Question

Consider the following electrode processes of a cell, $C{l}^{-}\to \frac{1}{2}C{l}_2+e^{-}$.
$[MCl+e^{-}\to M+C{l}^{-}]$
If EMF of this cell is $-1.140$V and $E^o$ 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)

${10}^{-10}$

The half cell reactions are:

$C{l}^{-}\to \frac{1}{2}C{l}_2+e^{-}$.

$[MCl+e^{-}\to M+C{l}^{-}]$

The net cell reaction is

$MCl\to M+\frac{1}{2}C{l}_2$

$K_{sp}=\left[M^{+}\right]\left[C{l}^{-}\right]$

$E=E^o-\frac{0.059}{n}log\frac{{P_{C{l}_2}}^{1/2}}{\left[M^{+}\right]\left[C{l}^{-}\right]}$

Assuming $P_{Cl2}=1$ atm.

$E=E^o-\frac{0.059}{n}log\frac{1^{1/2}}{\left[M^{+}\right]\left[C{l}^{-}\right]}$

$E=E^o+\frac{0.059}{n}log\left[M^{+}\right]\left[C{l}^{-}\right]$

$E=E^o+\frac{0.059}{n}log\text{Ksp}$

$-1.14=-0.55+\frac{0.059}{1}log\text{Ksp}$

$-0.59=\frac{0.059}{1}log\text{Ksp}$

$-10=log\text{Ksp}$

$\therefore$ $\text{Ksp}={10}^{-10}$.