A-LevelChemistryElectrochemistry9701/41 · Oct/Nov 2020 · Q311 Marks

Identify the substances liberated at the anode and at the cathode during the electrolysis of aqueous sodium sulfate, Na2SO4(aq). anode .......................................................................................................................................... cathode ........................................................................................................................................

(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 4.50A. Calculate the volume of chlorine and the mass of sodium that are liberated under room conditions.[4]
(c)(i)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 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]

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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] M1: Q = 1.5 × 60 × 60 × 4.5 = 24300 (C)
[1] M2: no. of F / moles of e¯ = 24300 / 96500 = 0.25(1813) ecf
[1] M3: volume of Cl2 = 24 × 0.252 / 2 = 3.02 dm³ ecf min 2sf
[1] M4: mass of Na = 0.252 × 23 = 5.79 (5.7917) g Na ecf min 2sf
[1] MnO4–, H+, Mn2+ in same beaker AND H+ in other beaker
[1] both electrodes Pt(s) (ALLOW graphite)
[1] one solute clearly identified as 1M / 1 mol dm–3
[1] 298 K OR 1 atm
[1] voltmeter / potentiometer labelled (or circled V)
[1] salt bridge labelled (must touch the solution)
[1] a good delivery system for H2(g)
[1] H2(g)
[4] mark as two correct points = 1 mark
[1] F2 OR S2O82– OR H2O2 OR HOCl OR Co3+ OR Pb4+
[1] 2Mn2+ + 8H2O + 5F2 → 2MnO4– + 16H+ + 10F- OR 2Mn2+ + 5S2O82–+ 8H2O → 2MnO4– + 16H+ + 10SO42– OR Mn2+ + 4H2O + 5Co3+ → 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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