Question

The electrochemical cell shown below is a concentration cell.
$M \mid M ^{2+}$ (saturated solution of a sparingly soluble salt, $\left. MX _{2}\right)\left| M ^{2+}\left(0.001 mol dm ^{-3}\right)\right| M$ 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 $298 K$ is $0.059 V$ The solubility product $\left( K _{ sp } ; mol ^{3} dm ^{-9}\right)$ of $MX _{2}$ at $298 K$ based on the information available for the given concentration cell is (take $2.303 \times R \times 298 / F =0.059 V )$

• 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. $) \| M ^{2+}(0.001 M )$

$\left( K _{ m p }=?\right)$

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)

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

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