The half-cell reaction for the oxidation of hydrogen is: $\frac{1}{2} \text{H}_2(g) \rightarrow \text{H}^+(aq) + e^-$ According to the Nernst equation, the oxidation potential is given by: $E_{\text{ox}} = E^\circ_{\text{ox}} - \frac{0.0591}{n} \log \frac{[\text{H}^+]}{(P_{\text{H}_2})^{1/2}}$ For this reaction, $n = 1$. The standard oxidation potential of hydrogen $E^\circ_{\text{ox}} = 0 \text{ V}$ and $P_{\text{H}_2} = 1 \text{ atm}$: $E_{\text{ox}} = 0 - 0.0591 \log [\text{H}^+]$ We know that $\text{pH} = -\log [\text{H}^+]$, so: $E_{\text{ox}} = 0.0591 \times \text{pH}$ Given $\text{pH} = 10$, substituting the value: $E_{\text{ox}} = 0.0591 \times 10 = 0.591 \text{ V} \approx 0.59 \text{ V}$.