Question

Three Faradays of electricity are passed through molten $Al _{2} O _{3}$, aqueous solution of $CuSO _{4}$ and molten $NaCl$ taken in three different electrolytic cells. Then the mole ratio of $A l, C u$ and $N a$ 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)

$\because A l^{3+}+3 e^{-} \rightarrow A l$
$\therefore 3 F$ of electricity will deposite $1$ mole of $AI$.
$\because C u^{2+}+2 e^{-} \rightarrow C u$
$\therefore 3 F$ of electricity will deposite $1.5$ mole of $Cu$.
$\because N a^{+}+e^{-} \rightarrow N a$
$\therefore 3 F$ 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 $2: 3: 6$.