The emf of the cell $A g | A g I | K I (0.05 M) | | A g N O_{3} (0.05 M) A g$ is $0.788 V$ . Calculate the solubility product of $A g I$ .

25 X 10^-13

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1.10 x 10^-16

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8 x 10^-11

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6.6 x 10^-19

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Solution

The correct option is B 1.10 x 10^-16
$K_{S P}$ of $A g I = [{A g}^{+}] [I^{-}] = [A g^{+}] [0.05]$
For a given cell, $E_{C e l l} = E_{O x i d i s e d} + E_{R e d u c e d}$
Now using Nernst equation,
$E_{C e l l} = E_{O P A g / {A g}^{+}} - \frac{0.059}{1} l o g [{A g}^{+}]_{L H S} + E_{R P {A g}^{+} / A g} + \frac{0.059}{1} l o g [A g^{+}]_{R H S}$
$0.788 = \frac{0.059}{1} l o g \frac{[A g^{+}]_{R H S}}{[A g^{+}]_{L H S}} 0.788 = 0.059 l o g \frac{0.05}{[A g^{+}]_{L H S}}$
$∴ [A g^{+}]_{L H S} = 2.203 \times 10^{- 15} K_{S P} = 2.203 \times 10^{- 15} \times 0.05 = 1.10 \times 10^{- 16}$
Hence the solubility product of $A g I$ is $1.10 \times 10^{- 16}$

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