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The reduction potential of hydrogen electrode when placed in a buffer solution is found to be -0.413 V. The pH of the buffer is

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answer is C.

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Detailed Solution

$E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{1}{[H^{+}]} E_{cell} = 0 - \frac{0.0591}{1} (- \log [H^{+}]) \log [H^{+}] = \frac{- 0.413}{0.0591} = - 7 - \log [H^{+}] = 7 = p^{H}$

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