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 three different electrolytic cells. Then the mole ratio of $Al,Cu$ and $Na$ deposited on the cathode will be

• A. $3:4:6$
• B. $2:1:6$
• C. $3:2:1$
• D. $2:3:6$

Answer 1

Answer: (D)

$\because A{l}^{3+}+3e^{-}\to Al$
$\therefore 3F$ of electricity will deposite $1$ mole of $AI$.
$\because C{u}^{2+}+2e^{-}\to Cu$
$\therefore 3F$ of electricity will deposite $1.5$ mole of $Cu$.
$\because N{a}^{+}+e^{-}\to Na$
$\therefore 3F$ of electricity will deposite $3$ mole of $Na$.
Hence, the mole ratio of $Al,Cu$ and $Na$ deposited on the cathode will be $1:1.5:3$ or $2:3:6$.