Question

How long (approximate) should water be electrolysed by passing through 100 amp current so that the oxygen released can completely burn 27.6 g of diborane? (Atomic wt. of B = 10.8)

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

Answer 1

Answer: (C)

3.2 hours

Answer 2

The balanced equation for the combustion of diborane ($\text{B}_2\text{H}_6$) is:

$\text{B}_2\text{H}_6\text{(g)} + 3\text{O}_2\text{(g)} \longrightarrow \text{B}_2\text{O}_3\text{(s)} + 3\text{H}_2\text{O(l)}$

The molar mass of diborane ($\text{B}_2\text{H}_6$) is calculated using the given atomic weight of B = 10.8 and the atomic weight of H = 1 (approximately).

Molar mass of $\text{B}_2\text{H}_6 = 2 \times \text{Atomic wt. of B} + 6 \times \text{Atomic wt. of H} = 2 \times 10.8 + 6 \times 1 = 21.6 + 6 = 27.6$ g/mol.

The number of moles of 27.6 g of diborane is:

Moles of $\text{B}_2\text{H}_6 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{27.6 \text{ g}}{27.6 \text{ g/mol}} = 1 \text{ mole}$.

From the balanced combustion equation, 1 mole of $\text{B}_2\text{H}_6$ requires 3 moles of $\text{O}_2$ for complete combustion.
So, we need 3 moles of $\text{O}_2$.

Oxygen is produced by the electrolysis of water. The half-reaction for the production of oxygen at the anode is:

$2\text{H}_2\text{O(l)} \longrightarrow \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4\text{e}^-$

This equation shows that 1 mole of $\text{O}_2$ is produced by the transfer of 4 moles of electrons.
To produce 3 moles of $\text{O}_2$, the number of moles of electrons required is $3 \times 4 = 12$ moles.

According to Faraday's laws of electrolysis, the charge of one mole of electrons is equal to one Faraday (F), which is approximately 96500 Coulombs.
The total charge required to produce 3 moles of $\text{O}_2$ is $Q = \text{Number of moles of electrons} \times F = 12 \times 96500 \text{ C}$.

The relationship between charge ($Q$), current ($I$), and time ($t$) is $Q = I \times t$.
We are given the current $I = 100$ A. We need to find the time $t$.

$12 \times 96500 = 100 \times t$
$t = \frac{12 \times 96500}{100} \text{ seconds}$
$t = 12 \times 965 \text{ seconds}$
$t = 11580 \text{ seconds}$.

To convert the time from seconds to hours, we divide by 3600 (number of seconds in an hour).

$t \text{(in hours)} = \frac{11580}{3600} = \frac{1158}{360} = \frac{193}{60}$ hours.

Now, we convert the fraction to a decimal or mixed number:

$\frac{193}{60} = \frac{180 + 13}{60} = 3 + \frac{13}{60}$ hours.
$\frac{13}{60}$ hours $= \frac{13}{60} \times 60$ minutes $= 13$ minutes.
So, the time is 3 hours and 13 minutes.

In decimal form, $\frac{193}{60} = 3.2166...$ hours.
The question asks for the approximate time. Comparing this value with the given options:

The value 3.2166... hours is closest to 3.2 hours.