Question
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
✓
Answer
For hydrogen electrode, $\text{H}^++\text{e}^-\rightarrow\frac{1}{2}\text{H}_2$ it is given that pH = 10
$\therefore\ [\text{H}_+]=10\ \text{M}$
Now, using Nernst equation:
$\text{H}_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}=\text{E}^\ominus_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}-\frac{\text{RT}}{\text{nF}}\text{In}\frac{1}{[\text{H}^+]}$
$\text{E}^\ominus_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}-\frac{0.0591}{1}\text{log}\frac{1}{[\text{H}^+]}$
$=0-\frac{0.0591}{1}\text{log}\frac{1}{[10^{-10}]}$
= -0.0591 log 1010
= -0.591 V
$\therefore\ [\text{H}_+]=10\ \text{M}$
Now, using Nernst equation:
$\text{H}_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}=\text{E}^\ominus_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}-\frac{\text{RT}}{\text{nF}}\text{In}\frac{1}{[\text{H}^+]}$
$\text{E}^\ominus_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}-\frac{0.0591}{1}\text{log}\frac{1}{[\text{H}^+]}$
$=0-\frac{0.0591}{1}\text{log}\frac{1}{[10^{-10}]}$
= -0.0591 log 1010
= -0.591 V
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