The reduction electrode potential, E of $0.1$ M solution of $M^{+} i o n s : (E_{R P} = - 2.36 V)$ is

- 4.82 V

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- 2.41 V

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+ 2.41 V

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None of these

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Solution

The correct option is A $- 2.41 V$
$E = E_{R P}^{\circ} + \frac{0.0591}{n} log [M^{+}]$
Given, $E_{R P}^{\circ} = - 2.36 V, [M^{+}] = 0.1 M$
$n = 1$ (for $M^{+} \to M)$
$E = E_{R P}^{\circ} + \frac{0.0591}{n} log [M^{+}]$
$= - 2.36 + \frac{0.0591}{1} log 0.1$
$= - 2.36 + 0.0591 \times (- 1)$
$= - 2.36 - 0.0591$
$= - 2.419 V$

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