IGCSE Chemistry 0620 Paper 4: Electrolysis, Metal Extraction and Rates of Reaction

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 75 minutes | Total: 100 marks | Parts: 30

Paper at a glance

Paper	Time	Marks	Calculator
Paper 4 (Extended, structured)	1 h 15 min	75	Allowed

Questions

Question 1 (16 marks) – Principles of electrolysis and electrode products

(a) Explain what electrolysis is and state two conditions that must be met for electrolysis to occur. (3 marks)

(b) Write the electrode equations for the electrolysis of molten sodium chloride at the cathode and at the anode. (4 marks)

(c) Solid sodium chloride does not conduct electricity. Explain why. (3 marks)

(d) When concentrated sodium chloride solution is electrolysed, chlorine rather than oxygen forms at the anode. Explain why. (3 marks)

(e) In the same electrolysis hydrogen, not sodium, forms at the cathode. Explain why. (3 marks)

Question 2 (17 marks) – Electroplating and charge calculations

(a) During copper plating a current of 2.00 A passes for 30.0 minutes. Calculate the charge passed, in coulombs. (3 marks)

(b) From the previous part, calculate the theoretical mass of copper deposited at the cathode (Cu = 63.5, one mole of electrons = 96500 C). (4 marks)

(c) The same charge is used to plate silver (Ag+ + e- -> Ag, Ag = 107.9). Calculate the mass of silver deposited and compare it with the mass of copper. (4 marks)

(d) The mass of copper obtained is usually slightly less than the calculated value. Suggest two reasons. (3 marks)

(e) When a piece of iron is electroplated, should it be connected to the positive or the negative terminal? Explain. (3 marks)

Question 3 (17 marks) – Extraction of aluminium

(a) Explain why aluminium cannot be extracted by reducing its oxide with carbon, as iron is. (3 marks)

(b) Before electrolysis, bauxite is treated with sodium hydroxide solution. State the purpose of this step. (3 marks)

(c) Write the electrode equations for the electrolysis of molten aluminium oxide at the cathode and at the anode. (4 marks)

(d) What is the purpose of adding cryolite to the electrolysis cell? Explain its effect on energy costs. (4 marks)

(e) The carbon anodes are gradually used up during electrolysis. Explain why. (3 marks)

Question 4 (16 marks) – Collision theory and factors affecting rate

(a) Define the rate of a chemical reaction and state two ways of measuring it. (3 marks)

(b) Use collision theory to explain why increasing the concentration of a reactant increases the rate. (3 marks)

(c) How does a catalyst increase the rate of a reaction? State what happens to its mass and chemical properties. (4 marks)

(d) Replacing a lump of calcium carbonate with the same mass of powder makes the reaction with hydrochloric acid much faster. Explain why. (3 marks)

(e) A rise of 10 C usually roughly doubles the rate of a reaction. Explain this using collision theory. (3 marks)

Question 5 (17 marks) – Calculating average rate from experimental data

(a) Marble chips react with 1.0 mol/dm3 hydrochloric acid. The volume of carbon dioxide collected is 0 cm3 at 0 s, 45 cm3 at 30 s, 78 cm3 at 60 s, 96 cm3 at 90 s, 104 cm3 at 120 s and 108 cm3 at 150 s. Calculate the average rate over the first 30 s. (3 marks)

(b) Calculate the average rate between 30 s and 60 s and explain why it is slower than over the first 30 s. (4 marks)

(c) Explain why the curve becomes flat after about 150 s and state the approximate final gas volume. (3 marks)

(d) If 2.0 mol/dm3 acid were used with the same mass of calcium carbonate, state how the initial slope and the final volume would change. (4 marks)

(e) If the same mass of calcium carbonate were ground to powder, how would the curve differ? (3 marks)

Question 6 (17 marks) – Energy and environmental issues in industrial electrolysis

(a) Explain why extracting a metal by electrolysis is expensive and state one siting consideration that reduces the cost. (3 marks)

(b) An aluminium smelter runs at 15000 A for 24 hours without a break. Calculate the charge passed. (4 marks)

(c) From the previous part, calculate the theoretical mass of aluminium, in kg, that the plant could produce in 24 hours (one mole of electrons = 96500 C, Al = 27). (4 marks)

(d) What are the environmental benefits of recycling aluminium compared with extracting it from bauxite? (3 marks)

(e) In the chlor-alkali industry, state one safety or environmental risk linked to chlorine and one control measure. (3 marks)

Answers and marking notes

Question 1 – Principles of electrolysis and electrode products

(a) Electrolysis is the breakdown of an ionic compound, when molten or in aqueous solution, by the passage of direct current (one mark); freely moving ions must be present (one mark); a direct current supply must provide the energy (one mark). [3 marks]

(b) Cathode: Na+ + e- -> Na (two marks, one for the reactant and one for the product); anode: 2Cl- -> Cl2 + 2e- (two marks, one for the reactant and one for the product). [4 marks]

(c) In the solid the ions are held in a fixed lattice (one mark); the ions can only vibrate and cannot move from place to place (one mark); without mobile charged particles the solid cannot conduct (one mark). [3 marks]

(d) The concentration of chloride ions in the solution is much higher than that of hydroxide ions (one mark); chloride ions lose electrons more readily and are oxidised (one mark); so chlorine is the main product at the anode (one mark). [3 marks]

(e) Hydrogen ions from the water accept electrons more readily than sodium ions (one mark); sodium ions are not reduced in aqueous solution (one mark); so hydrogen is released at the cathode (one mark). [3 marks]

