Some electrode potentials are shown in Table 3.1. Table 3.1 electrode reaction E/V V2+ + 2e-v -1.20 V3+ + e 2+ -0.26 VO2+ + 2H+ + e¯ = V3+ + H₂O +0.34 VO2+ + 2H+ + e¯ ⇌ VO2+ + H₂O +1.00 Fe2+ + 2e-Fe -0.44 Fe3+ + 3e Fe -0.04 Fe3+ + eFe2+ +0.77 2H+ + 2e¯ H2 0.00 ClO® + H2O + 2e¯⇌ Cl¯ + 2OH- +0.89
- (a)(i)Complete the diagram to show a standard hydrogen electrode. Label your diagram. Identify all substances. You do not need to state standard conditions. [Figure for diagram][1]
- (a)(ii)An electrochemical cell is set up using an Fe3+/Fe2+ electrode and a standard hydrogen electrode. Identify the positive electrode in the electrochemical cell and the direction of electron flow in the external circuit. positive electrode .......................................................................................................... Electrons flow from the ............................ electrode to the ............................ electrode.[1]
- (b)(i)The vanadium-containing species in the electrode reactions given in Table 3.1 are V, V2+, V3+, VO2+ and VO2+. Identify one vanadium-containing species that does not react with Fe2+ ions under standard conditions. Use data from Table 3.1 to explain your answer.[1]
- (b)(ii)Identify all the vanadium-containing species that will react with Fe2+ ions under standard conditions.[1]
- (b)(iii)Write an equation for one of the possible reactions identified in (ii).[1]
- (c)(i)Another electrochemical cell is set up using an Fe3+/Fe2+ electrode and an alkaline ClO¯/Cl- electrode. The concentration of Fe3+ is 1000 times greater than the concentration of Fe2+ in the Fe3+/Fe2+ electrode. All other conditions are standard. Use the Nernst equation to calculate the E value of the Fe3+/Fe2+ electrode. Show your working. E = ............................ V[2]
- (c)(ii)Write an equation for the reaction that occurs in the cell, under these conditions.[1]
- (d)Another electrochemical cell is set up using an Fe2+/Fe electrode and an alkaline ClO¯/Cl- electrode under standard conditions. Calculate the value of ΔG° for the cell. ΔG° = ............................ kJ mol-1[3]
- (e)A solution of iron(II) sulfate, FeSO4(aq) is electrolysed with iron electrodes. Under the conditions used, no gas is evolved at the cathode. A current of 0.640A is passed for 17.0 minutes. The mass of the cathode increases by 0.185g. Use these results to calculate an experimental value for the Avogadro constant, L. Show your working. L = ............................ mol-1[3]
- (f)(i)Iron(II) chloride, FeCl2, is oxidised by chlorine to form iron(III) chloride, FeCl3, under standard conditions. 2FeCl2(s) + Cl2(g) → 2FeCl3(s) ΔH° = –128 kJ mol-1 Table 3.2 species S°/JK-1 mol-1 Cl2(g) 223 FeCl2(s) 120 FeCl3(s) 142 Use Table 3.2 and other data to calculate the Gibbs free energy change, ΔG°, for this reaction. Show your working. ΔG° = ............................ kJ mol-1[3]
- (f)(ii)Predict whether this reaction becomes more or less feasible at a higher temperature. Explain your answer. The reaction becomes ............................ feasible. explanation ..........................................................................................................................[1]
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 2023). The mark or mark type is shown in brackets.
- [1] (a)(i) hydrogen, delivery system, H+, platinum,
- [1] (a)(ii) iron..... hydrogen . . . . . iron
- [1] (b)(i) (for specified V2+, V3+ or VO2+) Eº is more positive than / above –0.44 AND more negative than / below 0.77 V
- [1] (b)(ii) V and VO2+
- [1] (b)(iii) V + Fe2+ → V2+ + Fe OR VO2+ + 2H+ + Fe2+ → VO2+ + H2O + Fe3+
- [1] (c)(i) Nernst: E = 0.77 + (0.059/z)log[ox] / [red]
- [1] (c)(i) 0.947
- [1] (c)(ii) 2Fe3+ + Cl- + 2OH- → 2Fe2+ + CIO- + H2O
- [1] (d) Ecell = 1.33 V
- [1] (d) ΔG° = -nE°cellF
- [1] (d) -257
- [1] (e) 0.64 × 17 × 60 = 653 / 652.8 Coulombs
- [1] (e) 652.8 ÷ 1.6x10–19 = 4.08 × 1021 (number of electrons) 4.08 × 1021 ÷ 2 = 2.04 × 1021 (number of atoms Fe)
- [1] (e) 0.185 ÷ 55.8 = 3.31 × 10-3 (number of moles Fe atoms 2.04 × 1021 ÷ 3.31 × 10–3 = L = 6.153 × 1023)
- [1] (f)(i) ΔS = -179
- [1] (f)(i) ΔG = ΔH -TΔS
- [1] (f)(i) -74.7
- [1] (f)(ii) less AND ΔG becomes more positive
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