These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: F = 96,485 C mol⁻¹, and these standard reduction potentials at 298 K.
| Half-reaction | E° (V) |
|---|---|
| Br₂(l) + 2e⁻ → 2Br⁻(aq) | +1.07 |
| Ag⁺(aq) + e⁻ → Ag(s) | +0.80 |
| Fe³⁺(aq) + e⁻ → Fe²⁺(aq) | +0.77 |
| I₂(s) + 2e⁻ → 2I⁻(aq) | +0.54 |
| Cu²⁺(aq) + 2e⁻ → Cu(s) | +0.34 |
| 2H⁺(aq) + 2e⁻ → H₂(g) | 0.00 |
| Fe²⁺(aq) + 2e⁻ → Fe(s) | −0.44 |
| Cr³⁺(aq) + 3e⁻ → Cr(s) | −0.74 |
| Zn²⁺(aq) + 2e⁻ → Zn(s) | −0.76 |
| Al³⁺(aq) + 3e⁻ → Al(s) | −1.66 |
| Mg²⁺(aq) + 2e⁻ → Mg(s) | −2.37 |
Question 1 (multiple choice · foundation)
A galvanic cell runs the reaction 2Cr(s) + 3Cu²⁺(aq) → 2Cr³⁺(aq) + 3Cu(s) under standard conditions. What is E°cell?
- (A) −1.08 V
- (B) −0.40 V
- (C) +1.08 V
- (D) +2.50 V
Answer and explanation
Answer: (C). Cu²⁺ is reduced (cathode) and Cr is oxidised (anode). E°cell = E°(cathode) − E°(anode) = (+0.34) − (−0.74) = +1.08 V.
- (A) subtracts the wrong way round (anode − cathode). A cell that runs by itself must have a positive E°cell.
- (B) adds the two reduction potentials, 0.34 + (−0.74), without reversing the anode half-reaction.
- (D) multiplies by the coefficients: 3(0.34) − 2(−0.74) = 2.50 V. E° values are never multiplied.
Question 2 (multiple choice · core)
For the reaction 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(s), what is ΔG° at 298 K?
- (A) −44.4 kJ mol⁻¹
- (B) −22.2 kJ mol⁻¹
- (C) +44.4 kJ mol⁻¹
- (D) −4.44 × 10⁴ kJ mol⁻¹
Answer and explanation
Answer: (A). Fe³⁺ is reduced and I⁻ is oxidised: E° = 0.77 − 0.54 = +0.23 V. Two electrons transfer in the equation as written (2 Fe³⁺ each gain 1; 2 I⁻ lose 1 each), so n = 2. ΔG° = −(2)(96,485 C mol⁻¹)(0.23 V) = −44,400 J mol⁻¹ = −44.4 kJ mol⁻¹.
- (B) uses n = 1, the electrons in a single Fe³⁺ half-reaction, not in the balanced equation.
- (C) has the wrong sign. A positive E° must give a negative ΔG°.
- (D) is the answer in joules labelled as kilojoules; it is 1000 times too large.
Question 3 (multiple choice · core)
Which reaction is thermodynamically unfavoured under standard conditions, so that it would need an externally applied potential to proceed?
- (A) Mg(s) + Fe²⁺(aq) → Mg²⁺(aq) + Fe(s)
- (B) Br₂(l) + 2I⁻(aq) → 2Br⁻(aq) + I₂(s)
- (C) Cu(s) + 2H⁺(aq) → Cu²⁺(aq) + H₂(g)
- (D) Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)
Answer and explanation
Answer: (C). For each, E° = E°(species reduced) − E°(species oxidised):
| Reaction | E° (V) |
|---|---|
| (A) | −0.44 − (−2.37) = +1.93 |
| (B) | +1.07 − (+0.54) = +0.53 |
| (C) | 0.00 − (+0.34) = −0.34 |
| (D) | 0.00 − (−0.76) = +0.76 |
Only (C) has a negative E° (so a positive ΔG°). Copper is above hydrogen on the ladder, so H⁺ cannot oxidise it under standard conditions.
