Question

Electrode potential of hydrogen electrode is $18mV$. Then $\left[H^{+}\right]$ is:

• A. $0.2$
• B. $1$
• C. $2$
• D. $5$

Answer 1

The correct option is C $2$

$H^{+}+e^{-}\to \frac{1}{2}{H}_2$

$E_{cell}=E^0-\frac{0.059}{n}log\left[\frac{{\left(P_{H_2}\right)}^{1/2}}{\left(H^{+}\right)}\right]$

$0.018=0-\frac{0.059}{1}log\left[\frac{1}{\left(H^{+}\right)}\right]$

$\frac{0.018}{0.059}=log\left[\left(H^{+}\right)\right]$

$log\left[H^{+}\right]=0.305$

$\left[H^{+}\right]=2$