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S6 Chemistry: Chemical Cells and Electrolysis (Paper 3: challenge)
S6 Chemistry: Chemical Cells and Electrolysis (Paper 3: challenge)
Ten challenge items on the energy changes of the two cell types, oxidation at the anode, the definition of an electrolyte, copper from 19300 C, aluminium from molten oxide, the mole ratio and volume of hydrogen and oxygen, current efficiency, silver in a series cell, the charge to remove all copper from a 0.500 M solution, and a confusion between amperes and coulombs.
Direct answers Answers in order: electrical to chemical and chemical to electrical, oxidation, whether it conducts when dissolved or molten, 6.35 g, 2.70 g, 0.100 mol and 2.40 cubic decimetres, 89.8 percent, 4.32 g, 96500 C, and one ampere being one coulomb per second.
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: Chemical Cells and Electrolysis |
| Level | S6 |
02 Questions (10 items)
- FoundationWhat are the energy conversions in an electrolytic cell and in a galvanic cell?
- FoundationIs oxidation or reduction taking place at the anode during electrolysis?
- FoundationWhat distinguishes an electrolyte from a non-electrolyte?
- IntermediateCopper(II) sulphate solution is electrolysed with carbon electrodes and 19300 C of charge passes. What mass of copper deposits at the cathode (F = 96500, Cu = 63.5)?
- IntermediateDuring the electrolysis of molten aluminium oxide, 0.300 mol of electrons passes. What mass of aluminium forms at the cathode (Al = 27.0)?
- IntermediateIn the electrolysis of water, 0.0500 mol of oxygen is collected at the anode. How many moles and what volume of hydrogen form at the cathode at a molar volume of 24.0 cubic decimetres per mol?
- ChallengeAn electrolysis should give 1.27 g of copper but only 1.14 g is obtained. What is the current efficiency?
- ChallengeTwo cells are in series and carry the same current. Cell A deposits 0.0200 mol of copper and cell B contains silver nitrate solution. What mass of silver deposits in cell B (Ag = 108)?
- ChallengeHow much charge in coulombs is needed to deposit all the copper ions from 1.00 litre of 0.500 M copper(II) sulphate solution (F = 96500)?
- Error redoA student writes 1 A as 1 C. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The electrolytic cell converts electrical energy into chemical energy and the galvanic cell does the reverse | An electrolytic cell needs an external supply of electrical energy while a galvanic cell discharges on its own. |
| 2 | Oxidation | The anode is always where oxidation occurs, and in an electrolytic cell it is joined to the positive terminal. |
| 3 | Whether it conducts when dissolved in water or when molten | An electrolyte supplies free ions when dissolved or molten whereas a non-electrolyte does not. |
| 4 | 6.35 g | 19300 divided by 96500 gives 0.200 mol of electrons, so there is 0.100 mol of copper and that times 63.5 is 6.35 g. |
| 5 | 2.70 g | Each aluminium ion needs three electrons, so 0.300 divided by three gives 0.100 mol and that times 27.0 is 2.70 g. |
| 6 | 0.100 mol, a volume of 2.40 cubic decimetres | The mole ratio of hydrogen to oxygen is two to one, and 0.100 times 24.0 gives 2.40 cubic decimetres. |
| 7 | 89.8 percent | 1.14 divided by 1.27 times 100 gives 89.8 percent, and the shortfall comes from side reactions and current losses. |
| 8 | 4.32 g | Cells in series pass the same number of electrons, so 0.0200 mol of copper means 0.0400 mol of electrons and 0.0400 mol of silver, which times 108 is 4.32 g. |
| 9 | 96500 C | There is 0.500 mol of copper ions which needs 1.00 mol of electrons, and that times 96500 gives 96500 C. |
| 10 | One ampere is one coulomb per second, so current is a rate of charge flow | Charge is current times time, and treating amperes as coulombs drops the time factor from every calculation. |
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 first whether the cell is galvanic or electrolytic, name the products at each electrode, then work through charge, moles of electrons and atomic mass in order.Tutor Registration →
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