The emf of the cell, $P t | H_{2} (1 atm), | H^{+} (0.1 M, 40 mL) | | A g^{+} (0.8 mM) | A g$ is $0.9 V$ . Calculate the emf when $60 mL$ of $0.05 M N a O H$ is added to the anodic compartment.
Given: $\frac{2.303 R T}{F} = 0.06, log 2 = 0.3$

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Solution

The cell given here is,
$P t | H_{2} (1 atm), | H^{+} (0.1 m, 40 mL) | | A g^{+} (0.8 M) | A g$
So according to Nernst equation,
$E_{c e l l} = E_{c e l l}^{\circ} - \frac{2.303 R T}{n F} log \frac{[H^{+}]}{[A g^{+}]}$
$\Rightarrow 0.9 = E_{c e l l}^{\circ} - \frac{0.06}{1} l o g \frac{0.1}{0.8}$
$\Rightarrow 0.9 = E_{c e l l}^{\circ} + 0.054$
$\Rightarrow E_{c e l l}^{\circ} = 0.846 V$
The molarity of $H^{+}$ ions after addition of $N a O H$
$= \frac{M_{1} V_{1} - M_{2} V_{2}}{V_{1} + V_{2}} = \frac{0.1 \times 40 - 0.05 \times 60}{40 + 60}$
= $\frac{(4 - 3) \times 10^{- 3}}{100} = 0.01 \times 10^{- 3} M$
Now, EMF of cell-
$E_{c e l l} = - E_{cell}^{\circ} - \frac{0.06}{1} log \frac{(0.01)}{(0.8)}$
$\Rightarrow E_{cell} = 0.846 - 0.06 log \frac{(0.01)}{(0.8)}$
$\Rightarrow E_{cell} = 0.846 + 0.06 log \frac{(0.8)}{(0.01)}$
$\Rightarrow E_{cell} = 0.846 + 0.06 log 80$
$\Rightarrow E_{cell} = 0.846 + 0.06 \times 1.90$
$\Rightarrow E_{cell} = 0.96 V$

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