A Level Chemistry: Electrochemistry — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Electrochemistry revision notes

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Section A — short answer

1. State the standard conditions for measuring an electrode potential. [3]

2. Explain why the standard hydrogen electrode is assigned a value of 0.00 V. [2]

3. Write the conventional cell diagram for a cell made from Zn²⁺/Zn (−0.76 V) and Ag⁺/Ag (+0.80 V), and calculate its e.m.f. [3]

4. State the relationship between ΔG and E_cell, and give the condition for a reaction to be feasible. [2]

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Section B — structured

5. Consider the two half-equations:

Fe3+ + e-   <=>  Fe2+     E = +0.77 V
I2 + 2e-    <=>  2I-      E = +0.54 V

(a) Deduce which species is the stronger oxidising agent, giving a reason. [2]

(b) Write the overall equation for the spontaneous reaction and calculate E_cell. [3]

(c) Calculate ΔG for this reaction. (F = 96 500 C mol⁻¹) [2]

(d) The calculated ΔG is negative, yet under certain conditions no observable reaction occurs. Give two possible reasons. [2]

6. A student electrolyses concentrated aqueous sodium chloride using inert electrodes.

(a) Write the half-equation at the cathode and identify the product. [2]

(b) Explain why hydrogen, not sodium, is produced at the cathode. [2]

(c) State the product at the anode, and explain how the product would differ if the solution were very dilute. [3]

7. A current of 0.500 A is passed through molten aluminium oxide for 2.00 hours.

(a) Calculate the charge passed. [1]

(b) Write the half-equation for the formation of aluminium. [1]

(c) Calculate the mass of aluminium deposited. (A_r(Al) = 27.0) [3]

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Section C

8. The electrode potential of a Cu²⁺/Cu half-cell depends on the concentration of Cu²⁺.

(a) State the Nernst equation, and use Le Chatelier’s principle to explain what happens to E if the concentration of Cu²⁺ is increased. [3]

(b) This Cu²⁺/Cu half-cell, with [Cu²⁺] below standard concentration, is paired with a standard Zn²⁺/Zn half-cell. State, with a reason, whether the resulting cell e.m.f. is higher or lower than the standard value. [2]

9. Dilute sulfuric acid and concentrated copper(II) sulfate solution are electrolysed separately using inert electrodes.

(a) State the general rule for which cation is discharged at the cathode in aqueous electrolysis. [2]

(b) Predict the cathode product in each solution, explaining your answer. [3]

(c) State the general rule for which anion is discharged at the anode, and predict the anode product in both solutions. [3]

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Answers

1. 298 K [1]; solution concentrations of 1 mol dm⁻³ [1]; pressure of 100 kPa for any gases [1].

2. An electrode potential can only be measured as a potential difference between two half-cells [1], so one must be chosen as an arbitrary reference against which all others are measured [1]. Answers saying hydrogen “has no potential” score nothing — the value is a convention, not a measurement.

3. Zn | Zn²⁺ ‖ Ag⁺ | Ag [1] — single line for a phase boundary, double for the salt bridge [1]. E_cell = +0.80 − (−0.76) = +1.56 V [1].

4. ΔG = −nFE_cell [1]. The reaction is feasible when E_cell is positive, so that ΔG is negative [1].

5. (a) Fe³⁺ [1], because it has the more positive E⦵, so it is more readily reduced [1].

(b) 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂ [1] (correct balancing [1]). E_cell = +0.77 − (+0.54) = +0.23 V [1].

(c) n = 2, so ΔG = −(2)(96 500)(0.23) [1] = −44 390 J mol⁻¹ ≈ −44.4 kJ mol⁻¹ [1].

(d) Any two: the activation energy is high, so the rate is negligible [1]; the conditions are not standard, so the actual electrode potentials differ from E⦵ [1]. “Feasible” means thermodynamically possible, not fast — examined nearly every series.

6. (a) 2H₂O + 2e⁻ → H₂ + 2OH⁻ (or 2H⁺ + 2e⁻ → H₂) [1]; hydrogen [1].

(b) Sodium is more reactive than hydrogen / Na⁺ has a more negative electrode potential [1], so hydrogen is preferentially reduced [1].

(c) Chlorine [1]. In very dilute solution oxygen would be produced instead [1], because the chloride concentration is too low for chloride to be discharged in preference to hydroxide [1].

7. (a) Q = It = 0.500 × 7200 = 3600 C [1].

(b) Al³⁺ + 3e⁻ → Al [1].

(c) n(e⁻) = 3600 ÷ 96 500 = 0.0373 mol [1]. n(Al) = 0.0373 ÷ 3 = 0.01243 mol [1]. m = 0.01243 × 27.0 = 0.336 g [1]. Forgetting to divide by 3 is the standard error here.

8. (a) At 298 K, E = E⦵ + (0.059 ÷ z) log([oxidised] ÷ [reduced]) [1] — the 0.059 V constant applies specifically at 298 K, and pure solid Cu (the reduced species here) has a fixed activity of 1, so only [Cu²⁺] appears in the ratio. Increasing [Cu²⁺] (the oxidised species) shifts the half-equilibrium Cu²⁺ + 2e⁻ ⇌ Cu towards reduction, by Le Chatelier’s principle [1], making E more positive [1].

(b) The cell e.m.f. is lower than the standard value [1], because a below-standard [Cu²⁺] makes the copper electrode potential less positive than +0.34 V, reducing the difference between the two electrode potentials [1].

9. (a) The less reactive cation — the one with the more positive E⦵ — is discharged in preference, though at very low concentration this simple rule can be overridden (see question 6(c)); electrode material and overpotential effects can also shift which species is actually discharged in practice [2].

(b) In dilute sulfuric acid, the only cation present is H⁺, so hydrogen is discharged at the cathode [1]. In concentrated copper(II) sulfate, Cu²⁺ has a more positive E⦵ than H⁺, so copper is discharged instead of hydrogen [2].

(c) At reasonable concentration, Cl⁻, Br⁻ and I⁻ are discharged in preference to hydroxide; otherwise oxygen is produced from OH⁻ [1]. This rule does not extend to F⁻, which (like OH⁻ itself) is harder to oxidise than water, so fluoride solutions still produce oxygen at the anode; and, as in question 6(c), even Cl⁻ reverts to giving oxygen if the solution is dilute enough. In both solutions in this question the anion present is sulfate, which is never discharged in preference to hydroxide, so oxygen is produced at the anode in both cases [2].

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Where marks are usually lost

• Subtracting the wrong way round and reporting a negative e.m.f. for a spontaneous cell.
• Treating “feasible” as “will be observed”.
• Using the electron count from a half-equation instead of the overall equation in ΔG = −nFE.
• Forgetting to convert hours to seconds.
• Not dividing by the number of electrons when calculating mass deposited.
• Applying standard E⦵ values to a non-standard concentration without adjusting via the Nernst equation.
• Assuming the halide-vs-hydroxide anode rule also applies to sulfate, which is never discharged in preference to hydroxide.