1. Write the relevant half-reactions

Given standard reduction potentials:

$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}, \qquad E^\circ = +1.51\,\text{V}$

$\mathrm{2CO_2 + 2H^+ + 2e^- \rightarrow H_2C_2O_4}, \qquad E^\circ = -0.49\,\text{V}$

In the given overall reaction, oxalic acid is oxidized to $\mathrm{CO_2}$, so we reverse the second half-reaction:

$\mathrm{H_2C_2O_4 \rightarrow 2CO_2 + 2H^+ + 2e^-}$

Its oxidation potential is:

$E^\circ_{\text{ox}} = +0.49\,\text{V}$

2. Calculate standard cell potential

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} + E^\circ_{\text{anode(ox)}}$

$E^\circ_{\text{cell}} = 1.51 + 0.49 = 2.00\,\text{V}$

3. Balance electrons to find $n$

Reduction half:

$\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}$

Oxidation half:

$\mathrm{H_2C_2O_4 \rightarrow 2CO_2 + 2H^+ + 2e^-}$

LCM of electrons $=10$, so multiply:

• Mn half by $2$
• Oxalic acid half by $5$

Thus, for the overall reaction,

$n = 10$

Balanced overall reaction:

$\mathrm{2MnO_4^- + 5H_2C_2O_4 + 6H^+ \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2}$

4. Use relation between $E^\circ$ and equilibrium constant

At $298\,\text{K}$,

$E^\circ_{\text{cell}} = \frac{0.0591}{n} \log K$

So,

$\log K = \frac{nE^\circ_{\text{cell}}}{0.0591}$

Substitute $n=10$ and $E^\circ=2.00$ V:

$\log K = \frac{10 \times 2.00}{0.0591} = \frac{20}{0.0591}$

$\log K \approx 338.41$

Given $K_{eq} = 10^x$, we have

$x \approx 338.41$

Nearest integer:

$\boxed{x = 338}$

5. Comparison with stored answer

Stored correct answer = $338$

Our derived answer matches the stored answer.