The equation representing the standard electrode potential, E°, for the reduction of MnO₄⁻(aq) to Mn²⁺(aq) in acid solution is given. MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇋ Mn²⁺(aq) + 4H₂O(l) E° = +1.52V
- (a)Identify the substances liberated at the anode and at the cathode during the electrolysis of aqueous sodium sulfate, Na₂SO₄(aq). anode cathode[1]
- (b)When molten sodium chloride is electrolysed, chlorine is liberated at the anode and sodium is liberated at the cathode. A sample of molten sodium chloride is electrolysed for 1.50 hours using a current of 4.50 A. Calculate the volume of chlorine and the mass of sodium that are liberated under room conditions. volume of chlorine = dm³ mass of sodium = g[4]
- (c)(i)Draw a diagram of the apparatus that would be used to measure the E° value of this half-cell. Your diagram should be fully labelled to identify all apparatus, substances and conditions.[4]
- (c)(ii)Use the Data Booklet to identify a substance that could be used to oxidise Mn²⁺ ions to MnO₄⁻ ions under standard conditions. Write an equation for the reaction.[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [1] (anode =) oxygen / O2 AND (cathode =) hydrogen/H2 BOTH [1]
- [4] M1: Q = 1.5 × 60 × 60 × 4.5 = 24300 (C) [1]
- M2: no. of F / moles of e⁻ = 24300 / 96500 = 0.25(1813) [1] ecf
- M3: volume of Cl2 = 24 × 0.252 / 2 = 3.02 dm³ [1] ecf min 2sf
- M4: mass of Na = 0.252 × 23 = 5.79 (5.7917) g Na [1] ecf min 2sf
- [4] MnO4-, H+, Mn2+ in same beaker AND H+ in other beaker
- both electrodes Pt(s) (ALLOW graphite)
- one solute clearly identified as 1M/1 mol dm-3
- 298 K OR 1 atm
- voltmeter / potentiometer labelled (or circled V)
- salt bridge labelled (must touch the solution)
- a good delivery system for H2(g)
- H2(g)
- mark as two correct points = 1 mark [4]
- [2] F2 OR S2O82- OR H2O2 OR HOCI OR Co³+ OR Pb4+ [1]
- 2Mn2+ + 8H2O + 5F2 → 2MnO4¯ + 16H+ + 10F-[1]
- OR 2Mn2+ + 5S2O82⁻+ 8H2O → 2MnO4¯ + 16H+ + 10SO42-
- OR Mn2+ + 4H2O + 5C03+ → MnO4 + 8H+ + 5Co2+
- OR 2Mn2+ + 8H2O + 5Pb4+ → 2MnO4¯ + 16H+ + 5Pb2+
- OR 2Mn2+ + 5H2O2 → 2MnO4¯ + 6H+ + 2H2O
- OR 2Mn2+ + 10HOCl → 2MnO4¯ + 6H+ + 5Cl2 + 2H2O
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