At $25^{o} C$ , calculate electrode potential of $M g^{+} / M g$ electrode in which concentration of $M g^{+}$ is $0.1 M$ .
( $E_{M g^{+} / M g}^{o} = - 2.36 V$ , $R = 8.14 J / K m o l^{- 1}$ $1 F = 96500$ coulomb/mol)

- 4.75 V

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- 2.90 V

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- 2.41 V

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- 3.95 V

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Solution

The correct option is D $- 2.41 V$
$M g^{+} + e^{-} \to M g, E^{0} c e l l = - 2.36 V E_{c e l l} = E_{c e l l}^{0} - \frac{R T}{n F} l n \frac{1}{[M g^{+}]} E_{c e l l} = - 2.30 V - \frac{8.14 \times 298 K}{1 \times 96500} = l n \frac{1}{0.1} E_{c e l l} = - 2.41 V$

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