Question

The quantity of electricity needed to electrolyse completely $1M$ solution of $CuS{O}_4$, $B{i}_2{\left(S{O}_4\right)}_{3^{\prime }}AlC{l}_3$ and $AgN{O}_3$ each will be

• A. 2 F, 6 F, 3 F and 1 F respectively
• B. 6 F, 2 F, 3 F and 1 F respectively
• C. 2 F, 6 F, 1 F and 3 F respectively
• D. none of the above.

Answer 1

Answer: (A)

$CuS{O}_4+2e^{-}\to Cu+S{O}_4^{2-}$
$B{i}_2{\left(S{O}_4\right)}_3+6e^{-}\to 2Bi+3S{O}_4^{2-}$
$AlC{l}_3+3e^{-}\to Al+3C{l}^{-}$
$AgN{O}_3+e^{-}\to Ag+N{O}_3^{-}$
Since, in $1MCuS{O}_4$ solution, $1MB{i}_2{\left(S{O}_4\right)}_3$ solution, $1MAlC{l}_3$ solution and $1MAgN{O}_3$ 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 $=1F$ electricity That’s why number of Faraday required to deposit $1M$ of each $CuS{O}_4$, $B{i}_2{\left(S{O}_4\right)}_3$, $AlC{l}_3$and $AgN{O}_3$ solution are $2F,6F,3F$ and $F$ respectively.