Electrochemistry Result Question 25
Passage
The electrochemical cell shown below is a concentration cell. $M \mid M^{2+}$ (saturated solution of a sparingly soluble salt, $\left.M X_2\right)|| M^{2+}\left(0.001 \mathrm{~mol} \quad \mathrm{dm}^{-3}\right) \mid M$. The emf of the cell depends on the difference in concetration of $M^{2+}$ ions at the two electrodes. The emf of the cell at $298 K$ is $0.059 V$ .
(2012)
25. The solubility product $\left(K_{\mathrm{sp}}: \mathrm{mol}^3 \mathrm{dm}^{-9}\right)$ of $M X_2$ at 298 based on the information available the given concentration cell is (take $2.303 \times R \times 298 / \mathrm{F}=0.059 \mathrm{~V}$ )
(a) $1 \times 10^{-15}$
(b) $4 \times 10^{-15}$
(c) $1 \times 10^{-12}$
(d) $4 \times 10^{-12}$
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Answer:
Correct Answer: 25. ( b )
Solution:
For the given concentration cell, the cell reaction are $M \longrightarrow M^{2+}$ at left hand electrode.
$ \begin{array}{r} M^{2+} \longrightarrow M \text { at right hand electrode } \\ \Rightarrow M^{2+} \text { (RHS electrode) } \longrightarrow M^{2+} \text { (LHS electrode) } \\ E^{\circ}=0 \end{array} $
Applying Nernst equation
$ \begin{aligned} & E_{\text {cell }}=0.059=0-\frac{0.059}{2} \log \frac{\left[M^{2+}\right] \text { at LHS electrode }}{0.001} \\ & \Rightarrow \log \frac{\left[M^{2+}\right] \text { at LHS electrode }}{0.001}=-2 \\ & \Rightarrow\left[M^{2+}\right] \text { at LHS electrode }=10^{-2} \times 0.001=10^{-5} \mathrm{M} \end{aligned} $
The solubility equilibrium for $M X_2$ is
$ M X_2(s) \rightleftharpoons M^{2+}(a q)+2 X^{-}(a q) $
Solubility product, $\left.K_{\mathrm{sp}}=\right]\left[M^{2+}\right]\left[X^{-}\right]^2$
$ =10^{-5} \times\left(2 \times 10^{-5}\right)^2=4 \times 10^{-15} $
$\left[\because\right.$ In saturated solution of $\left.M X_2,\left[X^{-}\right]=2\left[M^{2+}\right]\right]$
BETA
