Question

Three faradays of electricity are passed through molten $A{l}_2{O}_3,$ aqueous solution of $CuS{O}_4$ and molten $NaCl$ taken in different electrolytic cells. The amount of $Al,Cu$ and $Na$ deposited at the cathodes will be in the ratio of:
[Assume: $Al,Cu\text{and}Na$ are depositing at cathode]

• A. 1 mole : 2 mole : 3 mole
• B. 3 mole : 2 mole : 1 mole
• C. 1 mole : 1.5 mole : 3 mole
• D. 1.5 mole : 2 mole : 3 mole

Answer 1

The correct option is C 1 mole : 1.5 mole : 3 mole

In $A{l}_2{O}_3,$ $Al$ has charge 3+
Thus, the reduction reaction will be
$A{l}^{3+}\left(l\right)+3e^{-}\to Al\left(s\right)$
Thus, 3 Faraday are required for 1 mole of $A{l}^{3+}$.
So, $3F$ will produce $1mol$ of $A{l}_2{O}_3$.

In $CuS{O}_4,$ $Cu$ has charge 2+
$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu\left(s\right)$
Thus, 2 Faraday are required for 1 mole of $C{u}^{2+}$.
So, $3F$ will produce $1.5mol$ of $CuS{O}_4$

In $NaCl,Na$ has charge 1+
$N{a}^{+}\left(l\right)+e^{-}\to Na\left(s\right)$
Thus, 1 Faraday is required for 1 mole of $N{a}^{+}$.
So, $3F$ will produce $3mol$ of $NaCl$

Hence, ratio will be
$1mol:1.5mol:3mol$