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प्रश्न
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
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उत्तर
For the hydrogen electrode,
\[\ce{H+ + e- -> 1/2 H2}\]
From the Nernst equation,
\[\ce{E_{H^+/H_2} = E^\circ_{H^+/H_2} -\frac{0.0591}{n} log \frac{1}{[H+]}}\]
= \[\ce{0 -\frac{0.0591}{1} log \frac{1}{10^{-10}}}\] ...[∵ H+ = 1 × 10−pH]
= − 0.0591 × 10
= − 0.591 V
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