Complete the q...

Solution
a) Faraday’s two laws of electrolysis are as follows:
(i) First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).
(ii) Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation).

b) 
$$\begin{gathered} E_{cell}^0=\frac{0.059}{n}\log K_{eq} \\ {E}_{C{u}^{2+}/Cu}^0-E_{Z{n}^{2+}/Zn}^0=\frac{0.059}{2}\log K_{eq} \\ 0.34-\left(-0.76\right)=\frac{0.059}{2}\log K_{eq} \\ \log K_{eq}=\frac{1.1\times 2}{0.059} \\ \log K_{eq}=37.29 \\ {K}_{eq}=1.95\times {10}^{37} \end{gathered}$$