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(D) The reaction is $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$.Here,the number of electrons transferred,$n = 2$.The rela

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$e^{160}$

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(D) The reaction is $Zn_{(s)} + Cu^{2+}_{(aq)} ightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$.Here,the number of electrons transferred,$n = 2$.The relationship between standard cell potential and equilibrium constant is given by $\Delta G^{\circ} = -nFE^{\circ}_{cell} = -RT \ln K$.Rearranging for $\ln K$,we get $\ln K = \frac{nFE^{\circ}_{cell}}{RT}$.Substituting the given values: $n = 2$,$F = 96000 \ C \ mol^{-1}$,$E^{\circ}_{cell} = 2 \ V$,$R = 8 \ J \ K^{-1} \ mol^{-1}$,and $T = 300 \ K$.$\ln K = \frac{2 	imes 96000 	imes 2}{8 	imes 300}$.$\ln K = \frac{384000}{2400} = 160$.Therefore,$K = e^{160}$.

If the standard electrode potential for a cell is $2 \ V$ at $300 \ K,$ the equilibrium constant $(K)$ for the reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Zn^{2+}_{(aq)} + Cu_{(s)}$ at $300 \ K$ is approximately $(R = 8 \ J \ K^{-1} \ mol^{-1}, F = 96000 \ C \ mol^{-1})$

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