Electrochemistry 2 Question 26
«««< HEAD
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 $2.303 \times R \times 298 / \mathrm{F}=0.059 \mathrm{~V}$ )
======= ####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 $2.303 \times R \times 298 / \mathrm{F}=0.059 \mathrm{~V}$ )
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $1 \times 10^{-15}$
(b) $4 \times 10^{-15}$
(c) $1 \times 10^{-12}$
(d) $4 \times 10^{-12}$
Show Answer
Solution:
- For the given concentration cell, the cell reaction are $M \longrightarrow M^{2+}$ at left hand electrode.
$$ M^{2+} \longrightarrow M \text { at right hand electrode } $$
$\Rightarrow M^{2+}$ (RHS electrode) $\longrightarrow M^{2+}$ (LHS electrode)
$$ E^{\circ}=0 $$
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, $\left.K_{\text {sp }}=\right]\left[M^{2+}\right]\left[X^{-}\right]^{2}$
$$ =10^{-5} \times\left(2 \times 10^{-5}\right)^{2}=4 \times 10^{-15} $$
$\left[\because\right.$ In saturated solution of $\left.M X_{2},\left[X^{-}\right]=2\left[M^{2+}\right]\right]$
