1. 1
   The iron content of an iron supplement tablet is found by titrating dissolved Fe²⁺ ions with acidified potassium manganate(VII):
   MnO₄⁻(aq) + 8H⁺(aq) + 5Fe²⁺(aq) → Mn²⁺(aq) + 5Fe³⁺(aq) + 4H₂O(l)
   Standard electrode potentials: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ ⇌ Mn²⁺(aq) + 4H₂O(l), E⦵ = +1.51 V; Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq), E⦵ = +0.77 V. Faraday constant F = 96 500 C mol⁻¹.

   (a)

   What is the change in oxidation state of manganese in this reaction?

   [1 mark]

   • A+8 to +2
   • B+6 to +2
   • C+7 to +2
   • D+2 to +7

   (b)

   What is the standard cell potential, E⦵cell, for the reaction?

   [1 mark]

   • A+2.28 V
   • B+0.74 V
   • C−0.74 V
   • D−2.34 V

   (c)

   Determine the standard Gibbs energy change, ΔG⦵, in kJ mol⁻¹, for the reaction as written.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   A student is given the following standard electrode potentials:
   Zn²⁺(aq) + 2e⁻ ⇌ Zn(s), E⦵ = −0.76 V
   Ni²⁺(aq) + 2e⁻ ⇌ Ni(s), E⦵ = −0.26 V
   Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), E⦵ = +0.34 V
   I₂(aq) + 2e⁻ ⇌ 2I⁻(aq), E⦵ = +0.54 V
   Ag⁺(aq) + e⁻ ⇌ Ag(s), E⦵ = +0.80 V
   Br₂(aq) + 2e⁻ ⇌ 2Br⁻(aq), E⦵ = +1.09 V

   (a)

   Which reaction is spontaneous under standard conditions?

   [1 mark]

   • ANi(s) + Cu²⁺(aq) → Ni²⁺(aq) + Cu(s)
   • BCu(s) + Zn²⁺(aq) → Cu²⁺(aq) + Zn(s)
   • C2Ag(s) + Ni²⁺(aq) → 2Ag⁺(aq) + Ni(s)
   • DI₂(aq) + 2Br⁻(aq) → 2I⁻(aq) + Br₂(aq)

   (b)

   Which species in the list is the strongest reducing agent?

   [1 mark]

   • AAg
   • BI⁻
   • CBr⁻
   • DZn

   (c)

   A voltaic cell is built from the Ni²⁺/Ni and Ag⁺/Ag half-cells under standard conditions. Determine E⦵cell and deduce which electrode is the negative electrode.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Aqueous sodium chloride is electrolysed using inert graphite electrodes, first as a concentrated solution (brine) and then as a very dilute solution. Relevant standard electrode potentials:
   Na⁺(aq) + e⁻ ⇌ Na(s), E⦵ = −2.71 V
   2H₂O(l) + 2e⁻ ⇌ H₂(g) + 2OH⁻(aq), E⦵ = −0.83 V
   O₂(g) + 4H⁺(aq) + 4e⁻ ⇌ 2H₂O(l), E⦵ = +1.23 V
   Cl₂(g) + 2e⁻ ⇌ 2Cl⁻(aq), E⦵ = +1.36 V

   (a)

   Deduce the product formed at the cathode, giving its half-equation, and explain why sodium is not formed.

   [3 marks]

   (b)

   Compare the products at the anode for the concentrated and the very dilute solutions, giving half-equations, and explain why the product from brine is not the one predicted by the E⦵ values alone.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A workshop copper-plates steel keys (mainly iron) by electrolysis in aqueous copper(II) sulfate, with the key as the cathode. In trial 1 the anode was a copper bar: the key gained 0.296 g, the copper bar lost 0.296 g and the blue colour of the solution did not change. In trial 2 the copper bar was replaced by an inert graphite anode: the key gained a similar mass, a colourless gas was given off at the anode, and the solution became paler blue and more acidic.
   Standard electrode potentials: Fe²⁺(aq) + 2e⁻ ⇌ Fe(s), E⦵ = −0.45 V; Cu²⁺(aq) + 2e⁻ ⇌ Cu(s), E⦵ = +0.34 V; O₂(g) + 4H⁺(aq) + 4e⁻ ⇌ 2H₂O(l), E⦵ = +1.23 V. Faraday constant F = 96 500 C mol⁻¹.

   (a)

   Explain all the observations in trials 1 and 2, deducing the half-equation at each electrode.

   [6 marks]

   (b)

   A trainee suggests saving electricity by simply dipping the steel keys into copper(II) sulfate solution. Evaluate this suggestion, including a calculation of E⦵cell and ΔG⦵ for the reaction that would occur, and compare the result with electroplating.

   [6 marks]

   Total for question 4: 12 marks

End of questions