Answer:
Option A
Explanation:
$MCl+e^{-} \rightarrow M+Cl^{-}$ cathode (reduction)
$Cl^{-} \rightarrow \frac{1}{2} Cl_{2}+e^{-}$ anode (oxidation)
$MCl \rightarrow M+ \frac{1}{2}Cl_{2}$
The $K_{c}$ of the cell reaction is calculated from
Nernst equation $E_{cell}= E^{0}_{cell}- \frac{0.059}{n} \log K_{c}$
$-1.140=-0.55-\frac{0.059}{1} \log c$
$-0.59=-0.059 \log K_{c}$
$\log K_{c}= \frac{0.59}{0.059} =10$
$\therefore K_{c}=10^{10}$
$K_{sp}$ is for $M+\frac{1}{2} Cl_{2} \rightarrow MCl \rightarrow M^{+}+Cl^{-}$
$\therefore$ $K_{sp}= \frac{1}{ K_{c}}$
$=\frac{1}{10^{10}}=10^{-10}$