- (A), (B) and (D) all have positive E° values, so they are favoured. In each, the reducing agent (Mg, I⁻, Zn) belongs to a half-reaction lower on the table than the species it reduces.
Question 4 (calculation · core)
A galvanic cell has a zinc electrode in 1 M Zn(NO₃)₂ and a platinum electrode in a solution that is 1 M in both Fe³⁺ and Fe²⁺. The cell runs under standard conditions.
(a) Identify the anode and write the balanced overall equation. (b) Calculate E°cell. (c) Calculate ΔG° for the equation in (a), in kJ mol⁻¹.
Worked solution
(a) Fe³⁺/Fe²⁺ (+0.77 V) is higher than Zn²⁺/Zn (−0.76 V), so Fe³⁺ is reduced at the platinum electrode and Zn is oxidised: zinc is the anode. Zn(s) → Zn²⁺(aq) + 2e⁻; Fe³⁺(aq) + e⁻ → Fe²⁺(aq) (× 2). Overall: Zn(s) + 2Fe³⁺(aq) → Zn²⁺(aq) + 2Fe²⁺(aq), n = 2.
(b) E°cell = (+0.77) − (−0.76) = +1.53 V.
(c) ΔG° = −(2)(96,485)(1.53) = −295,000 J mol⁻¹ = −295 kJ mol⁻¹.
Suggested mark points (4): 1 for zinc as the anode with a reason; 1 for the balanced equation with n = 2; 1 for +1.53 V (no coefficient multiplied); 1 for −295 kJ mol⁻¹ with the correct sign and unit.
Common error: using n = 1 because the iron half-reaction has one electron gives −148 kJ mol⁻¹.
Question 5 (constructed response · core)
A fictional metal Z forms Z²⁺ ions. A galvanic cell is built from Z in 1 M Z(NO₃)₂ and copper in 1 M Cu(NO₃)₂. The voltmeter reads +0.46 V, and the copper electrode gains mass.
(a) Calculate the standard reduction potential of Z²⁺/Z. (b) Calculate ΔG° for Z(s) + Cu²⁺(aq) → Z²⁺(aq) + Cu(s). (c) Predict, with a calculation, whether Z reacts with 1 M Fe²⁺ under standard conditions.
Worked solution and suggested Marlbridge rubric
(a) Copper gains mass, so Cu²⁺ is reduced: copper is the cathode and Z is the anode. 0.46 = E°(Cu²⁺/Cu) − E°(Z²⁺/Z) = 0.34 − E°(Z²⁺/Z), so E°(Z²⁺/Z) = −0.12 V.
(b) n = 2. ΔG° = −(2)(96,485)(0.46) = −88,800 J mol⁻¹ = −88.8 kJ mol⁻¹.
(c) For Z(s) + Fe²⁺(aq) → Z²⁺(aq) + Fe(s): E° = E°(Fe²⁺/Fe) − E°(Z²⁺/Z) = −0.44 − (−0.12) = −0.32 V. E° is negative (ΔG° positive), so the reaction is not thermodynamically favoured: Z does not displace iron under standard conditions.
| Point | What earns it |
|---|---|
| 1 | Copper identified as the cathode (from the mass gain) and Z as the anode |
| 1 | E°(Z²⁺/Z) = −0.12 V |
| 1 | ΔG° = −88.8 kJ mol⁻¹ with n = 2 |
| 1 | (c) E° = −0.32 V and the conclusion “not favoured”, linked to the negative sign |
Accept a value from (a) carried forward into (c).
Question 6 (constructed response · core)
For 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s), a student calculates E° and ΔG°. The student then writes the equation as 4Al(s) + 6Cu²⁺(aq) → 4Al³⁺(aq) + 6Cu(s) and claims: “Doubling the equation doubles both E° and ΔG°.”
(a) Calculate E° and ΔG° for the first equation. (b) Calculate E° and ΔG° for the doubled equation. (c) Evaluate the student’s claim, explaining what each quantity measures.
