Question

EMF of hydrogen electrode in term of $pH$ at ${25}^{o}C$ is

• A. $E_{H^{+}/H_2}=0.0591+pH$
• B. $E_{H^{+}/H_2}=0.0591-pH$
• C. $E_{H^{+}/H_2}=0.0591pH$
• D. $E_{H^{+}/H_2}=-0.0591pH$

Answer 1

The correct option is D $E_{H^{+}/H_2}=-0.0591pH$

For a hydrogen electrode,
Cathode cell reaction is
$H^{+}\left(aq\right)+e^{-}\to \frac{1}{2}{H}_2\left(g\right)$
According to Nernst equation,
$E_{H^{+}/H_2}=E_{H^{+}/H_2}^0-\frac{2.303RT}{nF}\text{log}\frac{p_{H_2}^{\frac{1}{2}}}{[H^{+}{]}^2}$

$E_{H^{+}/H_2}=E_{H^{+}/H_2}^0+\frac{2.303RT}{nF}log\frac{\left[H^{+}\right]}{p_{H_2}^{\frac{1}{2}}}$

Standard reductiom potential of hydrogen is zero

$E_{H^{+}/H_2}=+0.0591log\frac{\left[H^{+}\right]}{1}$


$E_{H^{+}/H_2}=-0.0591pH\because -log\left[H^{+}\right]=pH$