMF is so small that a change in concentration could make it negative [1] Examiner insight: In (b) use Fe²⁺/Fe for the iron metal and Fe³⁺/Fe²⁺ for the iron(III) ions; swapping them reverses the prediction.

Where marks are usually lost

• Writing IUPAC half-equations as oxidations.
• Multiplying E⦵ when a half-equation is scaled up.
• Giving a negative EMF for a cell written the right way round.
• Missing platinum or state symbols in a cell representation.
• Saying the salt bridge carries electrons.
• Treating feasibility as a statement about rate.
• Ignoring the charge on [CoO₂]⁻ when working out oxidation states.
• Explaining a concentration change without saying how the equilibrium shifts.
• Leaving electrons or spectator ions on both sides of an overall equation.

Next steps

• Revision notes for this section
• Study guide for this section
• Redox equations practice
• AQA A-Level Chemistry hub
• Printable AQA A-Level Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards, section 3.1.11 Electrode potentials and electrochemical cells (A-level only). Published by AQA.