The equilibrium constant of the reaction Cu_{ | vedclass

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Question

(A) The relationship between the standard cell potential $(E^{\circ})$ and the equilibrium constant $(K_{eq})$ is given by the Nernst equation at $298

Vedclass · Accepted Answer

$4.0 	imes 10^{15}$

Vedclass · Answer

(A) The relationship between the standard cell potential $(E^{\circ})$ and the equilibrium constant $(K_{eq})$ is given by the Nernst equation at $298 \ K$:$E^{\circ} = \frac{0.0591}{n} \log K_{eq}$Here,$n = 2$ (number of electrons transferred).Substituting the values:$0.46 = \frac{0.0591}{2} \log K_{eq}$$\log K_{eq} = \frac{0.46 	imes 2}{0.0591} \approx 15.566$$K_{eq} = 10^{15.566} = 10^{0.566} 	imes 10^{15} \approx 3.68 	imes 10^{15} \approx 4.0 	imes 10^{15}$.

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

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