Question

Consider the following electrode process of a cell,
$\mathrm{Cl}{}^{-1}\to \frac{1}{2}\mathrm{Cl}{}_2+\mathrm{e}{}^{-}$
$[\mathrm{MCl}+\mathrm{e}{}^{-}\to \mathrm{M}+\mathrm{Cl}{}^{-}]$
If EMF of this cell is $-1.140V$ and $E^0$ value of the cell is $-0.55V$ at $298K$, the value of the equilibrium constant of the sparingly soluble salt MCl is in the order of

• A. $10^{-10}$
• B. $10^{-8}$
• C. $10^{-7}$
• D. $10^{-11}$

Answer 1

Answer: (A)

$MCl+e^{-}⟶M+C{l}^{-}$ cathode (reduction)
$C{l}^{-}⟶\frac{1}{2}C{l}_2+e^{-}$ anode (oxidation)
$MCl\to M+\frac{1}{2}C{l}_2$
The $K_C$ of the cell reaction is calculated from
Nernst equation $E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.059}{n}\log K_c$
$-1.140=-0.55-\frac{0.059}{1}{\log }_C$
$-0.59=-0.059\log K_C$
$\log K_C=\frac{0.59}{0.059}=10$
$\therefore K_C=10^{10}$
$K_{Sp}$ is for $M+\frac{1}{2}C{l}_2⟶MCl⟶M^{+}+C{l}^{-}$
$\therefore K_{Sp}=\frac{1}{K_C}$
$=\frac{1}{10^{10}}=10^{-10}$