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S6 Chemistry: Electrochemical Cells (Paper 3: challenge)
S6 Chemistry: Electrochemical Cells (Paper 3: challenge)
Ten items on standard conditions, why platinum is used, a negative EMF meaning no spontaneous discharge, −212.3 kJ, −88.8 kJ, −611.8 kJ, −424.6 kJ for n = 4, aluminium-copper at 2.00 V, ranking three cells with a 1.10 V gap, and a misconception linking a more negative ΔG° to a smaller EMF.
Direct answers Answers in order: 298 K, 1 bar and 1.00 M, an inert conductor is needed, no, −212.3 kJ, −88.8 kJ, −611.8 kJ, −424.6 kJ which is twice the value, 2.00 V and feasible, zinc-silver then zinc-copper then copper-silver with a 1.10 V gap, and a more negative ΔG° means a larger E°.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S6 Chemistry: Electrochemical Cells |
| Level | S6 |
02 Questions (10 items)
- FoundationUnder what conditions are standard electrode potentials measured?
- FoundationWhy does an Fe³⁺/Fe²⁺ half cell need a platinum electrode?
- FoundationA cell has a negative standard EMF. Can it discharge on its own?
- IntermediateFor a zinc-copper cell with n = 2 and E° = 1.10 V, what is the standard free energy change in kJ mol⁻¹?
- IntermediateFor a copper-silver cell with n = 2 and E° = 0.46 V, what is the standard free energy change in kJ mol⁻¹?
- IntermediateFor a magnesium-silver cell with n = 2 and E° = 3.17 V, what is the standard free energy change in kJ mol⁻¹?
- ChallengeA reaction has n = 4 and E° = 1.10 V. What is ΔG° in kJ mol⁻¹ and how does it compare with n = 2?
- ChallengeWhat is the standard EMF of an aluminium-copper cell (Al³⁺/Al at −1.66 V and Cu²⁺/Cu at +0.34 V) and is it feasible?
- ChallengeRank the zinc-silver cell at 1.56 V, the zinc-copper cell at 1.10 V and the copper-silver cell at 0.46 V, and give the gap between largest and smallest.
- Error redoA student says a more negative ΔG° means a smaller standard EMF. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 298 K, 1 bar gas pressure and 1.00 mol dm⁻³ metal ion solution | Fixing the conditions puts every half cell on the same scale so the values can be compared. |
| 2 | Both species are ions in solution, so an inert conductor is needed to carry electrons | Platinum does not take part in the reaction and only provides a conducting surface, so it does not alter the potential. |
| 3 | No, an external supply is needed | A negative value means the free energy rises in that direction, so the reaction runs only when current is forced through. |
| 4 | −212.3 kJ mol⁻¹ | ΔG° equals minus n F E°, that is minus 2 times 96500 times 1.10 which is −212300 J or −212.3 kJ. |
| 5 | −88.8 kJ mol⁻¹ | Minus 2 times 96500 times 0.46 gives −88780 J, which is −88.8 kJ. |
| 6 | −611.8 kJ mol⁻¹ | Minus 2 times 96500 times 3.17 gives −611810 J or −611.8 kJ, a large driving force from a large EMF. |
| 7 | −424.6 kJ mol⁻¹, exactly twice the value for n = 2 | ΔG° is proportional to the moles of electrons, so doubling n doubles the free energy change while the EMF stays the same. |
| 8 | 2.00 V and feasible | 0.34 minus (−1.66) gives 2.00 V and a positive value means the free energy falls on discharge. |
| 9 | Zinc-silver 1.56 V, then zinc-copper 1.10 V, then copper-silver 0.46 V, a gap of 1.10 V | 1.56 minus 0.46 gives 1.10 V, exactly the EMF of the zinc-copper cell. |
| 10 | A more negative ΔG° means a larger E°, so the two move in opposite directions | From ΔG° = −nFE° a more negative free energy gives a larger EMF. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Compare the standard electrode potentials first, name the cathode and anode, subtract anode from cathode to get the EMF, then use the free energy change to judge feasibility.Tutor Registration →
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