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S6 Chemistry: Electrolysis and Cells (Paper 4: misconception drill)
S6 Chemistry: Electrolysis and Cells (Paper 4: misconception drill)
Ten drill items on the cathode on the negative terminal, oxidation at the anode, 1 800 C or 0.0187 mol of electrons from 3.00 A for 10 minutes, the 2 : 1 gas ratio, 965 C for 0.0100 mol of electrons, 15.9 g of copper from 0.500 mol of electrons, sodium from molten sodium chloride, chlorine from brine, opposite directions for cells and electrolysis, and minutes wrongly put into Q = I t.
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)
- FoundationWhich terminal does the cathode of an electrolytic cell connect to?
- FoundationWhat type of reaction happens at the anode of an electrolytic cell?
- FoundationHow many moles of electrons pass when 3.00 A flows for 10.0 minutes?
- IntermediateWhat is the volume ratio of the gases at the cathode and anode in water electrolysis?
- IntermediateWhat charge corresponds to 0.0100 mol of electrons?
- IntermediateWhat mass of copper can 0.500 mol of electrons deposit at the cathode?
- ChallengeWhat is obtained at the cathode when molten sodium chloride is electrolysed?
- ChallengeWhich gas is obtained at the anode when concentrated brine is electrolysed?
- ChallengeA student says a cell and an electrolytic cell run the same reaction, only wired differently. What is wrong?
- Error redoA student substitutes 10 minutes straight into Q = I t and gets 30 C for 3.00 A. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The negative terminal | The cathode receives electrons for reduction, so it connects to the negative terminal. |
| 2 | Oxidation | The anode always oxidises and the cathode always reduces, just as in a cell. |
| 3 | About 0.0187 mol | The charge is 3.00 times 600 seconds, that is 1 800 C, divided by 96 500 gives 0.01865 mol. |
| 4 | 2 : 1, hydrogen to oxygen | The half equations show 1 mol of hydrogen and 0.5 mol of oxygen per two moles of electrons. |
| 5 | 965 C | 0.0100 times 96 500 gives 965 C, a direct use of the Faraday constant. |
| 6 | About 15.9 g | Copper takes two electrons, so 0.500 mol of electrons gives 0.250 mol of copper, which is 15.875 g. |
| 7 | Sodium metal | Only sodium ions can be reduced in the melt, so sodium forms rather than hydrogen. |
| 8 | Chlorine | Chloride ions are oxidised to chlorine in concentrated brine, whereas dilute solutions mainly give oxygen. |
| 9 | The reactions run in opposite directions; a cell discharges spontaneously while electrolysis reverses it with a supply | Electrolysis drives the cell reaction backwards, which is why the two are reverse processes. |
| 10 | Time must be in seconds, so 10 minutes is 600 s and 3.00 A gives 1 800 C | The t in Q = I t must be in seconds, so using minutes makes the answer sixty times too small. |
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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