The equilibrium constant of the reaction:Cu_{ | vedclass

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(C) For a cell reaction in equilibrium at $298 \ K$,the relationship between the standard cell potential and the equilibrium constant is given by:$E_{

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$4.0 	imes 10^{15}$

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(C) For a cell reaction in equilibrium at $298 \ K$,the relationship between the standard cell potential and the equilibrium constant is given by:$E_{cell}^{\circ} = \frac{0.0591}{n} \log K_{C}$where $K_{C}$ is the equilibrium constant and $n$ is the number of electrons transferred in the balanced redox reaction.For the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} ightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$,the number of electrons transferred is $n = 2$.Given $E_{cell}^{\circ} = 0.46 \ V$,we substitute the values into the equation:$0.46 = \frac{0.0591}{2} \log K_{C}$$\log K_{C} = \frac{0.46 	imes 2}{0.0591} = \frac{0.92}{0.0591} \approx 15.567$$K_{C} = 10^{15.567} \approx 3.69 	imes 10^{15} \approx 4.0 	imes 10^{15}$

The equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is:

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The equilibrium constant of the reaction:$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is: