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S5 Chemistry Electrolysis, Paper 3: Faraday calculations and industrial cells
S5 Chemistry Electrolysis, Paper 3: Faraday calculations and industrial cells
Ten graded items on electrode products, quantitative Faraday masses and the industrial extraction of aluminium.
Focus Start from Q equals I t, divide by 96 500 for moles of electrons, then scale by the charge on the metal ion.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S5 Chemistry: Electrolysis |
| Level | Four basic, three advanced, two challenge and one error-repair item |
02 Questions (10 items)
- FoundationWhat forms at each electrode when molten sodium chloride is electrolysed?
- FoundationHow many moles of copper are deposited by 2 mol of electrons from copper sulphate solution?
- FoundationIs the anode of an electrolytic cell the site of oxidation or reduction?
- FoundationIn nickel plating of an iron object, which electrode holds the object and what is the anode made of?
- IntermediateCopper sulphate solution is electrolysed at 2.0 A for 1 930 s. Find the mass of copper deposited, with Cu as 63.5.
- IntermediateThe same 0.040 mol of electrons is used to electrolyse dilute sulphuric acid. Find the volume of hydrogen at room temperature, taking 24 dm3 per mole.
- IntermediateWhy is cryolite added to alumina in the extraction of aluminium?
- Challenge5.0 A is passed for 3 860 s through silver nitrate and then copper sulphate solution. Find the mass deposited in each case, with Ag as 107.9 and Cu as 63.5.
- ChallengeMolten alumina is electrolysed at 10.0 A for 9 650 s. Find the mass of aluminium produced, with Al as 27.0.
- Error redoA student says the anode is positive in both a primary cell and an electrolytic cell. Give the correct statement.
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Sodium at the cathode, chlorine at the anode | Sodium ions are reduced to the metal at the cathode and chloride ions are oxidised to chlorine at the anode. |
| 2 | 1 mol | Each copper ion needs two electrons, so the amount of copper is half the amount of electrons. |
| 3 | Oxidation, because anions lose electrons there | The anode is connected to the positive terminal, so electrons are removed and anions or the electrode itself are oxidised. |
| 4 | The iron is the cathode and the anode is pure nickel | Nickel ions are reduced onto the iron at the cathode while the nickel anode dissolves to replace them. |
| 5 | About 1.27 g | 3 860 coulombs is 0.040 mol of electrons, so 0.020 mol of copper forms and weighs about 1.27 grams. |
| 6 | 0.48 dm3 | One mole of hydrogen needs two moles of electrons, so 0.020 mol of gas forms, which is 0.48 dm3. |
| 7 | It lowers the melting point so the cell can run near 950 degrees and save energy | Cryolite lowers the melting point to about 950 degrees, so the cell runs far below the 2 045 degrees needed for pure alumina. |
| 8 | About 21.58 g of silver and 6.35 g of copper | 19 300 coulombs is 0.20 mol of electrons, giving 0.20 mol of silver but only 0.10 mol of copper. |
| 9 | 9.0 g | 96 500 coulombs is 1.00 mol of electrons, so a third of a mole of aluminium forms, which is 9.0 grams. |
| 10 | The anode is positive in an electrolytic cell but negative in a primary cell | In an electrolytic cell the anode is the positive electrode, but in a primary cell the anode is the negative one where electrons leave. |
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.
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