The potential for the given half cell at $298 \mathrm{~K}$ is (-) $\underline{\hspace{2cm}}$$\times 10^{-2} \mathrm{~V}$ $$\begin{aligned} & 2 \mathrm{H}_{(\mathrm{aq})}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_2(\mathrm{~g}) \\\\ & {\left[\mathrm{H}^{+}\right]=1 \mathrm{M}, \mathrm{P}_{\mathrm{H}_2}=2 \mathrm{~atm}} \end{aligned}$$(Given : $2.303 \mathrm{RT} / \mathrm{F}=0.06 \mathrm{~V}, \log 2=0.3$ )