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

Question

Dr. Abubacker · Accepted Answer

(d)
$AgI\ (s)\  + \ e^{-}\  \rightleftharpoons \ \ \ Ag\ (s) + \ I^{-}\$ ; $E^{{^\circ}}$ = 0.152 V
$\ \ \ \ \ \ \ \ \ \ \ Ag\ (s) \longrightarrow \ {Ag}^{+} + \ e^{-}$ $E^{{^\circ}}$ = - 0.8 V
[table] $AgI\ (s) \longrightarrow \ {Ag}^{+}\ \  + \ I^{-}$ ${\ \ E}^{{^\circ}}$ = - 0.952 [/table]
$\ \ \ \ \ E_{cell}^{{^\circ}} = \ \frac{0.059}{n}\log K_{sp}$
$- 0.952\  = \ \frac{0.059}{1}\log K_{sp}$
$\log K_{sp} = \ \frac{- 0.952}{0.059}\  = \  - \ 16.135$

Q. Given, the data at 25 ℃ , $Ag\ + \ I^{-}\ \longrightarrow \ AgI\ + \ e^{-};\ \ E^{{^\circ}}\ = \ 0.152\ V$ $Ag\ \rightarrow \ \ {Ag}^{+}\ \ + \ e^{-};\ \ \ \ \ \ \ \ {\ \ E}^{{^\circ}}\ = \ - 0.800V$ What is the value of log $K_{sp}$ for AgI? $\left( 2.303\ \frac{RT}{F} = 0.059\ V \right)$

Explanation

Reviewed by Dr. Abubacker

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