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(D) For the reaction,$A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$Using the relation $\Delta G^{\circ} = -RT \ln K_c = -nFE^{

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$1.0 	imes 10^{20}$

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(D) For the reaction,$A_{(s)} + 2B^{2+}_{(aq)} ightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$Using the relation $\Delta G^{\circ} = -RT \ln K_c = -nFE^{\circ}_{cell}$Here,$n = 2$ (number of electrons transferred).At $298 \ K$,the relation simplifies to $\log K_c = \frac{n E^{\circ}_{cell}}{0.0591}$.Substituting the values: $\log K_c = \frac{2 	imes 0.59}{0.059} = 20$.Therefore,$K_c = 10^{20} = 1.0 	imes 10^{20}$.Hence,option $(D)$ is the correct answer.

If the $E^{\circ}_{cell}$ of an equilibrium reaction $A_{(s)} + 2B^{2+}_{(aq)} \rightleftharpoons A^{2+}_{(aq)} + 2B_{(s)}$ at $298 \ K$ is $0.59 \ V$,the equilibrium constant $K_c$ is

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