S6 Chemistry Practice Paper: Electrochemistry, Industrial Processes And Green Chemistry (12 Questions, 100 Marks, 75 Minutes)

S6 Chemistry Practice Paper: Electrochemistry, Industrial Processes and Green Chemistry (12 questions, 100 marks, 75 minutes)

2026-11-06

S6 Chemistry: Electrochemistry, Industrial Processes and Green Chemistry

S6 chemistry practice paper: 12 questions on cell potentials, electrolysis, oxidation numbers, equilibrium, industrial processes, atom economy and green chemistry, with full answers.

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Section A: Electrochemistry and Cells (32 marks)

Answer all questions in Section A and label oxidation and reduction clearly.

#	Question	Answer and marking notes
1	(Cell potential) A zinc-copper cell is set up with zinc and copper electrodes in zinc sulphate and copper sulphate solutions, joined by a salt bridge. E(Zn2+/Zn) = -0.76 V and E(Cu2+/Cu) = +0.34 V; F = 96500 C mol-1. (a) Write the half-equations at the anode and the cathode. (b) Calculate the cell potential. (c) Calculate the standard free energy change per mole of reaction in kJ mol-1 and explain the sign.	Answer: (a) Anode (oxidation): Zn -> Zn2+ + 2e-; cathode (reduction): Cu2+ + 2e- -> Cu. (b) Cell potential = 0.34 – (-0.76) = 1.10 V. (c) Free energy change = -nFE = -2 x 96500 x 1.10 = -212300 J = -212.3 kJ mol-1; the negative sign shows the reaction is spontaneous and the cell can deliver electrical work. Mark points: both half-equations (2), cell potential (2), free energy calculation (3), meaning of the sign (1).
2	(Electrolysis) A current of 2.5 A is passed through copper(II) sulphate solution for 1800 s. (a) Calculate the charge passed. (b) Calculate the moles of copper deposited at the cathode. (c) Calculate the mass of copper deposited (M(Cu) = 63.5).	Answer: (a) Charge = 2.5 x 1800 = 4500 C. (b) Moles of electrons = 4500 / 96500 = 0.0466 mol; since Cu2+ + 2e- -> Cu, moles of copper = 0.0233 mol. (c) Mass = 0.0233 x 63.5 = 1.48 g. Mark points: charge (3), electron-copper relationship (2), mass (3).
3	(Oxidation numbers) In a methanol fuel cell, methanol is oxidised in alkaline conditions to carbon dioxide. (a) Calculate the oxidation number of carbon in CH3OH. (b) Calculate the oxidation number of carbon in CO2. (c) Hence state the number of electrons released per mole of methanol oxidised and write the electron balance.	Answer: (a) Let the oxidation number of carbon be x: x + 4(+1) + (-2) = 0, so x = -2. (b) In CO2 oxygen is -2, so carbon is +4. (c) The oxidation number rises from -2 to +4, an increase of 6, so 6 mol of electrons are released per mole of methanol; electron balance requires the electrons released by the oxidation half-reaction to equal those accepted by the oxygen reduction half-reaction. Mark points: methanol value (2), carbon dioxide value (2), electron count and balance (4).
4	(Water electrolysis) Dilute sulphuric acid is electrolysed with a current of 1.0 A for 965 s. Molar volume at room temperature is 24.0 dm3 mol-1. (a) Calculate the moles of electrons passed. (b) Calculate the moles of hydrogen produced at the cathode. (c) Calculate the volume of hydrogen in cm3. (d) Write the half-equation at the anode.	Answer: (a) Charge = 1.0 x 965 = 965 C; moles of electrons = 965 / 96500 = 0.0100 mol. (b) Cathode: 2H2O + 2e- -> H2 + 2OH-, so hydrogen = 0.0050 mol. (c) Volume = 0.0050 x 24.0 = 0.120 dm3 = 120 cm3. (d) Anode: 2H2O -> O2 + 4H+ + 4e-. Mark points: electron moles (2), hydrogen moles (2), volume conversion (2), anode half-equation (2).

Section B: Industrial Processes and Equilibrium (36 marks)

Answer all questions in Section B and show every step with units.

