Question

What is the electrode potential of hydrogen electrode in contact with a solution whose ph is 10?

Answer 1

Explanation:

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Answer 2

Explanation:

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For hydrogen electrode,

$\sf H^{+}+e^{-}\longrightarrow \dfrac{1}{2}H_2$

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Given, pH = 10

$\sf \therefore\:[H^{+}]\:=\:10^{-10}M$

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Now, using Nernst equation:

$\sf H_{\left(H^{+},\: \dfrac{1}{2}H_2\right)}</p><p>\:=\:E_{\left(H^{+},\: \dfrac{1}{2}H_2\right)}^{\Theta}- \dfrac{RT}{nF}ln \dfrac{1}{[H^{+}]}$

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= $\sf E_{\left(H^{+},\: \dfrac{1}{2}H_2\right)}^{\Theta}- \dfrac{0.0591}{1}log \dfrac{1}{[H^{+}]}$

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= $\sf 0- \dfrac{0.0591}{1}log \dfrac{1}{[10^{-10}]}$

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= $\sf -0.0591\,log\,10^{10}$

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= $\sf -0.591V$