A-LevelChemistryElectrochemistry9701/41 · Oct/Nov 2021 · Q19 Marks

When dilute sulfuric acid is electrolysed, water is split into hydrogen and oxygen. 2H2O(l) → 2H2(g) + O2(g) A current of xA is passed through the solution for 14.0 minutes. 462 cm³ of hydrogen are produced at the cathode, measured under room conditions.

(a)Calculate the number of hydrogen molecules produced during the electrolysis.[2]
(b)Calculate the total number of electrons transferred to produce this number of hydrogen molecules.[1]
(c)Calculate the quantity of charge, in coulombs, of the total number of electrons calculated in (b).[1]
(d)Calculate the current, x, passed during this experiment.[1]
(e)(i)The standard entropies, Sº, of three species are given in the table. [Table with species: H2O(l), H2(g), O2(g) and S°/JK⁻¹mol⁻¹ values: +70, +131, +205] Calculate ΔS® for the reaction 2H2O(l) → 2H2(g) + O2(g).[1]
(e)(ii)ΔΗ® for the reaction 2H2O(l) → 2H2(g) + O2(g) is +572kJmol⁻¹. Calculate ΔG for this reaction at 298 K.[2]
(e)(iii)Predict the effect of increasing temperature on the spontaneity of this reaction. Explain your answer.[1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

[1] moles of H2 = 462 / 24 000 = 0.01925
[1] molecules of H2 = 0.019 x 6.02 × 10^23 = 1.16 × 10^22 (1.1×10^22 / 1.2 × 10^22) min 2sf ecf M1
[1] number of electrons = 1.16 x 10^22 x 2 = 2.32 × 10^22 min 2sf ecf 1a
[1] = 2.32 ×10^22 × 1.6 x 10^-19 = 3.71 x 10^3 min 2sf ecf 1b
[1] X = 3.71 × 10^3 / (14 × 60) = 4.4 (A) min 2sf ecf 1c
[1] ΔS = 262 + 205 – 140 = (+) 327 (JK-1 mol-1)
[1] ΔG = ΔH – TΔS OR use of Gibbs equation
[1] ΔG = 572 – (298 × 0.327) = (+)474.6 (kJ mol-1) min 3sf ecf 1e(i)
[1] becomes more feasible / spontaneous as TΔS is more positive / -TΔS becomes more negative

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