The electrochemical cell shown below is a concentration cell $M\left|\underset{\left(SaturatedM{X}_2\right)}{M^{2+}}\right|\left|\underset{\left(0.001molar\right)}{M^{2+}}\right|M$  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 $(K_{sp};mo{l}^{3}d{m}^{-9})$ of  $M{X}_2$ at 298K based on the information available for the given concentration cell is $(take2.303R298/F=0.059V)$

${\text{M/M}}^{\text{2+}}\left({\text{sat.sol}}^{\text{n}}{\text{ofMX}}_{\text{2}}\right){\text{||M}}^{\text{2+}}\left({\text{0.001\hspace{0.17em}molL}}^{\text{-1}}\right)\text{||M}$

Anode ;  $\text{M}\to {\text{M}}^{+2}\left(Cmol{L}^{-1}\right)+2e^{-}$
Cathode  $M^{2+}\left(0.001mol{L}^{-1}\right)+2e^{-}\to M$
Cell reaction
 $M^{2+}\left(0.001mol{L}^{-1}\right)\to M^{2+}\left(Cmol{L}^{-1}\right);(n=2)$
For concentration cell : $E_{cell}^0=0$
${\text{E}}_{\text{cell}}^{\text{0}}\text{=-}\frac{\text{0.059}}{\text{n}}\text{log}\frac{{\left[{\text{M}}^{\text{2+}}\right]}_{\text{anode}}}{{\left[{\text{M}}^{\text{2+}}\right]}_{\text{cathode}}}$

$\text{0.059=-}\frac{0.059}{n}\log \left(\frac{c}{{10}^{-3}}\right)$

$$\begin{gathered} \log \left(\frac{c}{{10}^{-3}}\right)=-2=\log {10}^{-2} \\ C={10}^{-5}mol{L}^{-1} \\ M{X}_2\rightleftharpoons M^{2+}+2X^{-1} \end{gathered}$$

For saturated solution : $\left[{\text{M}}^{\text{2+}}\right]{\text{=10}}^{\text{-5}}{\text{mol\hspace{0.17em}L}}^{\text{-1}}$

$$\begin{gathered} \left[{\text{X}}^{\text{-}}\right]\text{=2}\left[{\text{M}}^{\text{2+}}\right]{\text{=2×10}}^{\text{-5}}\left(\text{M}\right) \\ {\text{K}}_{sp}=\left[M^{2+}\right]{\left[X^{-}\right]}^2=\left({10}^{-5}\right){(2\times {10}^{-5})}^2=4\times {10}^{-15} \end{gathered}$$