Question

When 1 F of electricity is passed through acidulated water, ${{O}_{2}}$ evolved is
A. 11.2$d{{m}^{3}}$
B. 5.6$d{{m}^{3}}$
C. 1.0 $d{{m}^{3}}$
D. 22.4 $d{{m}^{3}}$

Answer 1

In this question we have been asked to calculate the amount of oxygen at the end of the reaction. The given process is the process of electrolysis of water. Therefore, we will be writing the reaction of electrolysis of water. We shall then find the number of moles of oxygen. TO calculate the amount of oxygen liberated at the end of reaction.
Formula used:
$2{{H}_{2}}O\to 4{{H}^{+}}+2{{O}^{2-}}$

Complete answer:
We know that when electricity is passed through acidulated water then ions follow electrolysis. The process in which the ionic substances are broken down into simpler substances when electricity is passed, is known as Electrolysis.
Now we know that,
$2{{H}_{2}}O\to 4{{H}^{+}}+2{{O}^{2-}}$
We know,
$2{{O}^{-}}\to {{O}_{2}}+4{{e}^{-}}$
Similarly,
$4{{e}^{-}}+4{{H}^{+}}\to 2{{H}_{2}}$
Now, from above reaction we know that $n=4$
We know that 1 mole = 22.4 $d{{m}^{3}}$. So, 4 faraday charge will liberate 1 mole = 22.4 $d{{m}^{3}}$
Therefore, 1 F charge will liberate
$\dfrac{22.4}{4}=5.6d{{m}^{3}}$

So, the correct answer is “Option B”.

Additional Information:
In the 1800 Michael Faraday invented the process of electrolysis. He immersed two electrodes in water and allowed the current to flow through it. This caused the water to break down into hydrogen and oxygen.

Note:
Electrolysis of water means the decomposition of water in hydrogen and oxygen by passing the current through them. Electrolysis is an excellent option for production of hydrogen from renewable resources. This reaction takes place in a unit called electrolyser. Sulfuric acid is the best electrolyte for electrolysis of water.