Question

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 = 10.8u)
(A) 3.2 hours
(B) 1.6 hours
(C) 6.4 hours
(D) 0.8 hours

Answer 1

To solve this question, we first need to know what electrolysis is. When a direct electric current is passed to produce a chemical reaction that is otherwise non-spontaneous is known as electrolysis. It is an oxidation-reduction reaction.

Complete answer:
It is given to us that the atomic weight of borane is 10.8u and we know that the atomic weight of hydrogen is 1.008u.
So the molar mass of diborane ${{B}_{2}}{{H}_{6}}$ or mass of one mole would be 27.66g.
Now, let us write the balanced chemical equation of reaction between diborane and oxygen.
${{B}_{2}}{{H}_{6}}+3{{O}_{2}}\to {{B}_{2}}{{O}_{3}}+3{{H}_{2}}O$
Now, from the balanced chemical equation, we can see that 1 mole of diborane requires 3 moles of an oxygen molecule to form products 1 mole of diborane trioxide and 3 moles of water.
We know that during the electrolysis of water, electricity is used to dissociate water molecules into oxygen gas and hydrogen gas. It can be given by the equation
$2{{H}_{2}}O\to 2{{H}_{2}}+{{O}_{2}}$
Here hydrogen ions are reduced at the negatively charged cathode to give hydrogen gas.
$2{{H}^{+}}(aq)+2{{e}^{-}}\to {{H}_{2}}(g)$
Similarly, oxidation takes place at the positively charged anode to give oxygen gas.
$2{{H}_{2}}O(l)\to {{O}_{2}}(g)+4{{H}^{+}}(aq)+4{{e}^{-}}$
Now, we can see that 4 electrons are required to produce 1 mole of oxygen gas.
Or we can say that 4F (faraday) charge is required to produce 1 mole of oxygen gas.
So, 3 moles of oxygen gas required will be produced by 12F charge. Also,
$1Faraday=96500Coulombs$
Now, we know that according to Ohm's law,
$Q=I\times t$
Where Q denotes the charge, I gives the current passed and t is the time taken.
So, we know that the current passed I = 100A, and the charge required Q = 12F = $12\times 96500C$. So,
$t=\dfrac{Q}{I}=\dfrac{12\times 96500}{100}=11580s$
This can be converted into hours
$t=\dfrac{11580}{3600}\cong 3.22hr$

So, it takes approximately option (A) 3.2 hours for the water to be electrolyzed.

Note:
It should be taken into account that during electrolysis neutral molecules react at either one of the electrodes. Neutral atoms or molecules become ions when there is a gain or loss of electrons on the electrode's surface. They react with other ions by dissolving in the electrolyte.