Question

In passing 3 faraday of electricity through the three electrolytic cells connected in series containing Ag+, Ca2+ and Al+3 ion 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)

6 : 3 : 2

Answer 2

Faraday’s II Law, $\frac{W}{E} =$constant

So, gm eq of Ag+ = gm . eq of Ca+2 = gm. eq. of Al+3 = a

( Let )

Now, mole of Ag+ = a $(Ag^{+} + e^{-} \rightarrow Ag)$

mole of Ca+2 = $\frac{a}{2}(Ca^{+ 2} + 2e^{-} \rightarrow Ca)$

So, mole ratio of Ag+ : Ca+2 : Al+3

= $a:\frac{a}{2}:\frac{a}{3} = 6:3:2$