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}