Question

If 0.224L of \[{H_2}\] gas is formed at the cathode, the volume of \[\;{O_2}\] gas formed at the anode under identical conditions is:
A. 0.224L
B. 0.448L
C. 0.112L
D. 1.12L

Answer 1

Hint: We know that from Avogadro’s law that equal volumes of all gases contain equal numbers of molecules under similar conditions. Now since one mole of molecules of all gases contain the same number \[(6.022\; \times {10^{23}})\] of molecules, therefore they occupy the same volume under similar pressure and temperature. The mass of one mole of atoms is exactly equal to the atomic mass in grams of that element.

Complete step by step answer:
Loss of electrons is oxidation and gain of electrons is reduction, at cathode reduction takes place and at anode oxidation takes place. To find the Volume of oxygen we have to find the number of moles
$\Rightarrow$ At cathode reduction takes place: ${H_2}O + 2{e^ - } \to {H_{2(g)}} + 2O{H^ - }$
$\Rightarrow$ At anode oxidation takes place = $2{H_2}O \to {O_{2(g)}} + 4{H^ + } + 4{e^ - }$

\[{H_2}\] is liberated at cathode, number of electrons involved is 2 and n factor is 2 and \[\;{O_2}\] is liberated at anode number of electrons involved is 4 and n factor is 4. We know that equivalent of weight is moles$\times$n-factor and as explained above we conclude,

Equivalent of \[\;{O_2}\] = equivalent of \[{H_2}\] therefore,
$\Rightarrow$ Mole \[\;{O_2}\;\times\;\]n\[\;{O_2}\] = mole\[{H_2}\times\] n\[{H_2}\]
$\Rightarrow$ Mole \[\;{O_2}\] = x mole of \[{H_2}\]
Where we know that moles of \[{H_2}\] = $\dfrac{{0.224}}{{22.4}} = {10^{ - 2}}$.
So mole of \[\;{O_2}\]= $\dfrac{1}{2} \times {10^{ - 2}}$.
From the concept of mole and molar volume we know that volume equivalent is equal to its moles. Therefore at STP,
$\Rightarrow \dfrac{{{O_2}(volume)}}{{22.4}} = \dfrac{1}{2} \times {10^{ - 2}} $
$= 0.112L $

Hence options C is correct.

Note: Alternate method: by using stoichiometric equation it is shown as;
$\Rightarrow{H_2}O \to {H_2} + \dfrac{1}{2}{O_2}$
At ideal conditions, we know that on cathode \[{H_2}\] is liberated and on the anode \[\;{O_2}\] is liberated. Therefore from the reaction we can conclude that if 1 mole of \[{H_2}\] is liberated then $\dfrac{1}{2}$ mole of \[\;{O_2}\] is liberated. So at STP 22.4L of \[{H_2}\] is liberated and 11.2L of \[\;{O_2}\] is liberated from the relation that we see in the reaction. Therefore we are given 0.224L volume of hydrogen which will liberate, Volume of \[\;{O_2}\]= $\dfrac{{0.224}}{2} = 0.112L$.