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08/01/2022 6:38 pm
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Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
1 Answer
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08/01/2022 6:46 pm
For hydrogen electrode, H+ + e- → \(\frac{1}{2}H_2\)
it is given that pH = 10
∴ [H+] = 10-10M
Now, using Nernst equation:
\(H_{\frac{H^+}{\frac{1}{2}}H_2}\) = \(E^{\ominus}_{\frac{H^+}{\frac{1}{2}}H_2}\) - \(\frac{RT}{nF}\)ln \(\frac{1}{[H^+]}\)
= \(E^{\ominus}_{\frac{H^+}{\frac{1}{2}}H_2}\) - \(\frac{0.0591}{1}\)log \(\frac{1}{[H^+]}\)
= 0 - \(\frac{0.0591}{1}\)log \(\frac{1}{[10^{-10}]}\)
= -0.0591 log 1010
= -0.591 V