Given the data at $25^{\circ} \mathrm{C}$, $\mathrm{Ag}+\mathrm{I}^{-} \longrightarrow \mathrm{AgI}+\mathrm{e}^{-} ; \mathrm{E}^{\circ}=0.152 \mathrm{~V}$ $\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-} ; \quad \mathrm{E}^{\circ}=-0.800 \mathrm{~V}$ What is the value of $\log \mathrm{K}_{\mathrm{sp}}$ for AgI ? $\left(2.303 \frac{\mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}\right)$

JEE Main · 2006 · 30 Apr 2006 · Offline · Redox Reactions and Electrochemistry

Given the data at

2 5^{\circ} C

Ag + I^{-} ⟶ AgI + e^{-}; E^{\circ} = 0.152 V

Ag ⟶ Ag^{+} + e^{-}; E^{\circ} = - 0.800 V

What is the value of

lo g K_{sp}

for AgI ?

(2.303 \frac{RT}{F} = 0.059 V)

Answer: (D)

The solubility product of AgI belongs to

AgI (s) ⇌ A g^{+} + I^{-}

, which can be built from the two given half-reactions. Reversing the first gives

AgI + e^{-} \to Ag + I^{-}

with

E^{\circ} = - 0.152 V

, and adding the oxidation

Ag \to A g^{+} + e^{-}

with

E^{\circ} = - 0.800 V

yields the dissolution with

E^{\circ} = - 0.952 V

. Using

E^{\circ} = \frac{0.059}{n} lo g K_{s p}

with

n = 1

lo g K_{s p} = \frac{- 0.952}{0.059} = - 16.13

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option