What is the electrode potential (in V) of the following electrode at $25^{o}$ C?
$N i^{2 +} (0.1 M) | N i (s)$
[Standard reaction potential of $N i^{2 +} | N i$ is $- 0.25$ V, $\frac{2.303 R T}{F} = 0.06$ ]

- 0.28

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- 0.34

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- 0.82

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- 0.02

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Solution

The correct option is A $- 0.28$ V
${N i}^{+ 2} + 2 e^{-} ⟶ N i$

From Nernst equation,

$E = E^{0} - \frac{2.303 R T}{F} log K_{c} = E^{0} - \frac{0.059}{n} log K_{c}$

Where
$E$ =Electrode potential
$E^{0}$ =Standard Electrode potiential = $- 0.25 V$
$n$ = no. of electrons transfered
$K_{c}$ = equilibrium constant $= \frac{[N i]}{[{N i}^{2 +}]} = 0.1$

$E = E^{0} - \frac{0.059}{n} log K_{c} = - 0.25 - (- 0.06 / 2) \times log \frac{1}{0.1} = - 0.28 V$

Hence, the correct option is $A$

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What is the electrode potential (in V) of the following electrode at 25oC? Ni2+(0.1M)|Ni(s) [Standard reaction potential of Ni2+|Ni is −0.25V, 2.303RTF=0.06]

What is the electrode potential (in V) of the following electrode at $25^{o}$ C?
$N i^{2 +} (0.1 M) | N i (s)$
[Standard reaction potential of $N i^{2 +} | N i$ is $- 0.25$ V, $\frac{2.303 R T}{F} = 0.06$ ]