In acidic solution, bromate ions, BrO₃⁻(aq), oxidize iodide ions, I⁻(aq).

BrO₃⁻(aq) + 6H⁺(aq) + 6e⁻ \( \rightleftharpoons \) Br⁻(aq) + 3H₂O(l)

2I⁻(aq) \( \rightleftharpoons \) I₂(s) + 2e⁻

Formulate the equation for the redox reaction.

The change in the free energy for the reaction under standard conditions, ΔG^Θ, is −514 kJ at 298 K.

Determine the value of E^Θ, in V, for the reaction using sections 1 and 2 of the data booklet.

Calculate the standard electrode potential, in V, for the BrO₃⁻/Br⁻ reduction half‑equation using section 24 of the data booklet.

BrO₃^–(aq) + 6H⁺(aq) + 6I^–(aq) \( \rightleftharpoons \) Br^–(aq) + 3I₂(s) + 3H₂O(l)

Accept → for \( \rightleftharpoons \).

[1 mark]

n = 6

«ΔG^Θ = –nFE^Θ»

«\({E^\Theta } =  - \frac{{\Delta {G^\Theta }}}{{{\text{n}}F}} = \frac{{514 \times {{10}^3}{\text{ J mo}}{{\text{l}}^{ - 1}}}}{{6 \times 9.65 \times {{10}^4}{\text{ C mo}}{{\text{l}}^{ - 1}}}} = \)» 0.888 «V»

[2 marks]

«E^Θ = E^Θ(BrO₃^–/Br^–) – E^Θ(I₂/I^–)»

«E^Θ(BrO₃^–/Br^–) = E^Θ + E^Θ(I₂/I^–) = 0.888 + 0.54 =» «+» 1.43 «V»

[1 mark]