OxfordAQA A-Level Chemistry: Electrode potentials and electrochemical cells (9620) – 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 3.1.9, Electrode potentials and electrochemical cells (3.1.9.1 and 3.1.9.2), of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. It is International A2 content, so every question is International A-level only. The specification places it in Unit 3, and Unit 5 can assess all content, including required practical 6. Use only the E⦵ data given.

Revise first with the study guide or the revision notes. The OxfordAQA A-Level Chemistry hub and printable checklist cover the rest of the course, and the free diagnostics show where to focus.

Questions

1. Define the standard electrode potential of a half-cell. [3]

2. A student writes these three electrode reactions as oxidations. Rewrite each one using the IUPAC convention.

Na → Na⁺ + e⁻ Cr²⁺ → Cr³⁺ + e⁻ 2SO₄²⁻ → S₂O₈²⁻ + 2e⁻ [3]

3. Data: Ni²⁺ + 2e⁻ ⇌ Ni, E⦵ = −0.26 V; Au³⁺ + 3e⁻ ⇌ Au, E⦵ = +1.52 V.

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

4. Data: V³⁺ + e⁻ ⇌ V²⁺, E⦵ = −0.26 V; Co²⁺ + 2e⁻ ⇌ Co, E⦵ = −0.28 V.

A student writes the cell as Pt(s) | V²⁺(aq), V³⁺(aq) || Co²⁺(aq) | Co(s).

(a) Calculate the EMF of the cell as the student has written it. [1] (b) Explain what your answer shows, and write the correct representation. [2] (c) Write the overall equation for the reaction when the cell delivers a current. [1]

5. A student sets up a reference electrode: hydrogen gas at 100 kPa is bubbled over a copper electrode dipping into 1.00 mol dm⁻³ sulfuric acid at 298 K.

(a) Identify two errors in this set-up and explain why each is an error. [4] (b) A correctly built standard hydrogen electrode is connected to a standard cadmium half-cell. The voltmeter reads 0.40 V and the hydrogen electrode is positive. State E⦵ for Cd²⁺/Cd. [1]

6. Data:

S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻, E⦵ = +2.01 V VO₂⁺ + 2H⁺ + e⁻ ⇌ VO²⁺ + H₂O, E⦵ = +1.00 V Cd²⁺ + 2e⁻ ⇌ Cd, E⦵ = −0.40 V Cr³⁺ + e⁻ ⇌ Cr²⁺, E⦵ = −0.41 V

(a) Identify the strongest reducing agent in the data. [1] (b) Deduce whether cadmium metal can reduce Cr³⁺ ions to Cr²⁺ ions under standard conditions. Explain. [2] (c) Show that Cr²⁺ ions can reduce VO₂⁺ ions. Calculate the EMF and write the overall equation. [3] (d) Explain how increasing the concentration of Cr³⁺ ions could change your answer to (b). [2]

7. In required practical 6, a student builds the cell Ni(s) | Ni²⁺(aq) || Sn²⁺(aq) | Sn(s). Data: Ni²⁺/Ni −0.26 V; Sn²⁺/Sn −0.13 V. Three readings are 0.116 V, 0.120 V and 0.118 V.

(a) Calculate the expected EMF under standard conditions. [1] (b) Calculate the mean of the readings. [1] (c) Suggest two reasons why the mean is lower than expected. [2] (d) Explain why a high-resistance voltmeter is used. [1]

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

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

(a) Write the overall equation for discharge. [1] (b) Explain how these electrode reactions generate an electric current. [3] (c) Write the equation for the reaction at the cobalt-containing electrode when the cell is recharged. [1] (d) Give one benefit and one risk to society of using rechargeable lithium cells. [2]

9. A non-rechargeable zinc–air cell uses these half-equations:

[Zn(OH)₄]²⁻ + 2e⁻ ⇌ Zn + 4OH⁻, E⦵ = −1.20 V O₂ + 2H₂O + 4e⁻ ⇌ 4OH⁻, E⦵ = +0.40 V

(a) Deduce the reaction at each electrode during discharge. [2] (b) Write the overall equation. [2] (c) Deduce the EMF of the cell. [1] (d) Oxygen is taken in from the air continuously. Explain why this cell is still not a fuel cell. [1]

