Question

The reduction potential of hydrogen electrode having $pOH=4$ at $25^\circ C$ is

• A. Zero volt
• B. $-0.59V$
• C. $0.59V$
• D. $59mV$

Answer 1

Answer: (B)

$2H^{+}\left(\right.aq\left.\right)+2e^{-} \rightarrow H_{2}\left(\right.g\left.\right)$
$E=E^{0}-\frac{0 . 059}{n}log\frac{\left(PH\right)_{2}}{\left(H^{+}\right)^{2}}$
$E=0-\frac{0 . 059}{2}log\frac{1}{\left(\left(10\right)^{- 10}\right)^{2}}$ because $pOH=4$
So $pH=14-4=10\Rightarrow \left[H^{+}\right]=10^{- 10}molI^{- 1}$
Solving $E=-0.59V$
OR
$S.R.P$ .(hydrogen) $=-0.0591\times pH$
$pH=14-pOH$