Question:
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Solution:
For hydrogen electrode, $\mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \frac{1}{2} \mathrm{H}_{2}$, it is given that $\mathrm{pH}=10$
∴[H+] = 10−10 M
Now, using Nernst equation:
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= _17-11-08_Utpal_12_Chemistry_3_15_html_m2be9aa1b.gif){kind=small}
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_17-11-08_Utpal_12_Chemistry_3_15_html_7be69ac0.gif){kind=small}
_17-11-08_Utpal_12_Chemistry_3_15_html_m2bd4c672.gif){kind=small}
= −0.0591 log 1010
= −0.591 V
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