Question

Consider the following electrode process of a cell,
$\ce{Cl^{-1} -> \frac{1}{2} Cl_2 + e^{-}}$
$\ce{[MCl + e^{-} -> M + Cl^{-} ] }$
If EMF of this cell is $-1.140\, V$ and $E^0$ value of the cell is $-0.55\,V$ at $298\,K$, 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)

$M Cl + e ^{-} \longrightarrow M+ Cl ^{-}$ cathode (reduction)
$Cl ^{-} \longrightarrow \frac{1}{2} Cl _{2}+e^{-}$ anode (oxidation)
$M Cl \rightarrow M+\frac{1}{2} Cl _{2}$
The $K_{C}$ of the cell reaction is calculated from
Nernst equation $E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log K_{c}$
$-1.140=-0.55-\frac{0.059}{1} \,\log _{C}$
$-0.59=-0.059 \log K_{C}$
$\log K_{C}=\frac{0.59}{0.059}=10$
$\therefore K_{C}=10^{10}$
$K_{S p}$ is for $M+\frac{1}{2} Cl _{2} \longrightarrow M C l \longrightarrow M^{+}+ Cl ^{-}$
$\therefore K_{S p}=\frac{1}{K_{C}}$
$=\frac{1}{10^{10}}=10^{-10}$