Question

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

• A. $0.177V$
• B. $-0.177V$
• C. $0.087V$
• D. $0.548V$

Answer 1

The correct option is B $-0.177V$

The reaction at the hydrogen electrode is:
$H^{+}+e^{-}⟶\frac{1}{2}{H}_2\left(g\right)$
As, $pH=3,\left[H^{+}\right]={10}^{-3}M$
$E_{red}=E_{red}^{\circ }-\frac{0.0591}{n}log\frac{p_{H_2}^{\frac{1}{2}}}{\left[H^{+}\right]}$
$E_{red}=E_{red}^{\circ }-\frac{0.591}{1}log\frac{1^{\frac{1}{2}}}{{10}^{-3}}$
$=0+\frac{0.059}{1}log\left[{10}^{-3}\right]$
$=0.059\times -3=-0.177V$
Hence, (b) is correct.