Electrochemistry 1 Question 7
7. How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn $27.66 \mathrm{~g}$ of diborane?
(Atomic weight of $B=10.8 \mu$ )
(2018 Main)
(a) 6.4 hours
(b) 0.8 hours
(c) 3.2 hours
(d) 1.6 hours
Show Answer
Answer:
Correct Answer: 7. (c)
Solution:
- Given that, $i=100 \mathrm{amp}$. also, $27.66 \mathrm{~g}$ of diborane $\left(\mathrm{B}{2} \mathrm{H}{6}\right)$
Molecular mass of $\mathrm{B}{2} \mathrm{H}{6}=10.8 \times 2+6=27.6$
Number of moles of $\mathrm{B}{2} \mathrm{H}{6}$ in $27.66 \mathrm{~g}=\frac{\text { Given mass }}{\text { Molar mass }}=\frac{27.66}{27.6} \approx 1$
Now consider the equation
$$ \mathrm{B}{2} \mathrm{H}{6}+3 \mathrm{O}{2} \longrightarrow \mathrm{B}{2} \mathrm{O}{3}+3 \mathrm{H}{2} \mathrm{O} $$
From the equation we can interpret that 3 moles of oxygen is required to burn 1 mole (i.e. $27.6 \mathrm{~g}$ ) $\mathrm{B}{2} \mathrm{H}{6}$ completely.
Also consider the electrolysis reaction of water i.e.
$$ \begin{aligned} & \mathrm{H}{2} \mathrm{O} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{O}^{–} \ & 2 \mathrm{H}^{+} \xrightarrow[\text { Cathode }]{\stackrel{+2 e^{-}}{\longrightarrow}} 2 \mathrm{H} \longrightarrow \mathrm{H}{2} \uparrow \ & \mathrm{O}^{–} \xrightarrow[-2 \mathrm{e}^{-}]{\text {Anode }} \mathrm{O} \xrightarrow[\text { atoms }]{2 \text { such }} \mathrm{O}_{2} \uparrow \end{aligned} $$
From the above equation it can be easily interpreted that in electrolysis of water for the production of 1 mole of oxygen from 1 mole of $\mathrm{H}{2} \mathrm{O}$ at anode 4 moles electrons are required. Likewise for the production of 3 moles of $\mathrm{O}{2} 12(3 \times 4)$ moles of electrons will be needed.
So, the total amount of charge required to produce 3 moles of oxygen will be $12 \times F$ or $12 \times 96500$
We know $Q=i t$
So, $\quad 12 \times 96500=100 \times t$ in seconds
or
$$ \frac{12 \times 96500}{100 \times 3600}=\text { tin hours }=3.2 \text { hours } $$