10. An alkaline hydrogen–oxygen fuel cell is based on these half-equations:

O₂ + 2H₂O + 4e⁻ ⇌ 4OH⁻ 2H₂O + 2e⁻ ⇌ H₂ + 2OH⁻

(a) Write the reaction at the negative electrode and at the positive electrode. [2] (b) Write the overall equation and show that hydroxide ions are not used up. [2] (c) Explain why the cell does not need to be electrically recharged. [1] (d) State one risk to society of using hydrogen as the fuel. [1]

11. Data:

Ca²⁺ + 2e⁻ ⇌ Ca, E⦵ = −2.87 V Cd²⁺ + 2e⁻ ⇌ Cd, E⦵ = −0.40 V VO₂⁺ + 2H⁺ + e⁻ ⇌ VO²⁺ + H₂O, E⦵ = +1.00 V S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻, E⦵ = +2.01 V

(a) Identify the pair of half-cells that would give the largest EMF and calculate it. Suggest why a calcium half-cell with an aqueous solution is not practical. [3] (b) A cell is made from the cadmium and VO₂⁺/VO²⁺ half-cells. Write its conventional representation, calculate its EMF and write the overall equation. [5] (c) Predict, with a reason, the effect on the EMF of the cell in (b) of adding more acid to the vanadium half-cell. [2]

Answers

1. The EMF of a cell made from the half-cell and the standard hydrogen electrode [1]; measured at 298 K and 100 kPa [1]; with 1.00 mol dm⁻³ solutions of ions [1] [3] Examiner insight: A definition listing only the conditions, without the hydrogen electrode, is incomplete.

2. Na⁺ + e⁻ ⇌ Na [1]; Cr³⁺ + e⁻ ⇌ Cr²⁺ [1]; S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻ [1] [3] Examiner insight: Flipping only the arrow leaves an oxidation; the electrons must move to the left.

3. (a) Ni is more negative, so it is the negative electrode on the left [1]; Ni(s) | Ni²⁺(aq) || Au³⁺(aq) | Au(s) [1] (b) EMF = 1.52 − (−0.26) = +1.78 V [1] (c) Ni × 3 and Au × 2 to balance 6e⁻ [1]; 3Ni + 2Au³⁺ → 3Ni²⁺ + 2Au [1] Examiner insight: The half-equations are scaled to match electrons but the E⦵ values are not; scaling them gives a wrong EMF in (b).

4. (a) EMF = E(right) − E(left) = −0.28 − (−0.26) = −0.02 V [1] (b) A negative EMF means the cell is the wrong way round: cobalt is negative [1]; Co(s) | Co²⁺(aq) || V³⁺(aq), V²⁺(aq) | Pt(s) [1] (c) Co + 2V³⁺ → Co²⁺ + 2V²⁺ [1] Examiner insight: With close E⦵ values, check the order; a working cell always has a positive EMF.

5. (a) Copper electrode [1]; copper is not inert and could set up its own Cu²⁺/Cu equilibrium, so platinum is needed [1]; 1.00 mol dm⁻³ sulfuric acid [1]; it releases more than one H⁺ per formula unit, so [H⁺] is above 1.00 mol dm⁻³; use, for example, 1.00 mol dm⁻³ HCl [1] (b) −0.40 V [1] Examiner insight: 298 K and 100 kPa are correct here; each real error needs a reason as well as a name.

6. (a) Cr²⁺ (its half-equation is the most negative, −0.41 V) [1] (b) Cr³⁺/Cr²⁺ (−0.41 V) is less positive than Cd²⁺/Cd (−0.40 V) [1]; so the reaction is not feasible under standard conditions; EMF would be −0.01 V [1] (c) VO₂⁺/VO²⁺ (+1.00 V) is more positive than Cr³⁺/Cr²⁺ (−0.41 V), so feasible [1]; EMF = 1.00 − (−0.41) = +1.41 V [1]; VO₂⁺ + 2H⁺ + Cr²⁺ → VO²⁺ + H₂O + Cr³⁺ [1] (d) Higher [Cr³⁺] shifts Cr³⁺ + e⁻ ⇌ Cr²⁺ to the right, making its potential more positive [1]; it could become more positive than −0.40 V, so cadmium could then reduce Cr³⁺ [1] Examiner insight: In (a) the reducing agent is a right-hand species; choosing Cd because it is a metal misreads the table.

