What is achieved by Faraday using the principles of electrolysis?

Two electrolytic cells are connected in series containing $CuSO_4$ solution and molten $AlCl_3$. If in electrolysis $0.4$ moles of $Cu$ are deposited on the cathode of the first cell,the number of moles of $Al$ deposited on the cathode of the second cell is: (in $moles$)

Consider the following reaction:
$MnO_4^- + 8H^{+} + 5e^- \rightarrow Mn^{2+} + 4H_2O$,$E^\circ = 1.51 \ V$
The quantity of electricity required in Faraday to reduce five moles of $MnO_4^-$ is ..... .

What is the number of faraday required to form $1 \ mol \ H_2$ by reduction of $H^{+}$ ions?

Three voltameters containing aqueous solutions of $H_{2}SO_{4}$,$CuSO_{4}$,and $AgNO_{3}$ are connected in series as shown in the figure. $A$ current was passed for $10 \ hours$. $10.8 \ g$ of $Ag$ was deposited at the cathode in the $(III)$ electrolytic cell during electrolysis,given that the current efficiency is $50 \%$. If $Z_{1}$,$Z_{2}$,and $Z_{3}$ are the electrochemical equivalents for the formation of $H_{2}$,$Cu$,and $Ag$ respectively,then the ratio $Z_{1} : Z_{2} : Z_{3}$ is:

When electricity is passed through a solution of $AlCl_3$,$13.5 \ g$ of $Al$ is deposited. What is the number of Faradays used?