Question

Calculate the potential of hydrogen electrode in contact with a solution of $5\times {10}^{-3}M$ of $Ba(OH{)}_2$ at 1 atm pressure and 298K temperature:

• A. -0.059V
• B. 0.059V
• C. 0.59V
• D. -0.59V

Answer 1

The correct option is D -0.59V

$H^{+}+e^{-}\to \frac{1}{2}{H}_2$
$E_{red}=E_{red}^0-\frac{0.0591}{1}\left[log(P_{H_2}{)}^{1/2}/\left[H^{+}\right]\right]$
$=0-\frac{0.0591}{1}log\frac{1}{H^{+}}$
$=-0.0591\times pH$
$=-0.0591\times 10=-0.591V$
$pOH=-log{10}^{-2}\left[O{H}^{-}\right]=2\times 0.5\times {10}^{-3}={10}^{-2}$
$[pOH=2pH=10$