Q10 / 160 · TS EAMCET 2017 · Chemistry
Consider the following electrode processes of a cell,
$$
\mathrm{Cl}^{-} \rightarrow \frac{1}{2} \mathrm{Cl}_2+e^{-} \quad\left[\mathrm{MCl}+e^{-} \rightarrow M+\mathrm{Cl}^{-}\right]
$$
If EMF of this cell is $-1.140 \mathrm{~V}$ and $E^{\circ}$ value of the cell is $-0.55 \mathrm{~V}$ at $298 \mathrm{~K}$, the value of the equilibrium constant of the sparingly soluble salt $M \mathrm{Cl}$ is in the order of
- Correct answerA$10^{-10}$
- B$10^{-8}$
- C$10^{-7}$
- D$10^{-11}$
Solution
$\mathrm{MCl}+e^{-} \longrightarrow M+\mathrm{Cl}^{-}$cathode (reduction) $\mathrm{Cl}^{-} \longrightarrow \frac{1}{2} \mathrm{Cl}_2+e^{-}$anode (oxidation)
$$
\mathrm{MCl} \rightarrow \mathrm{M}+\frac{1}{2} \mathrm{Cl}_2
$$
The $K_C$ of the cell reaction is calculated from
Nernst equation $E_{\text {cell }}=E^{\circ}$ cell $-\frac{0.059}{n} \log K_c$
$$
\begin{aligned}
& -1.140=-0.55-\frac{0.059}{1} \log c \\
& -0.59=-0.059 \log K_c \\
& \log K_c=\frac{0.59}{0.059}=10 \\
& \therefore \quad K_C=10^{10} \\
& K_{s p} \text { is for } M+\frac{1}{2} \mathrm{Cl}_2 \longrightarrow M \mathrm{Cl} \longrightarrow M^{+}+\mathrm{Cl}^{-} \\
& \therefore K_{s p}=\frac{1}{K_c} \\
& \quad=\frac{1}{10^{10}}=10^{-10}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.