Question

How long (approximate) should water be electrolysed by passing through $100amperes$ current so that the oxygen released can completely burn $27.66g$ of diborane?
(Given: Atomic weight of $B=10.8u$)

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

Answer 1

The correct option is B $3.2hours$

Given that, $i=100amperes$ also, $27.66g$ of diborane ($B_2{H}_6$ )
Molecular mass of $B_2{H}_6=10.8\times 2+6=27.6$
Number of moles of $B_2{H}_6$ in $27.66g$ = $\frac{Givenmass}{Molarmass}=\frac{27.66}{27.6}=1$(approx)
Now consider the equation:
$B_2{H}_6$ + $3O_2$ $\to$ $B_2{O}_3$ + $3H_{2}O$
From the equation we can interpret that $3$ moles of oxygen is required to burn $1mole$ (i.e. $27.6g$) $B_2{H}_6$ completely.
Also consider the electrolysis reaction of water i.e.
$H_{2}O$ $\rightleftharpoons$ $2H^{+}$ + $O^{2-}$
$2H^{+}\to 2H\to H_2\uparrow$
$O^{2-}\to 2O+2e-\to O_2\uparrow$

From the above equation it can be easily interpreted that in electrolysis of water for the production of $1mole$ of oxygen from $2mole$ of $H_{2}O$ at anode $4moles$ electrons are required.
Likewise for the production of $3moles$ of $O_2$, $12(3\times 4)moles$ of electrons will be needed.
So, the total amount of charge required to produce $3moles$ of oxygen will be $12For2\times 96500$
We know; $Q=i\times t$
Given: $(i=100Amperes)$
So, = $12\times 96500=100\times t\left(inseconds\right)$
= $\frac{12\times 96500}{100\times 3600}=t\left(inhours\right)=3.2hours$