Question

In the electrolysis of ${{H}_{2}}O$, 11.2 litre of ${{H}_{2}}$ liberated at cathode at NTP. How much ${{O}_{2}}$ will be liberated at anode under the same conditions?
A. 11.2 litre
B. 22.4 litre
C. 32g
D. 5.6 litre

Answer 1

To attempt this question, first you will have to write a balanced chemical equation which is the prerequisite and then since, you have moles and you have to find out the volume of product, we can easily use the ideal gas equation which is $PV=nRT$,
Where P= pressure, V= volume, n= moles, R= gas constant and T is temperature.

Complete step by step answer:
This is the electrolysis of water looks like:
${{H}_{2}}O(l)\to {{H}_{2}}(g)+\dfrac{1}{2}{{O}_{2}}(g)$
We are already given that 11.2 litres of hydrogen are being liberated , then by applying Avogadro's law, which states at constant pressure and temperature, no. of moles will be directly proportional to the volume of the gas.
V ∝ n
Therefore, the volume of oxygen which will be liberated at anode would be half of the volume liberated of Hydrogen because there is $dfrac{1}{2}$ mole of oxygen present as compared to the Hydrogen gas.
Volume of oxygen liberated at anode:
$\Rightarrow \dfrac{Volume}{2}$
$\Rightarrow \dfrac{11.2}{2}=5.6lit$

Therefore, the answer of this question is D. 5.6 litres.

Note: Some students get confused between the NTP and STP which stands for Standard Temperature Pressure and Normal Temperature Pressure. There is not much difference as only the conditions of temperature and pressure changes a bit. In STP, the temp. is $0{}^\circ C$ and pressure is $100 kPa$, whereas in NTP, the temp. is $20{}^\circ C$ and pressure is $101.3 kPa$. These sometimes affect the reaction.