Question

The emf of the given cell is $0.788\text{V}$ at ${25}^{\circ }\text{C}.$
$Ag|AgI\text{in}KI\left(0.05\text{M}\right)||AgN{O}_3\left(0.05\text{M}\right)|Ag$

Calculate the value of solubility product of $AgI$ at ${25}^{\circ }\text{C}$

Take:
${10}^{-14.63}\approx 2.34\times {10}^{-15}$

• A. $1.17\times {10}^{-16}$
• B. $4.88\times {10}^{-5}$
• C. $2.34\times {10}^{-8}$
• D. $7.23\times {10}^{-20}$

Answer 1

The correct option is A $1.17\times {10}^{-16}$

$Ag|AgI\text{in}KI\left(0.05\text{M}\right)||AgN{O}_3\left(0.05\text{M}\right)|Ag$

For the given cell,
$E_{cell}=\frac{0.0591}{1}log\frac{[A{g}^{+}{]}_{\text{RHS}}}{[A{g}^{+}{]}_{\text{LHS}}}.....\left(1\right)$

In cathode, we have $0.05\text{M}$ solution of $AgN{O}_3$
Thus,
$[A{g}^{+}{]}_{\text{RHS}}=0.05\text{M}$

Given,
$E_{cell}=0.788\text{V}$

Substituting the values in equation (1), we get,
$0.788=\frac{0.0591}{1}log\frac{0.05}{[A{g}^{+}{]}_{\text{LHS}}}$

$0.788=\frac{0.0591}{1}log0.05-\frac{0.0591}{1}log[A{g}^{+}{]}_{\text{LHS}}$

$0.788=\frac{0.0591}{1}\times (0.7-2)-\frac{0.0591}{1}log[A{g}^{+}{]}_{LHS}$

$0.788=-\frac{0.0591}{1}\times 1.3-\frac{0.0591}{1}log[A{g}^{+}{]}_{\text{LHS}}$
$\frac{-0.788}{0.0591}-1.3=log[A{g}^{+}{]}_{\text{LHS}}$

$-14.63\approx log[A{g}^{+}{]}_{\text{LHS}}$

$[A{g}^{+}{]}_{\text{LHS}}=2.34\times {10}^{-15}$

$AgI$ is dissolved in $0.05\text{M}$ of $KI$ solution
$\therefore$ $\left[I^{-}\right]=0.05\text{M}$
For $AgI$,
$AgI\rightleftharpoons A{g}^{+}+I^{-}$
$K_{sp}=\left[A{g}^{+}\right]\left[I^{-}\right]$
$\Rightarrow K_{sp}=2.34\times {10}^{-15}\times 0.05$
$\Rightarrow K_{sp}=1.17\times {10}^{-16}$