Calculate the quantity of electricity required to produce $0.42 \ g$ of $Ag$ at the cathode during the electrolysis of an $AgNO_3$ solution. (Molar mass of $Ag = 108 \ g \ mol^{-1}$) (in $C$)

$0.592 \ g$ of copper is deposited in $60 \ minutes$ by passing $0.5 \ amperes$ current through a solution of copper$(II)$ sulphate. The electrochemical equivalent of copper$(II)$ (in $g \ C^{-1}$) is $\left(F=96500 \ C \ mol^{-1}\right)$

The electrochemical equivalent of a metal is $x \ g \ C^{-1}$. The equivalent weight of the metal is:

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 electric current is passed through acidified water for $1930 \,s$, $1120 \,mL$ of $H_2$ gas is collected (at $STP$) at the cathode. What is the current passed in amperes?

How much charge is required for the following reductions:
$(i)$ $1 \, mol$ of $Al^{3+}$ to $Al$.
$(ii)$ $1 \, mol$ of $Cu^{2+}$ to $Cu$.
$(iii)$ $1 \, mol$ of $MnO_{4}^{-}$ to $Mn^{2+}$.