The potential for the given half cell at \( 298 \, \text{K} \) is (−) _______ \( \times 10^{-2} \, \text{V} \).
\( 2\mathrm{H}^+(\mathrm{aq}) + 2e^- \rightarrow \mathrm{H}_2(\mathrm{g}) \)
\( [\mathrm{H}^+] = 1 \, \text{M}, \; P_{\mathrm{H}_2} = 2 \, \text{atm} \)
[Given : \( 2.303 \, \frac{RT}{F} = 0.06 \, \text{V}, \; \log 2 = 0.3 \)]
