Give the equation to calculate the equilibrium constant $K_C$ of a Daniell cell.
(N/A) For a Daniell cell,the overall cell reaction is: $Zn(s) + Cu^{2+}(aq) \rightleftharpoons Zn^{2+}(aq) + Cu(s)$.
At equilibrium,the cell potential $E_{cell} = 0$.
According to the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$.
At equilibrium,$Q = K_C$ and $E_{cell} = 0$,so $0 = E^{\circ}_{cell} - \frac{0.0591}{n} \log K_C$.
Rearranging this gives: $\log K_C = \frac{n E^{\circ}_{cell}}{0.0591}$.
For the Daniell cell,$n = 2$,so the equation is: $\log K_C = \frac{2 E^{\circ}_{cell}}{0.0591}$ or $K_C = 10^{\frac{n E^{\circ}_{cell}}{0.0591}}$.
At equilibrium,the cell potential $E_{cell} = 0$.
According to the Nernst equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log Q$.
At equilibrium,$Q = K_C$ and $E_{cell} = 0$,so $0 = E^{\circ}_{cell} - \frac{0.0591}{n} \log K_C$.
Rearranging this gives: $\log K_C = \frac{n E^{\circ}_{cell}}{0.0591}$.
For the Daniell cell,$n = 2$,so the equation is: $\log K_C = \frac{2 E^{\circ}_{cell}}{0.0591}$ or $K_C = 10^{\frac{n E^{\circ}_{cell}}{0.0591}}$.
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