Deduce the half-equation involving ethanedioic acid.

The standard electrode potential for the half-equation involving ethanedioic acid is ${E^\Theta } =  - 0.49V$. Using Table 14 of the Data Booklet, calculate the standard electrode potential for the equation on page 10.

Explain the sign of the calculated standard electrode potential.

Predict the sign of $\Delta {G^\Theta }$ for this reaction.

${{\text{(COOH)}}_2} \rightleftharpoons {\text{2C}}{{\text{O}}_2} + {\text{2}}{{\text{H}}^ + } + {\text{2}}{{\text{e}}^ - }$;

Accept either $\to$ or $\rightleftharpoons$.

Allow equation times 5.

Allow e instead of e^–.

${\text{MnO}}_4^ - {\text{(aq)}} + {\text{8}}{{\text{H}}^ + }{\text{(aq)}} + {\text{5}}{{\text{e}}^ - } \rightleftharpoons {\text{M}}{{\text{n}}^{2 + }}{\text{(aq)}} + {\text{4}}{{\text{H}}_2}{\text{O(l)}}$     ${E^\Theta } = 1.51{\text{ V}}$;

$\Delta {E^\Theta }{\text{ }}( = 1.51 + 0.49) = 2.00{\text{ V}}$;

First mark may be implied in the calculation.

Allow e instead of e–.

Accept either $\to$ or $\rightleftharpoons$.

positive sign, spontaneous reaction;

Allow ECF from (e) (i).

negative/$< 0$;

Do not allow ECF. This mark is independent of the answer in (e)(ii).

Even when candidates appeared to be getting (d) correct, electrons were omitted or the equation was unbalanced (usually electrons).

The standard electrode potential in (e) was given in standard form and 1.51 V was usually found in the Data Booklet but a significant minority gave the wrong manganese potential and many gave the answer as +1.02 V. Candidates should note that the sign of ${E^\Theta }$ should be given in any answer.

The standard electrode potential in (e) was given in standard form and 1.51 V was usually found in the Data Booklet but a significant minority gave the wrong manganese potential and many gave the answer as +1.02 V. Candidates should note that the sign of ${E^\Theta }$ should be given in any answer. The spontaneity and signs in part (e)(ii) caused little difficulty.

The standard electrode potential in (e) was given in standard form and 1.51 V was usually found in the Data Booklet but a significant minority gave the wrong manganese potential and many gave the answer as +1.02 V. Candidates should note that the sign of ${E^\Theta }$ should be given in any answer. The spontaneity and signs in part (f) caused little difficulty.