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Question

(A) The cell reactions are: Anode (oxidation): $Cl^{-} \rightarrow \frac{1}{2} Cl_2 + e^{-}$ Cathode (reduction): $MCl + e^{-} \rightarrow M + Cl^{-}$

Vedclass · Accepted Answer

$10^{-10}$

Vedclass · Answer

(A) The cell reactions are: Anode (oxidation): $Cl^{-} \rightarrow \frac{1}{2} Cl_2 + e^{-}$ Cathode (reduction): $MCl + e^{-} \rightarrow M + Cl^{-}$ Overall cell reaction: $MCl \rightarrow M + \frac{1}{2} Cl_2$ Using the Nernst equation at equilibrium ($E_{cell} = 0$ is not applicable here as the cell is not at equilibrium,but we use the relation $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$): Given $E_{cell} = -1.140 \ V$,$E^{\circ}_{cell} = -0.55 \ V$,$n = 1$: $-1.140 = -0.55 - 0.0591 \log K_c$ $-0.59 = -0.0591 \log K_c$ $\log K_c = \frac{0.59}{0.0591} \approx 10$ $K_c = 10^{10}$ Since the reaction $MCl \rightarrow M + \frac{1}{2} Cl_2$ is the reverse of the solubility product reaction $M^{+} + Cl^{-} \rightarrow MCl$,the equilibrium constant $K_c$ is the reciprocal of $K_{sp}$. $K_{sp} = \frac{1}{K_c} = \frac{1}{10^{10}} = 10^{-10}$

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