Question

How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane?
[Atomic weight of $B=10.8u$, $1F=96500C$]

• A. 3.2 hours
• B. 1.6 hours
• C. 6.4 hours
• D. 0.8 hour

Answer 1

The correct option is A 3.2 hours

The corresponding reactions are,
$\underset{1mol}{B_2{H}_6}+\underset{3mol}{3O_2}\to B_2{O}_3+3H_{2}O2H_{2}O\left(l\right)\to O_2\left(g\right)+4H^{+}\left(aq\right)+4e^{-}$
Again, $27.66g{B}_2{H}_6=\frac{27.66}{27.66}=1mol$
So oxygen required is 3 mol.

$\therefore$ Moles of $O_2\times 4=\frac{100\times t}{96500}$
$\Rightarrow 3\times 4=\frac{100\times t}{96500}$
$\Rightarrow t=\frac{96500\times 3\times 4}{3600\times 100}\text{hour}=\frac{965\times 12}{3600}=3.21\text{hours}$