Worked solution and suggested Marlbridge rubric
(a) E° = 0.34 − (−1.66) = +2.00 V. Six electrons transfer (2 Al × 3e⁻ = 3 Cu²⁺ × 2e⁻), so n = 6. ΔG° = −(6)(96,485)(2.00) = −1,158,000 J mol⁻¹ = −1158 kJ mol⁻¹ (−1.16 × 10³ kJ mol⁻¹).
(b) E° is still +2.00 V. Now n = 12, so ΔG° = −(12)(96,485)(2.00) = −2316 kJ mol⁻¹ (−2.32 × 10³ kJ mol⁻¹).
(c) The claim is half right. ΔG° doubles, because it is the free energy change for a stated amount of reaction, and twice as much reaction transfers twice as many electrons (n doubles). E° does not change, because it is energy per unit of charge (1 V = 1 J C⁻¹): doubling the reaction doubles the energy and the charge together, so their ratio stays the same.
| Point | What earns it |
|---|---|
| 1 | (a) E° = +2.00 V and n = 6 |
| 1 | (a) ΔG° = −1158 kJ mol⁻¹ |
| 1 | (b) E° unchanged and ΔG° = −2316 kJ mol⁻¹ (n = 12) |
| 1 | (c) ΔG° doubles because it depends on the amount of reaction (n) |
| 1 | (c) E° does not change because it is energy per unit charge (intensive) |
Question 7 (explanation · stretch)
A rechargeable cell uses a copper electrode in Cu²⁺(aq) and an inert graphite electrode in a solution containing Br₂ and Br⁻. When it supplies power, the reaction is Cu(s) + Br₂(l) → Cu²⁺(aq) + 2Br⁻(aq).
(a) Calculate E° and ΔG° for the discharge reaction, and state whether it is thermodynamically favoured. (b) To recharge the cell, a power supply drives the reverse reaction. Give E° and ΔG° for the recharging reaction, and identify the anode during recharging with its half-reaction. (c) Explain why the power supply must apply more than 0.73 V (under standard conditions) to recharge the cell.
Model answer and suggested Marlbridge rubric
(a) Br₂ is reduced, Cu is oxidised: E° = 1.07 − 0.34 = +0.73 V. n = 2, so ΔG° = −(2)(96,485)(0.73) = −141 kJ mol⁻¹. E° > 0 and ΔG° < 0, so discharge is thermodynamically favoured (galvanic).
(b) Recharging is the reverse reaction, so the signs reverse: E° = −0.73 V, ΔG° = +141 kJ mol⁻¹ (unfavoured, electrolytic). Now Br⁻ is oxidised, so the graphite electrode is the anode: 2Br⁻(aq) → Br₂(l) + 2e⁻. (Cu²⁺ is reduced at the copper electrode, the cathode.)
(c) The recharging reaction has a negative E°, so it will not run by itself. The power supply must push electrons “uphill” against the cell’s own potential of 0.73 V. Only an applied potential larger than 0.73 V supplies at least the +141 kJ of free energy per mole of reaction (ΔG° = −nFE°) that the unfavoured reaction needs; a smaller applied potential would be overcome by the cell, which would keep discharging.
| Point | What earns it |
|---|---|
| 1 | (a) E° = +0.73 V and ΔG° = −141 kJ mol⁻¹ |
| 1 | (a) “Favoured”, linked to the sign of E° or ΔG° |
| 1 | (b) E° = −0.73 V and ΔG° = +141 kJ mol⁻¹ for the reverse reaction |
| 1 | (b) Graphite is the anode during recharging, with 2Br⁻ → Br₂ + 2e⁻ |
| 1 | (c) Applied potential must exceed the cell potential to supply the free energy for the unfavoured reaction |
How did you do?
- Q1 or Q3 wrong: re-read “Calculating a standard cell potential” and Figure 1 in the study guide.
- Q2 or Q4(c) wrong: check n (electrons that cancel in the overall equation) and the J-to-kJ step (Worked example 1).
- Q5 wrong: practise working backwards from a measured value (Worked example 3).
- Q6 or Q7 incomplete: your reasoning needs the why: E° is per unit charge, and the sign of E° sets the sign of ΔG°.
Then tick off the topic checklist.