Question

In passing $3$ faraday of electricity through the three electrolytic cells connected in series containing $A{g}^{+},C{u}^{2+}$ and $A{l}^{+3}$ respectively. The molar ratio in which the three metal ions are liberated at the electrode is

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

Answer 1

Answer: (C)

$A{g}^{+}+e^{-}\to Ag$
$C{a}^{2+}+2e^{-}\to Ca$
$A{l}^{3+}+3e^{-}\to Al$
$3$ faradays liberate $1$ mol of $Al$, $3$ moles of $Ag$ and $3/2$ moles of $Cu$. Thus, molar ratio of
$Ag:Cu:Al$ is $3:3/2:1$ or $6:3:2$