Discussion Forum

The electrochemical cell shown below is a concentration cell $M|M^{2+}$ (saturated solution of a sparingly  soluble salt, $MX_{2})$|| $M^{2+}$ (0.001 mol dm^-3)|M. The emf of the cell depends on the difference in concentration of $M^{2+}$ ions at the two electrodes. The emf of  the cell at 298 is 0.059 V. The solubility product $ (k_{sp};mol^{3}dm^{-9})$  of MX, at 298 based on the information available the given concentration cell is (take 2.303 x R x 298/F= 0.059 V)

Answer:

Explanation:

For the given centration of the cell, the cell reactions are  $M \rightarrow M^{2+}$ at left hand electrode

$M^{2+} \rightarrow  M$  at right hand electrode

$\Rightarrow$ $M^{2+}$(RHS elecrode)

                     $\rightarrow  M^{2+}$ (LHS elelctrode)

$E^{0}=0$

Applying Nernst equation

 $E_{cell}=0.059=0-\frac{0.059}{2}$

$\log\frac{[M^{2+}]at LHS electrode}{0.001}$

$\Rightarrow$   $\log\frac{[M^{2+}]at LHS electrode}{0.001}=-2$

$\Rightarrow$  $[M^{2+}]$ at LHS electrode 

   =$10^{-2} \times 0.001=10^{-5} M$

The solubility equilibria for  $MX_{2}$ is 

$MX_{2}\rightleftarrows M^{2+}(aq)+2X^{-}(aq)$

 Solubility product , $K_{sp}=[M^{2+}][X^{-}]^{2}$

=  $10^{-5} \times (2 \times 10^{-5})^{2}=4 \times 10^{-15}$

  [$ \because  $  In satuarated solution of $MX_{2}$,$[X^{-}]$=2$[M^{2}]]$