Question

How much will be the reduction potential of a hydrogen Electrode change when its solution initially at pH = 0 Is neutralized to pH = 7?

• A. Increases by 0.059 V
• B. Decreases by 0.059 V
• C. Increases by 0.41V
• D. Decreases by 0.41 V

Answer 1

The correct option is D Decreases by 0.41 V

$H^{+}+e^{-}\to \frac{1}{2}{H}_2$
$E_{red}=E_{red}^0-\frac{0.0591}{2}log\left(\frac{1}{\left[H^{+}\right]}\right)$
$=E_{red}^0+0.0591log\left[H^{+}\right]$
$E_{red}^0=0$, when pH = O, $\left[H^{+}\right]={10}^0=1M$
$\therefore E_{red}=0$
When, $pH=7,\left[H^{+}\right]={10}^{-7}M$
$E_{red}=0.0591log{10}^{-7}$
$=-0.4137$
Thus, $E_{red}$ decreases by 0.41V