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
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Answer:
Correct Answer: 7. (c)
Solution:
- Given that, $i=100 \mathrm{amp}$. also, $27.66 \mathrm{g}$ of diborane $(\mathrm{B} _{2} \mathrm{H} _{6})$
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 } $
