Calculate the cell potential ( $E_{c e l l}$ ) of the following cell at $298 K$ .
$A g (s) | A g N O_{3} (0.01 M) | | A g N O_{3} (1.0 M) | A g (s)$

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Solution

As we know that,
$E_{c e l l} = {E^{\circ}}_{c e l l} - \frac{0.0591}{n} log \frac{C_{1}}{C_{2}}$
Given:- ${A g}_{(s)} | {A g N O_{3}}_{(0.001 M)} | | {A g N O_{3}}_{(1.0 M)} | {A g}_{(s)}$
Here,
$C_{1} = 10^{- 2} M$
$C_{2} = 1 M$
$n = 1$
When both electrode are of same metal,
${E^{\circ}}_{c e l l} = 0$
$∴ E_{c e l l} = 0 - \frac{0.0591}{1} log \frac{10^{- 2}}{1}$
$\Rightarrow E_{c e l l} = - 0.0591 \times (- 2) log 10$
$\Rightarrow E_{c e l l} = 0.1182 V$
Hence the cell potential of given cell is $0.1182 V$ .

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