#	Question	Answer and marking notes
5	(Haber process) 1 mol of nitrogen and 3 mol of hydrogen react in a 1 dm3 closed vessel; at equilibrium 1.2 mol of ammonia is present. (a) Calculate the moles of nitrogen and hydrogen at equilibrium. (b) Calculate Kc at this temperature. (c) Calculate the conversion of nitrogen. (d) Explain the effect of raising the temperature on the ammonia yield.	Answer: (a) Forming 1.2 mol of ammonia means 0.6 mol of nitrogen and 1.8 mol of hydrogen have reacted; at equilibrium nitrogen = 0.4 mol and hydrogen = 1.2 mol. (b) Kc = [NH3]2 / ([N2][H2]3) = 1.44 / (0.4 x 1.728) = 2.08. (c) Conversion of nitrogen = 0.6 / 1.0 x 100% = 60%. (d) The forward reaction is exothermic, so raising the temperature shifts the equilibrium to the left and lowers the ammonia yield. Mark points: equilibrium moles (3), Kc expression and value (3), conversion (2), temperature effect (1).
6	(Contact process) 200 mol of sulphur dioxide and 150 mol of oxygen enter a reactor and 96% of the sulphur dioxide is converted. (a) Calculate the moles of sulphur trioxide formed. (b) Calculate the moles of oxygen left. (c) Calculate the mass of sulphur trioxide in kg (M = 80). (d) Explain why excess oxygen is used industrially.	Answer: (a) Sulphur dioxide reacted = 200 x 96% = 192 mol; from 2SO2 + O2 -> 2SO3 the sulphur trioxide formed is 192 mol. (b) Oxygen used = 192 / 2 = 96 mol, so 150 – 96 = 54 mol remains. (c) Mass = 192 x 80 = 15360 g = 15.36 kg. (d) Excess oxygen shifts the equilibrium towards the product, raising the conversion of sulphur dioxide and ensuring complete oxidation. Mark points: sulphur trioxide moles (3), remaining oxygen (2), mass (2), excess oxygen explanation (2).
7	(Atom economy) Limestone decomposes on heating: CaCO3 -> CaO + CO2, M(CaCO3) = 100 and M(CaO) = 56. (a) Calculate the atom economy for calcium oxide as the target product. (b) Explain the difference between atom economy and percentage yield. (c) Suggest one way to improve the atom economy of this process.	Answer: (a) Atom economy = 56 / 100 x 100% = 56%. (b) Atom economy compares how much of the reactant atoms end up in the target product, ignoring operational losses, whereas percentage yield compares the actual amount obtained with the theoretical yield. (c) Collect the carbon dioxide by-product and use it as a feedstock, or choose a reaction route that produces no by-product. Mark points: atom economy calculation (3), difference (3), improvement (3).
8	(Activation energy) The activation energy of the forward reaction is 250 kJ mol-1 and the enthalpy change is -92 kJ mol-1. (a) Calculate the activation energy of the reverse reaction. (b) Describe the key features of the energy profile. (c) Explain how a catalyst affects the activation energy and the enthalpy change. (d) State whether a catalyst can change the position of equilibrium and explain.	Answer: (a) Reverse activation energy = 250 – (-92) = 342 kJ mol-1. (b) The reactants lie 92 kJ mol-1 above the products (exothermic), and the peak is 250 kJ mol-1 above the reactants. (c) A catalyst provides an alternative route of lower activation energy but does not change the enthalpy change, because the energies of reactants and products are unchanged. (d) No; a catalyst speeds up the forward and reverse reactions equally, so the position of equilibrium and the equilibrium constant stay the same and only the time taken to reach equilibrium is reduced. Mark points: reverse activation energy (3), profile features (2), catalyst effect (2), equilibrium position (2).

Section C: Green Chemistry and Energy (32 marks)

Answer all questions in Section C and give units with every percentage and energy answer.

#	Question	Answer and marking notes
9	(Green chemistry) A plant replaces benzene with a non-toxic solvent and designs a catalytic step that greatly reduces by-products. (a) State which green chemistry principles these two measures follow. (b) Explain how atom economy reflects green chemistry. (c) Explain why preventing waste is better than treating it afterwards.	Answer: (a) Replacing benzene with a non-toxic solvent follows the principle of using safer chemicals; the catalytic step that reduces by-products follows the principle of waste prevention and raises atom economy. (b) A high atom economy means most reactant atoms finish in the target product, so little waste is produced and resources are used efficiently, which is exactly the aim of green chemistry. (c) Treating waste afterwards needs extra energy, chemicals and equipment and may cause secondary pollution, so preventing waste at source saves resources and is more sustainable. Mark points: two principles (3), atom economy explanation (3), waste prevention reason (2).
10	(Carbon emissions) A power station burns 1.0 x 105 mol of methane completely. (a) Write the equation for complete combustion. (b) Calculate the moles of carbon dioxide produced. (c) Calculate the mass of carbon dioxide produced in kg.	Answer: (a) CH4 + 2O2 -> CO2 + 2H2O. (b) The mole ratio is 1:1, so 1.0 x 105 mol of carbon dioxide is produced. (c) Mass = 1.0 x 105 x 44 = 4.4 x 106 g = 4400 kg. Mark points: equation (2), mole ratio (2), mass with units (4).
11	(Yield) The theoretical yield of a reaction is 250 g but only 205 g of the target product is obtained. (a) Calculate the percentage yield. (b) Give two reasons why the yield is below 100%. (c) Explain why atom economy must also be considered when judging how green a process is.	Answer: (a) Percentage yield = 205 / 250 x 100% = 82.0%. (b) Side reactions consume reactants; the reaction is incomplete or given too little time; material is lost during transfer, filtration and purification. (c) A high yield with a low atom economy means most reactant atoms become waste, so the overall resource efficiency and environmental burden are still poor; both measures must be considered. Mark points: yield calculation (3), two reasons (3), atom economy link (2).
12	(Hydrogen production) Electrolysis of water produces 0.1 mol of hydrogen using 500 kJ of electrical energy; the heat of combustion of hydrogen is 286 kJ mol-1. (a) Calculate the energy released when this hydrogen burns completely. (b) Calculate the energy efficiency of the process. (c) Explain under what condition electrolytic hydrogen is low-carbon and why.	Answer: (a) Energy released = 0.1 x 286 = 28.6 kJ. (b) Efficiency = 28.6 / 500 x 100% = 5.72%. (c) Electrolytic hydrogen is low-carbon only when the electricity comes from renewable sources such as solar, wind or hydro power; if the electricity comes from burning fossil fuels, the emissions of making the hydrogen may exceed those of using the fuel directly. Mark points: energy calculation (3), efficiency (3), low-carbon condition and explanation (2).

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