The problems are on the electrolysis worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Electrolysis uses direct current to split a molten or dissolved ionic compound. These worked answers correct the 20 problems of the Chemistry 9-10 worksheet on electrolysis, from key words to spot the errors: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the method first and then the result in bold. Compare your half equations carefully, because a missing electron or a wrong sign loses marks. Where a calculation has several steps, check that each step uses the right ratio.
Level: Chemistry grades 9-10 (GCSE · IGCSE · O Level · Class 9-10) · Curriculum: GCSE C4
1. A molten or dissolved ionic compound that conducts because its ions move. 2. Also, the negative electrode, where positive ions gain electrons (reduction). 3. In addition, the positive electrode, where negative ions lose electrons (oxidation). Electrolyte = ions free to move; cathode negative; anode positive.
Problem 2 Why molten or dissolved?
1. In the solid the ions are held in fixed positions in the lattice, so no charge can flow. 2. When molten the ions are free to move, so positive ions go to the cathode and negative ions to the anode, carrying charge. Mobile ions are needed for current.
Problem 3 Molten potassium iodide
1. K⁺ (positive ions go to the negative cathode). 2. Cathode: potassium metal; anode: iodine. 3. 2I⁻ → I₂ + 2e⁻. Cathode: potassium; anode: iodine.
Problem 4 Reading a cell diagram
1. Electrode B. 2. Moreover, the long thin line of the battery symbol is the positive terminal and the short thick line is the negative terminal. B is joined to the short thick line, so B is negative, the cathode. 3. Positive ions move to B, to the right. B is the cathode.
Problem 5 Products from molten salts
1. Cathode zinc, anode chlorine. 2. Then cathode sodium, anode bromine. 3. Cathode magnesium, anode oxygen. Metal at the cathode, non-metal at the anode.
Problem 6 Complete the half equations
1. Na. 2. 2e⁻, giving 2Cl⁻ → Cl₂ + 2e⁻. 3. Cu. 4. Equations 1 and 3 gain electrons, so they are reductions at the cathode. Reductions: Na⁺ and Cu²⁺.
Problem 7 Three aqueous electrolytes
1. Copper is less reactive than hydrogen, so copper forms at the cathode; chloride is a halide, so chlorine forms at the anode. 2. Hydrogen at the cathode; no halide, so oxygen at the anode. 3. Sodium is more reactive than hydrogen, so hydrogen forms; nitrate is not a halide, so oxygen forms. 1: Cu and Cl₂; 2: H₂ and O₂; 3: H₂ and O₂.
Problem 8 Why hydrogen, not sodium?
1. Na⁺ and H⁺ (from water). 2. Sodium is more reactive than hydrogen, so H⁺ gains electrons more readily and hydrogen gas is released. 3. Na⁺ and OH⁻ remain, that is sodium hydroxide solution. Sodium hydroxide solution.
Problem 9 Gas volumes from acid
1. X is hydrogen (cathode, larger volume); Y is oxygen. 2. 2H⁺ + 2e⁻ → H₂. 3. Two electrons make one H₂, but four electrons are needed for one O₂ (4OH⁻ → O₂ + 2H₂O + 4e⁻). The same electrons therefore give two H₂ for every one O₂. X: hydrogen 18 cm³; Y: oxygen 9 cm³, ratio 2 : 1.
Problem 10 Matching products to uses
1. Hydrogen, chlorine and sodium hydroxide. 2. Chlorine: disinfectant and bleach; sodium hydroxide: soap making; hydrogen: fuel. 3. H⁺ ions are removed as hydrogen gas, leaving OH⁻ and Na⁺ in solution. H₂ fuel, Cl₂ disinfectant, NaOH soap.
Problem 11 Copper plating
1. Cu²⁺ + 2e⁻ → Cu. 2. Two electrons per atom: 0.20 ÷ 2 = 0.10 mol. 3. 0.10 × 64 = 6.4 g. 6.4 g of copper.
Problem 12 Refining copper
1. Anode: Cu → Cu²⁺ + 2e⁻. Cathode: Cu²⁺ + 2e⁻ → Cu. 2. 12.8 − 0.8 = 12.0 g. 3. 12.0 ÷ 12.8 × 100 = 93.75%. 12.0 g dissolves; the anode is 93.8% pure.
Problem 13 Silver-plated spoon
1. The cathode, so positive silver ions reach it. 2. Also, a bar of pure silver, which replaces the ions used. 3. Ag⁺ + e⁻ → Ag. Spoon = cathode; Ag⁺ + e⁻ → Ag.
1. It dissolves the oxide so the mixture melts near 950 °C, saving energy. 2. Oxygen formed at the anodes reacts with the hot carbon and burns it away. 3. C + O₂ → CO₂. C + O₂ → CO₂.
Problem 16 Current and mass
1. 0.40 g. 2. 1.0 A: 0.60 ÷ 30 = 0.020 g/min; 0.5 A: 0.30 ÷ 30 = 0.010 g/min. 3. 0.15 ÷ 0.010 = 15 min. 4. Current is the charge flowing per second, and each Cu²⁺ needs two electrons, so double the electrons per second deposits double the copper. 0.020 and 0.010 g/min; 15 min.
Problem 17 Masses in aluminium extraction
1. 54 000 ÷ 27 = 2000 mol. 2. Al₂O₃ gives 2Al, so 1000 mol Al₂O₃; M = 102, so 1000 × 102 = 102 000 g. 3. 2Al₂O₃ → 4Al + 3O₂ gives 1500 mol O₂ and each O₂ burns one C: 1500 × 12 = 18 000 g. 102 kg of Al₂O₃ and 18 kg of carbon.
Problem 18 Brine plant output
1. M = 58.5; 11 700 ÷ 58.5 = 200 mol. 2. 200 mol NaOH × 40 = 8000 g. 3. 100 mol Cl₂ × 24 = 2400 dm³. 8.0 kg NaOH and 2400 dm³ Cl₂.
Problem 19 Carbon or copper electrodes?
1. Oxygen from hydroxide ions. 2. The copper anode is oxidised and dissolves; the same amount of copper is deposited on the cathode, so the changes are equal. 3. Anode: Cu → Cu²⁺ + 2e⁻; cathode: Cu²⁺ + 2e⁻ → Cu. 4. With copper electrodes each Cu²⁺ removed is replaced, but with carbon the Cu²⁺ is used up. Copper dissolves at the anode and plates at the cathode.
Problem 20 Spot the errors
1. The anode is positive. 2. Hydrogen forms at the cathode because sodium is more reactive than hydrogen. 3. At the anode chloride ions lose electrons: 2Cl⁻ → Cl₂ + 2e⁻. 4. In a solid the ions cannot move, so the salt must be molten or dissolved. Four errors corrected.