7. (a) −0.13 − (−0.26) = 0.13 V [1] (b) (0.116 + 0.120 + 0.118) ÷ 3 = 0.118 V [1] (c) Any two of: oxide on an electrode; solutions not exactly 1.00 mol dm⁻³ or temperature not 298 K; the voltmeter draws some current [2] (d) So that almost no current flows and the reading equals the EMF [1] Examiner insight: In (c), “human error” is too vague; name a specific cause linked to this cell.

8. (a) Li + CoO₂ → Li⁺[CoO₂]⁻ [1] (b) Lithium is oxidised at the negative electrode, releasing electrons [1]; the electrons flow through the external circuit to the positive electrode, where CoO₂ is reduced (cobalt +4 to +3) [1]; Li⁺ ions move through the electrolyte to the positive electrode to balance the charge [1] (c) Li⁺[CoO₂]⁻ → Li⁺ + CoO₂ + e⁻ [1] (d) Benefit: reused over many charge cycles, so fewer cells are discarded [1]; risk: an overcharged or crushed cell can overheat and ignite [1] Examiner insight: In (b), say where electrons go and where ions go; they take different paths.

9. (a) Negative electrode: Zn + 4OH⁻ → [Zn(OH)₄]²⁻ + 2e⁻ [1]; positive electrode: O₂ + 2H₂O + 4e⁻ → 4OH⁻ [1] (b) Double the zinc equation to balance 4e⁻ [1]; 2Zn + O₂ + 2H₂O + 4OH⁻ → 2[Zn(OH)₄]²⁻ [1] (c) 0.40 − (−1.20) = +1.60 V [1] (d) The zinc is not supplied continuously; once it is used up, the cell stops [1] Examiner insight: In (b), 8OH⁻ on the left and 4OH⁻ on the right should cancel to leave 4OH⁻; check that charge balances (−4 each side).

10. (a) Negative: H₂ + 2OH⁻ → 2H₂O + 2e⁻ [1]; positive: O₂ + 2H₂O + 4e⁻ → 4OH⁻ [1] (b) Doubling the negative-electrode equation and adding gives 2H₂ + O₂ → 2H₂O [1]; the 4OH⁻ made at the positive electrode equals the 4OH⁻ used at the negative electrode [1] (c) Hydrogen and oxygen are supplied continuously, so the reactants are never used up [1] (d) Any one: hydrogen–air mixtures are explosive; it is bulky to store; producing it may use fossil-fuel energy [1] Examiner insight: Reverse the hydrogen half-equation for the negative electrode.

11. (a) Ca²⁺/Ca and S₂O₈²⁻/SO₄²⁻ [1]; EMF = 2.01 − (−2.87) = +4.88 V [1]; calcium reacts with water, so it would not stay in equilibrium with its ions [1] (b) Cadmium is the negative electrode on the left; the vanadium half-cell needs platinum [1]; Cd(s) | Cd²⁺(aq) || VO₂⁺(aq), H⁺(aq), VO²⁺(aq) | Pt(s) [1]; EMF = 1.00 − (−0.40) = +1.40 V [1]; vanadium equation × 2 [1]; Cd + 2VO₂⁺ + 4H⁺ → Cd²⁺ + 2VO²⁺ + 2H₂O [1] (c) H⁺ is on the left of the vanadium half-equation, so more acid shifts it right and its potential becomes more positive [1]; the EMF increases [1] Examiner insight: In (b), include H⁺ in the cell representation because it appears in the half-equation.

Where marks are usually lost

• Writing an electrode reaction as an oxidation instead of an IUPAC reduction.
• Scaling an E⦵ value when the half-equation is scaled.
• Leaving out Pt(s) or H⁺(aq) from a cell representation.
• Using sulfuric acid in the hydrogen electrode without noticing that [H⁺] is not 1.00 mol dm⁻³.
• Saying electrons travel through the electrolyte or the salt bridge.
• Leaving spare OH⁻ or electrons uncancelled in an overall equation.
• Explaining a concentration change without stating the direction of the equilibrium shift.

Next steps

• Revision notes for this section
• Study guide for this section
• Equilibria practice (Le Chatelier’s principle)
• OxfordAQA A-Level Chemistry hub
• Printable OxfordAQA A-Level Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, section 3.1.9 Electrode potentials and electrochemical cells (International A2). Published by OxfordAQA.