Explain the equilibrium state in a Daniell cell and derive its equilibrium constant. Given: $E^o_{cell} = 1.1 \ V$.

(N/A) The Daniell cell reaction is: $Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$.
At the anode,$Zn$ is oxidized to $Zn^{2+}$,increasing its concentration,while at the cathode,$Cu^{2+}$ is reduced to $Cu$,decreasing its concentration. Consequently,the cell potential decreases over time.
When the concentrations of $Zn^{2+}$ and $Cu^{2+}$ ions no longer change and the voltmeter reads $0 \ V$,the system has reached equilibrium.
At equilibrium: $Zn_{(s)} + Cu_{(aq)}^{2+} \rightleftharpoons Zn_{(aq)}^{2+} + Cu_{(s)}$.
The equilibrium constant is $K_C = \frac{[Zn^{2+}]}{[Cu^{2+}]}$.
Using the Nernst equation at equilibrium: $E_{cell} = E^o_{cell} - \frac{0.0591}{n} \log K_C = 0$.
Substituting $E^o_{cell} = 1.1 \ V$ and $n = 2$: $1.1 = \frac{0.0591}{2} \log K_C$.
$\log K_C = \frac{1.1 \times 2}{0.0591} \approx 37.22$.
$K_C = 10^{37.22} \approx 1.64 \times 10^{37}$.