Question 2 – Electroplating and charge calculations

(a) Charge equals current multiplied by time (one mark); 30.0 minutes is 1800 seconds (one mark); 2.00 multiplied by 1800 gives 3600 C (one mark). [3 marks]

(b) The amount of electrons is 3600 divided by 96500, that is 0.0373 mol (one mark); from Cu2+ + 2e- -> Cu the copper is half of this, 0.0187 mol (one mark); mass equals moles multiplied by molar mass (one mark); the answer is 1.18 g (one mark). [4 marks]

(c) The amount of silver equals the amount of electrons, 0.0373 mol (one mark); the mass is 0.0373 multiplied by 107.9 (one mark); the answer is 4.03 g (one mark); the mass of silver is greater than that of copper (one mark). [4 marks]

(d) Some of the current reduces hydrogen ions so hydrogen is released (one mark); the deposit may flake off or be lost by splashing (one mark); an oxidised or poorly cleaned surface prevents even deposition (one mark). [3 marks]

(e) The iron should be connected to the negative terminal, the cathode (one mark); metal cations gain electrons there and are reduced (one mark); the reduced metal is deposited on the surface of the iron (one mark). [3 marks]

Question 3 – Extraction of aluminium

(a) Aluminium is more reactive than carbon (one mark); aluminium oxide is very stable and carbon is not a strong enough reducing agent (one mark); electrolysis is therefore needed (one mark). [3 marks]

(b) It dissolves the aluminium oxide as soluble sodium aluminate (one mark); iron oxide and other impurities do not dissolve and are filtered off (one mark); purified aluminium oxide is then obtained for electrolysis (one mark). [3 marks]

(c) Cathode: Al3+ + 3e- -> Al (two marks); anode: 2O2- -> O2 + 4e- (two marks). [4 marks]

(d) Cryolite lowers the melting point of aluminium oxide from about 2045 C to about 950 C (two marks); the lower temperature saves the energy needed for heating (one mark); the melt also conducts better so less electrical energy is wasted (one mark). [4 marks]

(e) Oxygen produced at the anode reacts with the hot carbon (one mark); carbon dioxide is formed (one mark); the anodes therefore have to be replaced regularly (one mark). [3 marks]

Question 4 – Collision theory and factors affecting rate

(a) The rate is the amount of reactant used up, or product formed, per unit time (one mark); the volume of gas evolved can be measured against time (one mark); the loss in mass or a colour change can also be measured (one mark). [3 marks]

(b) There are more particles in a given volume (one mark); the frequency of collisions increases (one mark); more collisions per second have enough energy to react, so the rate rises (one mark). [3 marks]

(c) A catalyst provides an alternative route with a lower activation energy (one mark); more collisions are therefore successful and the rate increases (one mark); the mass of the catalyst is unchanged (one mark); its chemical properties are also unchanged (one mark). [4 marks]

(d) The powder has a much larger surface area (one mark); more particles are exposed to the acid (one mark); the frequency of collisions increases, so the rate increases (one mark). [3 marks]

(e) The average kinetic energy of the particles increases (one mark); far more collisions have energy at or above the activation energy (one mark); the collision frequency also rises slightly, so the rate increases markedly (one mark). [3 marks]

Question 5 – Calculating average rate from experimental data

(a) Average rate equals the change in volume divided by the time (one mark); 45 divided by 30 (one mark); the answer is 1.5 cm3/s (one mark). [3 marks]

(b) The change in volume is 78 minus 45, that is 33 cm3 (one mark); 33 divided by 30 gives 1.1 cm3/s (one mark); the concentration of the acid falls as the reaction proceeds (one mark); the lower concentration reduces the collision frequency, so the rate falls (one mark). [4 marks]

(c) The reactants have been used up, or one of them is completely used up (two marks); the final volume of carbon dioxide is about 108 cm3 (one mark). [3 marks]

(d) The initial slope would be steeper, that is the initial rate is greater (one mark); the doubled acid concentration raises the collision frequency (one mark); the amount of calcium carbonate is unchanged and it is the limiting reagent (one mark); the final volume of gas would be about the same (one mark). [4 marks]

(e) The curve would be steeper, so the reaction is faster (one mark); it would flatten sooner (one mark); the final volume of gas would be the same (one mark). [3 marks]

Question 6 – Energy and environmental issues in industrial electrolysis

(a) Very large currents are passed for long periods, so the energy demand is very high (one mark); electricity is the main cost (one mark); the plant can be sited where hydroelectric or geothermal power is cheap (one mark). [3 marks]

(b) Charge equals current multiplied by time (one mark); 24 hours is 86400 seconds (one mark); 15000 multiplied by 86400 (one mark); the answer is 1.296 times 10 to the power 9 C (one mark). [4 marks]

(c) The amount of electrons is 1.296 times 10 to the power 9 divided by 96500, that is 1.34 times 10 to the power 4 mol (one mark); from Al3+ + 3e- -> Al the aluminium is one third of this (one mark); the mass is 4477 multiplied by 27, that is 120870 g (one mark); about 121 kg (one mark). [4 marks]

(d) Recycling uses only about five per cent of the energy needed for electrolytic extraction (one mark); less bauxite has to be mined, so there is less damage to land (one mark); waste and the emissions linked to it are reduced (one mark). [3 marks]

(e) Chlorine is toxic, so a leak would harm workers and nearby residents (one mark); cells should be sealed and kept slightly below atmospheric pressure (one mark); waste gas is absorbed in alkali and monitored continuously (one mark). [3 marks]

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