S6 Chemistry Practice Paper: Redox, Electrochemistry And Chemical Cells (12 Questions, 100 Marks)

S6 Chemistry Practice Paper: Redox, Electrochemistry and Chemical Cells (12 questions, 100 marks)

2026-11-08

S6 Chemistry: Redox, Electrochemistry and Cells

S6 Chemistry: oxidation numbers and half-equation balancing, cell EMF, electrolysis stoichiometry and redox titration, applied to lead-acid cells, brine electrolysis and plating.

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Section A: Oxidation Numbers and Redox Reactions (4 questions, 25 marks)

Oxidation numbers: fix K, O and H first, then use a total charge of zero for a compound or the ionic charge for an ion; check atoms and charge after balancing.

#	Question	Answer and marking notes
1	State the oxidation number of chromium in K2Cr2O7 and of manganese in MnO4-, showing your working.	In K2Cr2O7, K = +1 and O = -2, so 2(+1) + 2x + 7(-2) = 0 gives x = +6 for chromium. In MnO4-, y + 4(-2) = -1 gives y = +7 for manganese. Both elements are at their highest oxidation state, which is why K2Cr2O7 and MnO4- are common oxidising agents.
2	Which of the following are redox reactions? Explain each one and name the species oxidised and the species reduced (write none if not applicable): (a) Zn + Cu2+ -> Zn2+ + Cu; (b) HCl + NaOH -> NaCl + H2O; (c) 2H2O2 -> 2H2O + O2.	(a) redox: Zn goes from 0 to +2 (oxidised) and Cu2+ from +2 to 0 (reduced). (b) not redox: H +1, Cl -1, Na +1 and O -2 are all unchanged; it is only a neutralisation. (c) redox and a disproportionation: oxygen is -1 in H2O2, -2 in H2O (reduced) and 0 in O2 (oxidised), so the same element is both oxidised and reduced.
3	Balance by the oxidation-number method: MnO4- + Fe2+ + H+ -> Mn2+ + Fe3+ + H2O, and give the six coefficients in the order MnO4-, Fe2+, H+, Mn2+, Fe3+, H2O.	Mn goes from +7 to +2, gaining 5 electrons; Fe goes from +2 to +3, losing 1 electron, so MnO4- : Fe2+ = 1 : 5. Balanced: 1 MnO4- + 5 Fe2+ + 8 H+ -> 1 Mn2+ + 5 Fe3+ + 4 H2O. Charge check: left = -1 + 10 + 8 = +17, right = +2 + 15 = +17; hydrogen 8 = 8 and oxygen 4 = 4. The coefficients are 1 : 5 : 8 : 1 : 5 : 4.
4	Use half-equations to balance the reaction in acidic medium: IO3- + I- + H+ -> I2 + H2O, and state what type of reaction it is.	Reduction: 2IO3- + 12H+ + 10e- -> I2 + 6H2O. Oxidation: 2I- -> I2 + 2e-, multiplied by 5 to give 10I- -> 5I2 + 10e-. Adding gives 2IO3- + 10I- + 12H+ -> 6I2 + 6H2O, that is 1 : 5 : 6 : 3 : 3. Iodine changes from +5 and -1 to 0, so it is both oxidised and reduced: a disproportionation.

Section B: Cells, Electrolysis and Electrode Reactions (4 questions, 33 marks)

Electrodes and charge: oxidation at the anode, reduction at the cathode; E cell = E cathode − E anode; Q = It, moles of electrons = Q / 96500, then use the electron count in the half-equation.

#	Question	Answer and marking notes
5	For the cell Zn(s) \| Zn2+(aq) \|\| Cu2+(aq) \| Cu(s), E(Zn2+/Zn) = -0.76 V and E(Cu2+/Cu) = +0.34 V. (a) Which electrode is the anode? (b) Calculate E cell. (c) Write the overall cell reaction and state the direction of electron flow in the external circuit.	(a) Zinc is the anode, where oxidation occurs. (b) E cell = E cathode – E anode = 0.34 – (-0.76) = 1.10 V. (c) Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s); electrons flow from zinc to copper in the external circuit while anions such as SO4^2- move towards the zinc half-cell in the salt bridge.
6	CuSO4(aq) is electrolysed with inert electrodes at a current of 2.0 A for 30 minutes. Take the charge on one mole of electrons as 96500 C and the molar mass of copper as 63.5 g/mol. (a) Calculate the charge passed. (b) Calculate the mass of copper deposited at the cathode.	(a) Q = It = 2.0 x 1800 = 3600 C. (b) n(e-) = 3600 / 96500 = 0.0373 mol; Cu2+ + 2e- -> Cu, so n(Cu) = 0.0373 / 2 = 0.01865 mol and mass = 0.01865 x 63.5 = 1.18 g. This is a direct application of Faraday laws of electrolysis.
7	In the electrolysis of concentrated brine: (a) state the product at the anode and write its electrode equation; (b) state the products at the cathode and write the electrode equation; (c) write the overall equation and name the alkali finally obtained in solution.	(a) Chlorine at the anode: 2Cl- -> Cl2 + 2e- (chloride oxidised). (b) Hydrogen and hydroxide ions at the cathode: 2H2O + 2e- -> H2 + 2OH-. (c) 2NaCl + 2H2O -> 2NaOH + H2 + Cl2. The sodium and hydroxide ions left in solution give sodium hydroxide, so brine electrolysis yields chlorine, hydrogen and caustic soda.
8	In nickel electroplating, which terminal is the workpiece connected to? What material is used as the anode? Explain why the concentration of Ni2+ in the electrolyte stays roughly constant.	The workpiece is the cathode, connected to the negative terminal, and the anode is a piece of nickel. At the cathode Ni2+ + 2e- -> Ni deposits nickel on the workpiece; at the anode Ni -> Ni2+ + 2e- dissolves nickel at the same rate, so the Ni2+ concentration stays roughly constant and plating can continue. The coating is thin, even, adherent and corrosion resistant.

