Question

One mole of electron passes through each of the solutions of $AgN{O}_3$, $CuS{O}_4$ and $AlC{l}_3$ when Ag, Cu, and Al are deposited at the cathode. The molar ratio of Ag, Cu and Al deposited are ________.

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

Answer 1

Answer: (B)

$6:3:2$

Given $1$ mole of $e^{-}$ passed through solution of $AgN{O}_3,CuS{O}_4,AlC{l}_3$.

The reactions involved are:-

$AgN{O}_3+e^{-}⟶A{g}^{+}+N{O}_3$

$CuS{O}_4+2e^{-}⟶C{u}^{2+}+{S{O}_4}^{2-}$

$AlC{l}_3+3e^{-}⟶A{l}^{3+}+3C{l}^{-}$

If only one mole of $e^{-}$ is passed in above equations then the deposited amount of elements is

$AgN{O}_3$ will give $1$ equivalent $Ag$ $[\because$ only $1e^{-}$ required $]$

$CuS{O}_4$ will give $\frac{1}{2}$ mole $Cu$ $[\because 2$ mole $e^{-}$ required $]$

$AlC{l}_3$ will give $\frac{1}{3}$ mole $Al$ $[\because 3$ mole $e^{-}$ required $]$

$\therefore$ The molar ratio of $Ag,Cu,Al$ deposited are:

$Ag:Cu:Al$

$=1:\frac{1}{2}:\frac{1}{3}$

$=6:3:2$