JEE Main 2018 — Electrochemistry Question with Solution

Required reaction :

B₂H₆ + 3O₂ $$\to$$ B₂ O₃ + 3 H₂ O

Here molar mass of B₂H₆ =10.8 $$\times$$ 2 + 6 = 27.6 gm

Given weight of B₂H₆ = 27.66 g

$$\therefore$$No of moles of B₂H₆ = $$\frac{27.6}{27.66}\simeq 1$$ mole.

For combustion of 1 mole B₂H₆ 3 moles O₂ required.

This 3 mole of O₂ is obtained by electrolysis of H₂O.

2H₂O($$l$$) $$\to$$ O₂ (g) + 4 H⁺ (aq) + 4 e^$$-$$

From Faradays law of electrolysis,

moles $$\times$$ n_f = $$\frac{It}{96500}$$

Here moles of O₂ = 3.

N_f of O₂ = 4 (in H₂ change of O = $$-$$2

and in O₂ change of 0 = O.

So change in charge = 2 .

for two atoms of O₂ change in charge = 2 $$\times$$ 2 = 4)

$$\therefore$$ 3 $$\times$$ 4 = $$\frac{100\times t}{96500}$$

$$\Rightarrow$$ t = 12 $$\times$$ 965 sec.

$$\Rightarrow$$ t = $$\frac{12\times 965}{60\times 60}$$ hr

= 3.2 hr