The equilibrium constant for a cell reaction, $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^4(\mathrm{aq})

Question

The equilibrium constant for a cell reaction, $\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^4(\mathrm{aq}) ightarrow \mathrm{Cu}^{24}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ is $4 	imes 10^{16} \cdot \mathrm{E}^0$ (cell) for the cell reaction is

Professor PC Thomas Classes · Answer

$E^0 = \frac{0.059}{n}\log K_c$. $n=2$, $K_c=4 	imes 10^{16}$. $E^0 \approx 0.0295 	imes 16.6 \approx 0.49$ V.