The hydrogen electrode is dipped in a solution of$\mathrm{pH}=3 \mathrm{at} {25}^0\mathrm{C}$. The potential of the cell would be (the value of 2.303 RT/F is 0.059 V)

Here we have pH=3 i.e $\log \frac{1}{{\mathrm{H}}^{+}}=3$ and ${\mathrm{E}}_{\mathrm{cell}}^0=0.0\mathrm{V}$

Reaction will be,

${\mathrm{H}}^{+}+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{H}}_2$  ,  n=1

By Nernst equation

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^0-\frac{0.0591}{\mathrm{n}}\log \frac{{\left[{\mathrm{H}}_2\right]}^{{\displaystyle 1/2}}}{\left[{\mathrm{H}}^{+}\right]}$

By putting values in equation we get

${\mathrm{E}}_{\mathrm{cell}}0=0-\frac{0.0591}{1}\log \frac{1}{\left[{\mathrm{H}}^{+}\right]}$

${\mathrm{E}}_{\mathrm{cell}}=-0.0591\times 3=-0.177\mathrm{V}$

${\mathrm{E}}_{\mathrm{cell}}=-0.177\mathrm{V}$

Hence option C is correct.