Section C: Quantitative Calculations, Redox Titration and Applications (4 questions, 42 marks)

Quantitative work: use n = cV (V in dm3) and the stoichiometric ratio; voltages add in series; compare batteries on energy density, cost, life and environmental impact.

#	Question	Answer and marking notes
9	Given E(Ag+/Ag) = +0.80 V, E(Cu2+/Cu) = +0.34 V and E(Zn2+/Zn) = -0.76 V, build a cell from any two of these half-cells. (a) Which pair gives the largest E cell? Calculate it. (b) Write the anode and cathode half-equations of that cell. (c) What is E cell for the copper-silver pair, and is that reaction spontaneous?	(a) Silver with zinc gives the largest value: E cell = 0.80 – (-0.76) = 1.56 V. (b) Anode (zinc): Zn -> Zn2+ + 2e-; cathode (silver): 2Ag+ + 2e- -> 2Ag. (c) For copper with silver, E cell = 0.80 – 0.34 = 0.46 V; it is positive, so the reaction is spontaneous with copper oxidised and silver ions reduced.
10	Aluminium is extracted by electrolysing molten Al2O3 at 10.0 A for 4.0 hours. Calculate (a) the moles of electrons passed; (b) the mass of aluminium formed (Al = 27.0 g/mol, 96500 C/mol). Explain why the anodes must be replaced regularly.	(a) Q = 10.0 x 14400 = 144000 C, so n(e-) = 144000 / 96500 = 1.492 mol. (b) Al3+ + 3e- -> Al, so n(Al) = 1.492 / 3 = 0.4974 mol and mass = 0.4974 x 27.0 = 13.4 g. Oxygen produced at the anode burns the carbon anodes away as CO2, so they must be replaced regularly.
11	25.0 cm3 of acidified Fe2+ solution is titrated with 0.0200 M KMnO4 and the average titre is 24.0 cm3. The reaction is MnO4- + 5Fe2+ + 8H+ -> Mn2+ + 5Fe3+ + 4H2O. (a) Calculate the moles of Fe2+ titrated. (b) Calculate the concentration of the original Fe2+ solution. (c) What is used as the indicator and how is the end point judged?	(a) n(MnO4-) = 0.0200 x 0.0240 = 4.8e-04 mol; from the 1 : 5 ratio, n(Fe2+) = 5 x 4.8e-04 = 2.4e-03 mol. (b) c = 2.4e-03 / 0.0250 = 0.096 M. (c) Potassium permanganate is its own indicator: MnO4- is purple and Mn2+ is almost colourless, so the end point is a lasting pale pink colour that does not fade within about half a minute.
12	A lead-acid cell discharges as PbO2 + Pb + 2H2SO4 -> 2PbSO4 + 2H2O, with E(PbO2/PbSO4) = +1.69 V and E(PbSO4/Pb) = -0.36 V. (a) Calculate E cell for one cell. (b) How many such cells in series make a 12 V car battery, and what is the actual total voltage? (c) Explain why the sulfuric acid concentration falls during discharge. (d) Give one advantage and one disadvantage of the lead-acid battery compared with a lithium-ion battery.	(a) E cell = 1.69 – (-0.36) = 2.05 V. (b) 12 / 2.05 = 5.85, so six cells are connected in series: 6 x 2.05 = 12.3 V. (c) Each 2 mol of electrons consumes 2 mol H2SO4 and produces 2 mol H2O, so the acid is used up and diluted; the falling density can be checked with a hydrometer. (d) Advantage: high current capability, low cost and mature technology. Disadvantage: low energy density, lead and acid need recycling, and service life is shorter; a lithium-ion battery has a much higher energy density but costs more and needs a protection circuit.

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