Question

3 faraday 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. 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

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

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

$3$ Faraday of electricity will produce $1molAl$.

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

$2$ Faraday electricity produces $1molCu$

$3$ Faraday electricity produces $\frac{3}{2}molCu$

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

$1$ Faraday electricity produces $1molNa$

$3$ Faraday electricity produces $3molNa$

Ratio $=1:\frac{3}{2}:3\Rightarrow 1:1.5:3$

Hence, option C is correct.