Question

In the electrolysis of ${H_2}O$,$11.2$ litre of ${H_2}$ liberated at cathode at STP, How much ${O_2}$ will be liberated at anode under the same conditions?

Answer 1

Hint:We can calculate the volume of substance liberated from the ideal gas equation.
The ideal gas equation is,
$PV = nRT$
Where P is the pressure in the atmosphere.
V is the volume of gas in a liter.
n is the number of moles.
R is a universal gas constant.
T is the temperature.

Complete step by step answer:
Write the balanced chemical equation for the reaction,
${H_2}O\left( l \right)\xrightarrow{{}}{H_2}\left( g \right) + \dfrac{1}{2}{O_2}\left( g \right)$
Given,
The volume of hydrogen liberated is $11.2L$.
We know that,
Avogadro’s Law:
At constant pressure and temperature the number of moles in the gas and the volume of the gas are related proportionally.
$\dfrac{{\text{V}}}{{\text{n}}}{\text{ = K}}$
Where V is the volume of the gas.
n is the number of moles.
K is the constant.
The number of moles or the volume of the gases at constant temperature and pressure can be calculated using the relation,
$\dfrac{{{{\text{V}}_{\text{1}}}}}{{{{\text{n}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{n}}_{\text{2}}}}}$
Where ${V_1}\& {n_1}$ are the volume and the number of moles of gases at initial.
${V_2}\& {n_2}$ Are the volume and the number of moles of gases at final.
One mole of water gives one mole of hydrogen gas and $0.5mol$ of oxygen. Thus the volume of oxygen would be half of the volume of hydrogen.
The volume of oxygen$ = \dfrac{{11.2}}{2} = 5.6L$
The volume of oxygen is $5.6L$.

Note: Now we can discuss about how NTP and STP differ from each other as,
Standard temperature and pressure condition is known as STP. The standard temperature value is \[0^\circ C\] and the standard pressure value is \[100{\text{ }}kPa\] or \[1{\text{ }}bar.\] Normal Temperature and Pressure is known as NTP the value of pressure at NTP is \[101.325{\text{ }}kPa\] and the temperature at NTP is \[20^\circ C\].