When the same quantity of electricity is passed through solutions of $Ag^{+}$,$Cu^{2+}$,and $Fe^{3+}$ ions,what is the ratio of the number of moles of each metal deposited?

The number of moles of electrons required to deposit $36 \ g$ of $Al$ from an aqueous solution of $Al(NO_3)_3$ is (At. wt. of $Al = 27$)

The same quantity of electricity was passed through solutions of salts of elements $X$,$Y$,and $Z$ with atomic masses $7$,$27$,and $48$ respectively. The masses of $X$,$Y$,and $Z$ deposited were $2.1 \ g$,$2.7 \ g$,and $7.2 \ g$ respectively. The valencies of $X$,$Y$,and $Z$ respectively are:

During the electrolysis of concentrated $H_2SO_4$,perdisulphuric acid $(H_2S_2O_8)$ and $O_2$ are formed in equimolar amounts. The amount of $H_2$ that will be formed simultaneously is $(2H_2SO_4 \rightarrow H_2S_2O_8 + 2H^+ + 2e^-)$

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:

Charge required for reduction of $0.01 \ mol$ of $MnO_4^-$ in neutral medium is ............ $C$.