Question

24. Three faraday of electricity is passed through three
electrolytic cells connected in series containing
Agt. Ca2+ and Al' ions respectively. The molar
ratio in which the three metal ions are liberated at
the electrodes is
(1) 1:2:3
(2) 3:2:1
(3) 6:3:2
(4) 3:4:2​

Answer 1

1 F will deposit one mole of Ag

Ag+ + e gives Ag

2F will deposit one mole of Ca

Ca2+ + 2e gives Ca

3F will deposit one mole of Al

Al3+ + 3e gives Al

here , 3F is available

3: 3/2: 1

6: 3: 2

Answer 2

Answer: (3)  6:3:2

Explanation:

According to Faraday's second law of electrolysis, when same quantity of electricity is passed through solutions of different electrolytes connected in series, the mass of the substances produced produced at the electrodes are directly proportional to their equivalent weights.

a) $Ag^++e^-\rightarrow Ag$

As 1 Faraday will deposit 1 mole of Ag

3 F will deposit =$\frac{1}{1}\times 3=3$

b) $Ca^{2+}+2e^-\rightarrow Ca$

As 2 Faraday will deposit 1 mole of Ca

3 F will deposit =$\frac{1}{2}\times 3=\frac{3}{2}$

c) $Al^{3+}+3e^-\rightarrow Al$

As 3 Faraday will deposit 1 mole of Al

3 F will deposit =$\frac{1}{3}\times 3=1$

Thus mole ratio of $Ag^+$, $Ca^{2+}$ and $Al^{3+}$ will be 3 : $\frac{3}{2}$ : 1 = 6: 3: 2

Thus the molar  ratio in which the three metal ions are liberated at  the electrodes is 6:3:2