Correct Option : (C)
Explanation :
${pH}=7$ for water.
$-\log \left[{H}^{+}\right]=7 \Rightarrow\left[{H}^{+}\right]=10^{-7}$
$2 {H}_{(a q)}^{+}+2 e^{-} \longrightarrow {H}_{2({~g})}$
$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{p_{{H}_2}}{\left[{H}^{+}\right]^2}$
$0=0-\frac{0.0591}{2} \log \frac{p_{{H}_2}}{\left(10^{-7}\right)^2}$
$\log \frac{p_{{H}_2}}{\left(10^{-7}\right)^2}=0 \Rightarrow \frac{p_{{H}_2}}{\left(10^{-7}\right)^2}=1 \quad[\because \log 1=0]$
$p_{{H}_2}=10^{-14} {atm}$