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

The correct option is C

6 : 3 : 2

$\begin{array}{c}A{g}^{+}+e^{-}\to Ag\\ C{a}^{2+}+2e^{-}\to Ca\\ A{l}^{3+}+3e^{-}\to Al\end{array}|\begin{array}{c}\text{For 1 mole of}A{g}^{+},C{a}^{2+},\\ A{l}^{3+}1\text{mole of}{e}^{-},\text{2 mole of}\\ e^{-}\text{and 3 mole of}{e}^{-}\\ \text{required respectively.}\end{array}$

for 1 mole $e^{-}=1$ faraday $\to 1$ mole of $A{g}^{+}$

for 1 mole $e^{-}=1$ faraday $\to \frac{1}{2}$ mole of $C{a}^{2+}$

for 1 mole $e^{-}=1$ faraday $\to \frac{1}{3}$ 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 $6:3:2$