The hydrogen electrode is dipped in a solution of pH = 3 at 25 ° C . The potential of the electrode will be…

Given,
pH =$3$

$$\begin{gathered} \Rightarrow - \log [ {\mathrm{H}}^{+}] = 3 \\ [ {\mathrm{H}}^{+}] ={10}^{-3} \end{gathered}$$

The half cell reaction is

$2{\mathrm{H}}_{(\mathrm{aq}.)}^{+}+2{\mathrm{e}}^{-}\to {{\mathrm{H}}_2}_{\left(\mathrm{g}\right)}$

We know,

the Nernst equation is
$$\begin{gathered} {\mathrm{E}}_{\mathrm{red}}={{\mathrm{E}}^0}_{\mathrm{red}}+0.059 \log \left[{\mathrm{H}}^{+}\right] \\ =0+0.059 \log {10}^{-3} \\ =+ 0.059 \times (-3) \\ =- 0.177 \mathrm{V} \end{gathered}$$

$= -17.7\times {10}^{-2}\mathrm{V}$.