Question

Electrode potential of hydrogen electrode is $18mV,$ then $[H^{+}]$ is

• A. 0.2
• B. 1
• C. 2
• D. 5

Answer 1

Answer: (C)

Given : $E_{{}_{H^{+}/H_2}}=18\times 10^{-3}V,\left[H^{+}\right]=?$

Applying Nernst equation,

$E_{{}_{H^{+}/H_2}}=E_{{}_{H^{+}/H_2}}^{\circ }-\frac{0.0591}{1}log\frac{1}{\left[H^{+}\right]}$

$18\times 10^{-3}V=0+0.0591log\left[H^{+}\right]$

$18\times 10^{-3}V=0.0591log\left[H^{+}\right]$

log $\left[H^{+}\right]=0.3046$

$\therefore \left[H^{+}\right]=$ antilog $\left(0.3046\right)=2.02=2.0$