The Nernst equation is given by E = E^o - fra | vedclass

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(C) At equilibrium,the cell potential $E$ becomes $0$.Substituting $E = 0$ and $Q = K_{eq}$ into the Nernst equation: $0 = E^o - \frac{RT}{nF} \ln K_{

Vedclass · Accepted Answer

$E = E^o$

Vedclass · Answer

(C) At equilibrium,the cell potential $E$ becomes $0$.Substituting $E = 0$ and $Q = K_{eq}$ into the Nernst equation: $0 = E^o - \frac{RT}{nF} \ln K_{eq}$.This implies $E^o = \frac{RT}{nF} \ln K_{eq}$.Rearranging for $K_{eq}$,we get $\ln K_{eq} = \frac{n E^o F}{RT}$,which leads to $K_{eq} = e^{\frac{n E^o F}{RT}}$.Therefore,the statement $E = E^o$ is incorrect because at equilibrium $E = 0$,not $E^o$.

The Nernst equation is given by $E = E^o - \frac{RT}{nF} \ln Q$. If $Q = K_{eq}$,then which of the following is not correct?

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