Question

The electrochemical cell shown below is a concentration cell.
$M\mid M^{2+}$ (saturated solution of a sparingly soluble salt, $M{X}_2)\left|M^{2+}\left(0.001mold{m}^{-3}\right)\right|M$ The emf of the cell depends on the difference in concentrations of $M^{2+}$ ions at the two electrodes. The emf of the cell at $298K$ is $0.059V$ The solubility product $\left(K_{sp};mo{l}^{3}d{m}^{-9}\right)$ of $M{X}_2$ at $298K$ based on the information available for the given concentration cell is (take $2.303\times R\times 298/F=0.059V)$

• A. $1\times 10^{-15}$
• B. $4\times 10^{-15}$
• C. $1\times 10^{-12}$
• D. $4\times 10^{-12}$

Answer 1

Answer: (B)

$M\mid M^{+}($ sat. $)\Vert M^{2+}(0.001M)$

$\left(K_{mp}=?\right)$

emf of concentration cell.

$E_{\text{cell }}=\frac{-0.059}{n}\log \frac{{\left[M^{+2}\right]}_a}{{\left[M^{+2}\right]}_c}$

$0.059=\frac{0.059}{2}\log \frac{[0.001]}{{\left[M^{+2}\right]}_a}$

${\left[M^{+2}\right]}_a=10^{-5}=S$

(solubility of salt in saturated solution)

$\underset{(s)}{M{X}_2}\underset{(S)}{M^{+2}}+2x^{-}(aq)$

$K_{sp}=4S^3=4\times {\left(10^{-5}\right)}^3=4\times 10^{-15}$