Calculate the maximum work that can be obtained from the given electrochemical cell constructed with two metals $M$ and $N$ .
$[E_{M^{+ 2} / M}^{o} = - 0.76 V, E_{N^{+ 2}}^{o} / N = + 0.34 V]$
The cell reaction is $M + N^{+ 2} \to M^{+ 2} + N$

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Solution

According to given cell reaction
$E_{a n o d e (M^{+ 2} / M)}^{o} = - 0.76 V$
$E_{C a t h o d e (N^{+ 2} / N)}^{o} = + 0.34 V$
$n = 2$
Now, $W_{e l e c t r i c a l} = - n F E_{c e l l}^{o}$
$= - 2 \times 96500 \times [0.34 - (- 0.76)]$
$= - 2 \times 96599 \times 1.1$
$= - 212.3 K$ Joule.

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