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)

$2H^{+}(aq)+2e^{-}\to H_2(g)$
$E=E^0-\frac{0.059}{n}log\frac{{\left(PH\right)}_2}{{\left(H^{+}\right)}^2}$
$E=0-\frac{0.059}{2}log\frac{1}{{\left({\left(10\right)}^{-10}\right)}^2}$ because $pOH=4$
So $pH=14-4=10\Rightarrow \left[H^{+}\right]=10^{-10}mol{I}^{-1}$
Solving $E=-0.59V$
OR
$S.R.P$ .(hydrogen) $=-0.0591\times pH$
$pH=14-pOH$