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Q.
The quantity of electricity needed to electrolyse completely $1 \,M$ solution of $CuSO_{4}$, $Bi_{2} \left(SO_{4}\right)_{3'} AlCl_{3}$ and $AgNO_{3}$ each will be
Electrochemistry
Solution:
$CuSO_{4}+2e^{-} \rightarrow Cu+SO_{4}^{2-}$
$Bi_{2} \left(SO_{4}\right)_{3}+6e^{-}\rightarrow 2Bi+3SO^{2-}_{4}$
$AlCl_{3}+3e^{-}\rightarrow Al+3Cl^{-}$
$AgNO_{3}+e^{-}\rightarrow Ag+NO_{3}^{-}$
Since, in $1 \,M \,CuSO_{4}$ solution, $1 \,M \,Bi_{2}\left(SO_{4}\right)_{3}$ solution, $1 \,M \,AlCl_{3}$ solution and $1 \,M \,AgNO_{3}$ solution, $2$ mole electron, $6$ mole electron, $3$ mole electron and $1$ mole electron are needed to deposit $Cu, Bi, Al$ and $Ag$ at the cathode respectively
But one mole electron $= 1 F $ electricity That’s why number of Faraday required to deposit $1 \,M$ of each $CuSO_{4}$, $Bi_{2}\left(SO_{4}\right)_{3}$, $AlCl_{3} $and $AgNO_{3}$ solution are $2 F, 6 F, 3 F$ and $F$ respectively.