The potential for the given half cell at 298 K is $-\ldots\ldots\ldots . . \times 10 ^ { - 2 } \mathrm {~V}$.\\
$2 \mathrm { H } _ { (\mathrm { aq }) } ^ { + } + 2 \mathrm { e } ^ { - } \rightarrow \mathrm { H } _ { 2 } (\mathrm {~g})$\\
$\mathrm { H } ^ { + } = 1 \mathrm { M } , \mathrm { P } _ { \mathrm { H } _ { 2 } } = 2 \mathrm {~atm}$\\
(Given $2.303 RT / F = 0.06$ V, $\log 2 = 0.3$)