The equilibrium constant of the reaction: $\mathrm{Cu}_{(s)}+2 \mathrm{Ag}_{(a q)}^{+} \rightarrow \mathrm{Cu}_{(a q)}^{2+}+2 \mathrm{Ag}_{(s)}$ $\mathrm{E}^{\mathrm{o}}=0.46 \mathrm{~V}$ at $298 \mathrm{~K}$ is:

Question

Aicharya · Accepted Answer

As $\mathrm{E}_{\mathrm{cell}}^0=\frac{0.0591}{n} \log \mathrm{K}_c$
$\begin{aligned}
& \therefore 0.46=\frac{0.0591}{2} \log \mathrm{K}_c \\
& \therefore \log \mathrm{K}_c=\frac{2 \times 0.46}{0.0591}=15.57 \\
& \text {or } \mathrm{K}_c=\text { Antilog } 15.57=3.7 \times 10^{15}
\end{aligned}$

The equilibrium constant of the reaction: $\mathrm{Cu}_{(s)}+2 \mathrm{Ag}_{(a q)}^{+} \rightarrow…

Solution