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S6 Chemistry: Electrochemical Cells and Fuel Cells (Paper 3: challenge)
S6 Chemistry: Electrochemical Cells and Fuel Cells (Paper 3: challenge)
Ten challenge items on coefficients not affecting the potential, spontaneity of the iron tin reaction, the Gibbs energy calculation, charge from hydrogen, discharge time at 10 A, mass of water at 1 A, three cells in series, concentration and the Nernst equation, salt bridge against a short circuit, and the sign change on reversal.
Direct answers Answers in order: 1.56 V, plus 0.30 V, minus 88.8 kJ per mol, 96500 C, 1930 s, 0.336 g, 3.30 V, the standard value holds only at standard conditions, a short circuit, and it should be minus 1.56 V.
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 the reaction two silver ions plus zinc giving two silver plus zinc ion?
- FoundationWhat is the E-cell of iron plus tin ion giving iron ion plus tin, with Sn as minus 0.14 V and Fe as minus 0.44 V?
- FoundationFor a copper silver cell with a potential of 0.46 V and two moles of electrons, find the standard Gibbs free energy change, with F as 96500 C per mol.
- IntermediateHow much charge can 0.50 mol of hydrogen deliver in a fuel cell, with F as 96500 C per mol?
- IntermediateA fuel cell discharges steadily at 10.0 A. How many seconds does 0.100 mol of hydrogen last?
- IntermediateA fuel cell runs at 1.00 A for one hour. What mass of water forms in theory, with F as 96500 C per mol?
- ChallengeThree identical zinc copper cells are joined in series. What is the total voltage?
- ChallengeWhy does changing the ion concentration make the measured voltage differ from E-cell?
- ChallengeWhat happens if the salt bridge is removed and a copper wire joins the two solutions instead?
- Error redoA student reverses a cell reaction but still writes E-cell as plus 1.56 V. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | 1.56 V | The potential is an intensive quantity and does not depend on the coefficients, so it stays 0.80 minus minus 0.76. |
| 2 | Plus 0.30 V, so the reaction is spontaneous | Minus 0.14 minus minus 0.44 gives plus 0.30 V, and a positive value means it proceeds spontaneously. |
| 3 | Minus 88.8 kJ per mol | The change is minus n F E, that is minus two times 96500 times 0.46, giving minus 88780 J or minus 88.8 kJ per mol. |
| 4 | 96500 C | Half a mole of hydrogen releases one mole of electrons, so the charge is 96500 C. |
| 5 | 1930 s, about 32 minutes | That is 0.200 mol of electrons or 19300 C, and dividing by 10.0 A gives 1930 s. |
| 6 | 0.336 g | One hour is 3600 C or 0.0373 mol of electrons, which is 0.0187 mol of water, and that times 18.0 gives 0.336 g. |
| 7 | 3.30 V | In series the voltages add, so three times 1.10 gives 3.30 V and the current capacity stays the same. |
| 8 | The standard value holds only at standard concentration, and the actual potential shifts with the Nernst equation | The standard potential is a fixed reference, while the measured voltage depends on the concentration ratio. |
| 9 | Electrons flow straight through the wire, the cell short circuits and no useful voltage appears | A salt bridge passes ions but not electrons, and a wire does exactly the opposite. |
| 10 | Reversing the reaction reverses the sign, so it should be minus 1.56 V and the reverse is not spontaneous | The Gibbs energy changes sign on reversal, so from that relation the potential must change sign as well. |
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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