Electrochemistry - Result Question 58
####7. 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 $B=10.8 \mu$ )
(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, $i=100 amp$. also, $27.66 g$ of diborane $\left(B _2 H _6\right)$
Molecular mass of $B _2 H _6=10.8 \times 2+6=27.6$
Number of moles of $B _2 H _6$ in $27.66 g=\frac{\text { Given mass }}{\text { Molar mass }}=\frac{27.66}{27.6} \approx 1$
Now consider the equation
$$ B _2 H _6+3 O _2 \longrightarrow B _2 O _3+3 H _2 O $$
From the equation we can interpret that 3 moles of oxygen is required to burn 1 mole (i.e. $27.6 g$ ) $B _2 H _6$ completely.
Also consider the electrolysis reaction of water i.e.
$$ \begin{aligned} & H _2 O \rightleftharpoons 2 H^{+}+O^{–} \ & 2 H^{+} \xrightarrow[\text { Cathode }]{\stackrel{+2 e^{-}}{\longrightarrow}} 2 H \longrightarrow H _2 \uparrow \ & O^{–} \xrightarrow[-2 e^{-}]{\text {Anode }} O \xrightarrow[\text { atoms }]{2 \text { such }} O _2 \uparrow \end{aligned} $$
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 $H _2 O$ at anode 4 moles electrons are required. Likewise for the production of 3 moles of $O _2 12(3 \times 4)$ 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
$$ \frac{12 \times 96500}{100 \times 3600}=\text { tin hours }=3.2 \text { hours } $$
