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S6 Chemistry: Electrochemical Cells and Fuel Cells (Paper 2: intermediate)
S6 Chemistry: Electrochemical Cells and Fuel Cells (Paper 2: intermediate)
Ten intermediate items on magnesium copper and iron silver cells, the metal oxidised, the meaning of a positive potential, the overall and anode equations of an alkaline fuel cell, energy efficiency, the silver zinc cell, moles of electrons and charge, and the sign of the cathode.
Direct answers Answers in order: 2.71 V, magnesium, 1.24 V, yes with a negative Gibbs energy change, two H2 plus O2 giving two H2O, H2 plus two OH minus giving two H2O and two electrons, 82.9 percent, 1.56 V, 1.00 mol and 96500 C, and the cathode is positive.
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 and Fuel Cells |
| Level | S6 |
02 Questions (10 items)
- FoundationWhat is the E-cell of a magnesium copper cell with Mg as minus 2.37 V and Cu as plus 0.34 V?
- FoundationWhich metal is oxidised in a magnesium copper cell?
- FoundationWhat is the E-cell of an iron silver cell with Fe as minus 0.44 V and Ag as plus 0.80 V?
- IntermediateIf a cell has a positive E-cell, is the reaction spontaneous?
- IntermediateWhat is the overall equation for an alkaline hydrogen oxygen fuel cell?
- IntermediateWhat is the anode equation in an alkaline hydrogen oxygen fuel cell?
- ChallengeBurning 1.00 mol of hydrogen releases 286 kJ, while the same amount in a fuel cell gives 237 kJ of electrical work. Find the efficiency.
- ChallengeWhat is the E-cell of a silver zinc cell with Ag as plus 0.80 V and Zn as minus 0.76 V?
- ChallengeWhen 0.500 mol of hydrogen is fully oxidised in a fuel cell, how many moles of electrons and how much charge move, with F as 96500 C per mol?
- Error redoA student says the cathode of a galvanic cell is the negative electrode. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 2.71 V | Magnesium is the anode, so 0.34 minus minus 2.37 gives 2.71 V. |
| 2 | Magnesium | Magnesium has the lowest electrode potential, so it loses electrons most readily. |
| 3 | 1.24 V | Iron is the anode, so 0.80 minus minus 0.44 gives 1.24 V. |
| 4 | Yes, a positive potential means the Gibbs free energy change is negative | The cell potential and the Gibbs free energy change carry opposite signs, so a positive value means spontaneous. |
| 5 | Two H2 plus O2 gives two H2O | Two hydrogen molecules combine with one oxygen molecule to give two water molecules while the electrons flow through the external circuit. |
| 6 | H2 plus two OH minus gives two H2O plus two electrons | In alkaline conditions hydrogen reacts with hydroxide ions and releases two electrons. |
| 7 | 82.9 percent | 237 divided by 286 and multiplied by 100 gives 82.9 percent. |
| 8 | 1.56 V | 0.80 minus minus 0.76 gives 1.56 V. |
| 9 | 1.00 mol of electrons and 96500 C | Each hydrogen gives two electrons, so 0.500 times two is 1.00 mol and that times 96500 gives 96500 C. |
| 10 | In a galvanic cell the cathode is the positive electrode where reduction happens | Anode and cathode are named after the reaction while positive and negative are named after the potential, so the two sets must not be mixed up. |
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.
Decide which electrode is oxidised first, then subtract the anode potential from the cathode potential, and only then do the charge and mass calculations.Tutor Registration →
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