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

1.10 \times 10^{- 16}

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1.10 \times 10^{- 14}

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2.10 \times 10^{- 16}

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2.10 \times 10^{- 14}

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Solution

The correct option is A $1.10 \times 10^{- 16}$
Ag|AgI(0.05)M KI || (0.05)M $A g N O_{3}$ | Ag
$K_{s p}$ of AgI = $[A g^{+}] [I^{-}] = [A g^{+}] [0.05]$
For the given cell
$E_{c e l l} = E_{o x i d i s e d p r o d u c t (A g)} + E_{r e d u c e d p r o d u c t (A g)}$
Using Nerst equation:
$E_{c e l l} = E_{A g / A g^{+}} - \frac{0.059}{1} l o g [A g^{+}]_{L H S} + E_{A g^{+} / A g} + \frac{0.059}{1} l o g [A g^{+}]_{R H S}$
$0.788 = \frac{0.059}{1} l o g \frac{[A g^{+}]_{R H S}}{[A g^{+}]_{L H S}}$
$0.788 = 0.059 l o g \frac{0.05}{[A g^{+}]_{L H S}}$
$[A g^{+}]_{L H S} = 2.203 \times 10^{- 15}$
$∴ K_{s p} (A g I) = 2.203 \times 10^{- 15} \times 0.05 = 1.10 \times 10^{- 16}$

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Similar questions

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 $A g I$ .

A g | A g I | K I (0.05 M) | | A g N O_{3} (0.05 M) | A g

0.788 V

25^{\circ} C .

A g | A g I in K I (0.05 M) | | A g N O_{3} (0.05 M) | A g

A g I

25^{\circ} C

10^{- 14.63} \approx 2.34 \times 10^{- 15}

A g | A g I (s) (0.05 M) K I | | 0.05 M A g N O_{2} | A g

0.788 V

A g I

A g | A g {B r}_{(s)}, K B r (0.01 M) ∥ K C l (0.002 M), A g {C l}_{(s)} | {A g}_{(s)}

25^{o} C

K_{s p (A g C l)} = 4.8 \times 10^{- 10} M^{2}, K_{s p (A g B r)} = 2.4 \times 10^{- 14} M^{2}

\frac{2.303 R T}{F} = 0.06

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