Question

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

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

Answer 1

Answer: (C)

for $1$ mole electrons $=1$ faraday $⟶1$ mole of $A{g}^{+}$
for $1$ mole electrons $=1$ faraday $⟶\frac{1}{2}$ mole of $C{a}^{2+}$
for $1$ mole electrons $=1$ faraday $⟶\frac{1}{3}$ mole of $A{l}^{3+}$
So for $3$ faraday $\to 1\times 3,\frac{1}{2}\times 3,\frac{1}{3}\times 3$
$3:\frac{3}{2}:1$ or $6:3:2$