Clarify the stoichiometry of the reactions occurring on the electrodes of an electrolytic cell and its relationship with the quantity of electricity.

The amount of electricity (or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction.
For example,in the reaction: $Ag_{(aq)}^{+} + e^{-} \rightarrow Ag_{(s)}$
One mole of electrons is required for the reduction of one mole of silver ions. The charge on one mole of electrons is calculated as:
$= N_{A} \times 1.6021 \times 10^{-19} \ C$
$= 6.022 \times 10^{23} \ mol^{-1} \times 1.6021 \times 10^{-19} \ C \approx 96487 \ C \ mol^{-1}$
This quantity of electricity is called $1$ Faraday $(F)$. For calculations,we use $1 \ F \approx 96500 \ C \ mol^{-1}$.
In electrolytic reactions,the total electricity $(Q)$ is given by $Q = I \times t$,where $I$ is current in amperes and $t$ is time in seconds.
Steps to calculate the required electricity:
$(i)$ Write the balanced half-reaction occurring on the electrode.
$(ii)$ Determine the number of moles of electrons $(n)$ required to produce $1 \ mol$ of the product.
$(iii)$ The required charge is $n \times F$.
Examples:
$(i)$ For $Ag$: $Ag_{(aq)}^{+} + e^{-} \rightarrow Ag_{(s)}$. $1 \ mol$ of $Ag$ requires $1 \ F$.
$(ii)$ For $Mg$: $Mg_{(aq)}^{2+} + 2e^{-} \rightarrow Mg_{(s)}$. $1 \ mol$ of $Mg$ requires $2 \ F$.
$(iii)$ For $Al$: $Al_{(aq)}^{3+} + 3e^{-} \rightarrow Al_{(s)}$. $1 \ mol$ of $Al$ requires $3 \ F$.