Question

What will be the volume of ${{O}_{2}}$ at N.T.P. liberated by 5A current flowing for 193 sec through acidulated water?
(A) 56mL
(B) 112mL
(C) 158mL
(D) 965mL

Answer 1

Acidulated water refers to water with some amount of acid dissolved in it. The acid produces hydronium ions in the water and helps the current to flow. 1 mole of a gas at N.T.P. gives 22400mL.

Complete step by step solution:

Let’s see what the answer of the given question is:
First we will calculate the number of moles of electrons.
Number of mole of electron$=\dfrac{current\,\times \,time}{ch\arg e}$
$=\dfrac{5\,\times \,193}{96500}$
Required number of moles of electron = 0.01
Now, when electricity is passed through acidulated water, then electrolysis of water takes place
The oxidation reaction in this case for the liberation of oxygen gas is as follows:
${{4}^{-}}OH\,\to \,2{{H}_{2}}O\,+\,{{O}_{2}}\,+\,4{{e}^{-}}$
Which means that 4 moles of electron liberates 1 mole of ${{O}_{2}}$
Therefore, 1 mole of electron will liberate $\dfrac{1}{4}$ mole of oxygen
Hence, 0.01 mole of electrons will liberate $\dfrac{1}{4}\,\times \,0.01\,$ moles of oxygen
Now, we will calculate the volume of oxygen gas liberated at N.P.T.
As we know that at N.P.T., 1 mole of oxygen liberates 22400L
Therefore, $\dfrac{1}{4}\,\times \,0.01\,$ moles of oxygen will liberate
$=\,\dfrac{1}{4}\times 0.01\,\times \,22400\,mL$
Volume of oxygen produced = 56 mL
Hence, the answer to the given question is option (A).

Note: The flow of electricity in a solution takes place through the movement of ions. If there are no ions present in water then it will not conduct electricity. Addition of an acid in water helps to increase the flow of current through water.