The Nernst equation for the reduction potential of a non-metal X when $[X^{n -}] = a$ is given by

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$E^{0} - \frac{0.0591}{n} \log a_{10}^{n}$

$E^{0} + \frac{0.0591}{n} \log a_{10}$

$E^{0} - \frac{0.0591}{n} \log a_{10}$

answer is B.

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Detailed Solution

$\begin{aligned} X_{s} + n \bar{e} \to X_{aq}^{n -} \\ E = E^{0} - \frac{0.0591}{n} \log_{10} (x^{n -}) \\ = E^{0} - \frac{0.0591}{n} \log a_{10} \end{aligned}$

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