Question

The emf of the cell $Ag|AgI|KI\left(0.05M\right)||AgN{O}_3\left(0.05M\right)Ag$ is $0.788V$. Calculate the solubility product of $AgI$.

• A. 25 X 10^-13
• B. 1.10 x 10^-16
• C. 8 x 10^-11
• D. 6.6 x 10^-19

Answer 1

The correct option is B 1.10 x 10^-16

$K_{SP}$ of $AgI=\left[{Ag}^{+}\right]\left[I^{-}\right]=\left[A{g}^{+}\right]\left[0.05\right]$
For a given cell, $E_{Cell}=E_{Oxidised}+E_{Reduced}$
Now using Nernst equation,
$E_{Cell}=E_{OPAg/{Ag}^{+}}-\frac{0.059}{1}log[{Ag}^{+}{]}_{LHS}+E_{RP{Ag}^{+}/Ag}+\frac{0.059}{1}log[A{g}^{+}{]}_{RHS}$
$0.788=\frac{0.059}{1}log\frac{[A{g}^{+}{]}_{RHS}}{[A{g}^{+}{]}_{LHS}}0.788=0.059log\frac{0.05}{[A{g}^{+}{]}_{LHS}}$
$\therefore [A{g}^{+}{]}_{LHS}=2.203\times {10}^{-15}{K}_{SP}=2.203\times {10}^{-15}\times 0.05=1.10\times {10}^{-16}$
Hence the solubility product of $AgI$ is $1.10\times {10}^{-16}$