The reduction electrode potential $E$ of $0.1$ M solution of $M^{+}$ ions $(E_{k p}^{o} = - 2.36 V)$ is :

- 2.41

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+ 2.41

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- 4.82

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None of the above

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Solution

The correct option is A $- 2.41$
Option $1$ is correct
Given that : [ $n = 1, E^{0} = 2.36 V; [m^{+}] = 0$ ]
$E = E^{0} - \frac{0.0591}{n} log \frac{1}{m^{+}}$
$E = - 2.36 - \frac{0.0591}{1} log \frac{1}{0.1}$
$E = - 2.36 - 0.0591$
$= - 2.41$

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Similar questions

0.1

M^{+} i o n s : (E_{R P} = - 2.36 V)

The single electrode potential of metal $E$ of $0.1$ M solution of M $^{+} / M$ ions [ $E$ $^{0}$ $= - 2.36 V$ ] is nearly:

The single electrode potential of $0.1 M$ solution of $M^{+} / M$ ion is:

( $E^{o} = 2.36 V$ )

M^{n +} / M

M^{n +}

E_{R P} = E_{R P}^{o} + \frac{0.059}{n} l o g [M^{n +}]

298

0.1

1

K_{a} =

1.8 \times 10^{- 8}

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