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S6 Chemistry: Electrolysis and Cells (Paper 3: challenge)
S6 Chemistry: Electrolysis and Cells (Paper 3: challenge)
Ten items on oxidation at both anodes, about 9.57 × 10⁴ C for 1.00 g of hydrogen, a 1 : 2 copper to silver mole ratio in series, 20.6 μm from 0.0100 mol of silver on 50.0 cm², about 11.8 g of copper from 5.00 A for two hours, copper before zinc, inert electrodes, copper mass at 0.294 of silver, equal current in series, and electrons not moving through the solution.
Direct answers Answers in order: both anodes oxidise, about 9.57 × 10⁴ C, 1 : 2, about 20.6 μm, about 11.8 g, copper first because of its higher potential, inert electrodes avoid side reactions, about 0.294 times, current is equal in series, and electrons stay in the external circuit.
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: Electrolysis and Cells |
| Level | S6 |
02 Questions (10 items)
- FoundationIn terms of redox, what do the anode of an electrolytic cell and the anode of a cell have in common?
- FoundationWhat charge is theoretically needed to obtain 1.00 g of hydrogen by electrolysing water?
- FoundationTwo electrolytic cells are in series. For the same charge, what is the mole ratio of copper to silver deposited?
- IntermediateIf 0.0100 mol of silver is plated onto an object of area 50.0 cm² (silver density 10.5 g cm⁻³), what is the average thickness?
- IntermediateWhat mass of copper can be obtained at the cathode by electrolysing copper(II) sulphate with 5.00 A for 2.0 hours?
- IntermediateA solution contains both copper(II) and zinc(II) ions. Which metal is deposited first at the cathode, and why?
- ChallengeWhy are inert electrodes such as platinum or carbon preferred in an electrolytic cell?
- ChallengeFor the same charge, the mass of copper deposited is roughly what fraction of the mass of silver deposited?
- ChallengeA student says the second of two cells in series carries a smaller current. What is wrong?
- Error redoA student says electrons swim through the solution from anode to cathode during electrolysis. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Oxidation happens at the anode in both cases | The role is the same; only the polarity differs, positive in electrolysis and negative in a cell. |
| 2 | About 9.57 × 10⁴ C | 1.00 g of hydrogen is 0.496 mol, needing 0.992 mol of electrons, which is about 95 700 C. |
| 3 | Copper to silver is 1 : 2 | Copper needs two electrons and silver one, so twice as many moles of silver are deposited for the same charge. |
| 4 | About 20.6 μm | 1.079 g of silver has a volume of 0.1028 cm³, and dividing by 50.0 cm² gives 0.00206 cm, that is 20.6 μm. |
| 5 | About 11.8 g | The charge is 5.00 times 7 200 seconds, that is 36 000 C or 0.373 mol of electrons, giving 0.187 mol of copper, about 11.84 g. |
| 6 | Copper, because its standard electrode potential is higher | Copper at +0.34 V is above zinc at −0.76 V, so it is reduced more readily and discharges first. |
| 7 | So the electrode itself does not take part in the redox reaction and confuse the product analysis | A reactive electrode dissolves or reacts, for example a copper anode dissolves so that oxygen is no longer the anode product. |
| 8 | About 0.294 times | Copper gives 31.75 g per mole of electrons against 107.9 g for silver, and 31.75 divided by 107.9 is about 0.294. |
| 9 | Current is the same everywhere in a series circuit, so both cells pass the same current and charge | A series circuit has one path, so current is not used up and any difference in mass comes only from electrons per ion and molar mass. |
| 10 | Electrons travel only in the external circuit and electrodes; the solution conducts by ion migration | There are no free electrons in the solution, which conducts by cations and anions moving in opposite directions. |
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 the electrodes and the ion discharge order first, then use Q = It, divide by the Faraday constant 96 500 C per mole to get moles of electrons, and finally convert with the half equation to mass or moles of gas.Tutor Registration →
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