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S6 Chemistry: Electrolysis and Cells (Paper 2: intermediate)
S6 Chemistry: Electrolysis and Cells (Paper 2: intermediate)
Ten items on ionic conduction in solution, discharge order from electrode potentials, 0.395 g of copper from 2.50 A for 8 minutes, 1 930 s for 0.0100 mol of copper at 1.00 A, cathode products in a melt versus a solution, brine products with sodium hydroxide, impure copper as anode in refining, oxidation at the positive anode, an uneven deposit from too much current, and 0.0100 mol of silver from 0.0100 mol of electrons.
Direct answers Answers in order: ion migration, the higher potential discharges first, about 0.395 g, 1 930 s, sodium from the melt and hydrogen from the solution, chlorine and hydrogen with sodium hydroxide formed, impure copper anode and pure copper cathode, the positive anode still oxidises, a coarse deposit at high current, and 0.0100 mol of silver.
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)
- FoundationHow does charge travel inside the solution of an electrolytic cell?
- FoundationWhat mainly decides the discharge order of cations in aqueous solution?
- FoundationWhat mass of copper is deposited at the cathode when 2.50 A flows through copper(II) sulphate solution for 8.0 minutes?
- IntermediateHow long must 1.00 A flow to deposit 0.0100 mol of copper?
- IntermediateHow do the cathode products differ between molten sodium chloride and concentrated sodium chloride solution?
- IntermediateWhat are the products at the two electrodes when concentrated brine is electrolysed, and how does the electrolyte change?
- ChallengeIn the electrolytic refining of copper, where do the impure and pure copper go?
- ChallengeA student says the anode of an electrolytic cell is positive and therefore reduction happens there. What is wrong?
- ChallengeA student claims a larger current always gives a more even plated layer. Why is that wrong?
- Error redoHow many moles of silver are deposited when 0.0100 mol of electrons passes?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | By ion migration; electrons only move in the electrodes and the external circuit | Electrons cannot exist freely in aqueous solution, so conduction there is carried by ions moving to the electrodes. |
| 2 | The standard electrode potential; the more positive one discharges more readily at the cathode | Copper(II) is above hydrogen ions and hydrogen ions above zinc ions, so the order is copper, then hydrogen, then zinc. |
| 3 | About 0.395 g | The charge is 2.50 times 480 seconds, that is 1200 C; dividing by 96 500 and by 2 gives 0.00622 mol of copper, which is 0.395 g. |
| 4 | 1 930 s | 0.0100 mol of copper needs 0.0200 mol of electrons, that is 1 930 C, which takes 1 930 seconds at 1.00 A. |
| 5 | Molten gives sodium metal while the solution gives hydrogen gas | There is no water in the melt so sodium ions must be reduced, but in solution hydrogen ions are reduced more readily. |
| 6 | Chlorine at the anode, hydrogen at the cathode, and sodium hydroxide builds up in the solution | Chloride ions are oxidised to chlorine while water is reduced to hydrogen, leaving hydroxide ions behind. |
| 7 | Impure copper is the anode and pure copper the cathode | Copper at the anode dissolves into ions which are reduced again on the cathode, leaving impurities behind. |
| 8 | The anode is indeed positive, but oxidation, not reduction, always happens at the anode | The terminal only fixes the electron flow; the roles are fixed, with oxidation at the anode and reduction at the cathode. |
| 9 | Too large a current reduces ions too fast, giving a loose or burnt deposit instead of an even one | Plating depends on current density and agitation, and excess current only makes the crystals coarse. |
| 10 | 0.0100 mol of silver | A silver ion carries a single positive charge, so one mole of electrons reduces one mole of silver. |
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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