Calculate the emf of the cell in which the following reaction take place
$N i (s) + 2 A g^{+} (0.002 M) \to N i^{2 +} (0.160 M) + 2 A g I$
Given that $E^{o}$ cell $= 1.05$ v.

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Solution

$N i (s) + 2 A g^{+} (0.002 H) \to N i^{2 +} (0.160 M) + 2 A g I$
$E_{c e l l}^{o} = 1.05 V$
$E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{n} l o g \frac{[P]}{[R]}$
$E_{c e l l} = 1.05 - \frac{0.0591}{2} l o g \frac{[N i^{2 +}]}{[A g^{+ 2}]}$
$= 1.05 - \frac{0.0591}{2} l o g \frac{160}{0.002}$
$= 1.05 - 0.02955 l o g \frac{80 \times 10^{3}}{2}$
$= 1.05 - 0.02955 l o g (4 \times 10^{4})$
$= 1.05 - 0.02955 (4.602)$
$E_{c e l l} = 1.05 - 0.1359 = 0.92$
$∴$ The emf of the cell is $0.92 V$

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N i (s) + 2 A g^{+} (0.002 M) \to N i^{2 +} (0.160 M) + 2 A g (s)

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