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?

• 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

Answer: (C)

$1$ mole $:1.5$ mole $:3$ mole

In $A{l}_2{O}_3$, $Al$ has charge $3+$

$A{l}^{3+}+3e^{-}\to Al$

Thus, 3 Faraday are required for $1$ mole. So, $3F$ will produce 1 mole of $A{l}_2{O}_3$

In $CuS{O}_4$, $Cu$ has charge $2+$

$C{u}^{2+}+2e^{-}\to Cu$

Thus, $2$ Faraday are required for $1$ mole. So, $3F$ will produce $1.5$ moles of $CuS{O}_4$

In $NaCl$, $Na$ has charge $1+$

$N{a}^{+}+e^{-}\to Na$

Thus, $1$ Faraday are required for $1$ mole, So, $3F$ will produce $3$ mole of $NaCl$

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