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S6 Chemistry: Electrochemical Cells (Paper 2: intermediate)
S6 Chemistry: Electrochemical Cells (Paper 2: intermediate)
Ten items on the anode joining the negative end, the zinc half equation, the cell diagram convention, zinc-silver at 1.56 V, iron-copper at 0.78 V, nickel-silver at 1.05 V, series total of 1.56 V, doubling coefficients still 1.10 V, copper-chlorine at 1.02 V, and a misconception that a larger area raises the EMF.
Direct answers Answers in order: anode to the negative end, Zn → Zn²⁺ + 2e⁻, anode left and cathode right with single lines for boundaries and a double line for the bridge, 1.56 V, 0.78 V, 1.05 V, 1.56 V the same as zinc-silver, still 1.10 V, 1.02 V with chlorine at the anode, and no dependence on area.
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)
- FoundationDuring discharge, is oxidation at the anode or the cathode, and which end of the external circuit does it connect to?
- FoundationWrite the half equation for the zinc half cell during discharge.
- FoundationWhat is the convention for writing a cell diagram?
- IntermediateWhat is the standard EMF of a zinc-silver cell (Zn²⁺/Zn at −0.76 V and Ag⁺/Ag at +0.80 V)?
- IntermediateWhat is the standard EMF of an iron-copper cell (Fe²⁺/Fe at −0.44 V and Cu²⁺/Cu at +0.34 V)?
- IntermediateWhat is the standard EMF of a nickel-silver cell (Ni²⁺/Ni at −0.25 V and Ag⁺/Ag at +0.80 V)?
- ChallengeA zinc-copper cell gives 1.10 V and a copper-silver cell 0.46 V. What is the total EMF in series, compared with a zinc-silver cell?
- ChallengeIf all the coefficients in the zinc-copper reaction are doubled, what happens to the standard EMF?
- ChallengeWhat is the standard EMF of a copper-chlorine cell (Cu²⁺/Cu at +0.34 V and Cl₂/Cl⁻ at +1.36 V) and where is chlorine released?
- Error redoA student says enlarging the electrodes raises the standard EMF. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The anode, connected to the negative end | The anode loses electrons in oxidation, so it is the negative electrode where electrons leave the cell. |
| 2 | Zn → Zn²⁺ + 2e⁻ | Zinc loses two electrons and is oxidised, keeping charge balanced with the zinc ion. |
| 3 | The anode is on the left, the cathode on the right, a single line for a phase boundary and a double line for the salt bridge | The fixed convention of anode on the left and cathode on the right prevents misreading the electron flow. |
| 4 | 1.56 V | 0.80 minus (−0.76) gives 1.56 V, with zinc as the anode and silver as the cathode. |
| 5 | 0.78 V | 0.34 minus (−0.44) gives 0.78 V, with iron oxidised and copper ions reduced. |
| 6 | 1.05 V | 0.80 minus (−0.25) gives 1.05 V because the potential difference fixes the EMF. |
| 7 | 1.56 V, the same as the zinc-silver cell | In series the EMFs add to 1.10 plus 0.46 which is 1.56 V, exactly the same as combining zinc and silver directly. |
| 8 | It stays at 1.10 V | Standard electrode potentials are intensive, so doubling the coefficients changes charge and free energy but not the EMF. |
| 9 | 1.02 V and chlorine forms at the anode, the copper side | Chlorine has the higher potential so it is the cathode side; copper is oxidised while chloride ions are oxidised to chlorine at the anode. |
| 10 | The standard EMF does not depend on electrode area, which only affects the current available | The EMF depends on the nature of the two half cells while area changes only the rate of discharge. |
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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