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  4. Electrode potential of hydrogen electrode is 18 mV, then [H+] is

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Q. Electrode potential of hydrogen electrode is $18\, mV,$ then $[H^+]$ is

AIIMSAIIMS 2013Electrochemistry

Solution:

Given : $E_{_{H^{+}/H_{2}}} = 18\times10^{-3}\,V,\, \left[H^{+}\right]=?$

Applying Nernst equation,

$E_{_{H^{+}/H_{2}}}=E^{\circ}_{_{H^{+}/H_{2}}}-\frac{0.0591}{1}log \frac{1}{\left[H^{+}\right]}$

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

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

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

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