Question

$2{{I}^{-}}+{{O}_{3}}+{{H}_{2}}O\to 2O{{H}^{-}}+{{I}_{2}}+{{O}_{2}}$
The above redox reaction is used to determine the amount of ozone ${{O}_{3}}$ present in polluted air. The liberated iodine is titrated with sodium thiosulphate solution using suitable indicators. A 2.24 L of a mixture of ${{O}_{2}}$ and ${{O}_{3}}$ at STP was allowed to react with excess KI solution. The liberated ${{I}_{2}}$ requires 60 ml of 0.2 M sodium thiosulphate solution to reach the end point. The % of ozone (by volume) in the 2.24 L mixture is:

Answer 1

Hint: Redox reactions are also known by reduction oxidation reactions. These are those chemical reactions in which the oxidation state of atoms is changed. In this type of reaction if one species undergoes the process of oxidation then another will undergo reduction.

Complete Solution :
According to the conditions given in the question when ozone is react with excess of KI then the reaction can be shown as:
${{O}_{3}}+KI+{{H}_{2}}O\to 2KOH+{{I}_{2}}+{{O}_{2}}$

- Now the liberated ${{I}_{2}}$ reacts with sodium thiosulphate reaction is shown as:
${{I}_{2}}+2N{{a}_{2}}{{S}_{2}}{{O}_{3}}\to N{{a}_{2}}{{S}_{4}}{{O}_{6}}+2NaI$
Millimoles of ${{O}_{3}}$ is given by
$\dfrac{1}{2}\times 40\times \dfrac{1}{10}=2$; 0.002 moles

- Moles of total mixture can be calculated from the formula given by ideal gas equation i.e.
$PV=nRT$
Here $n=\dfrac{PV}{RT}$
$n=\dfrac{1\times 1}{0.08\times 273}=0.045$(Consider ideal conditions)

Mole of ${{O}_{2}}$= $0.045-0.002=0.043$
Mass of ${{O}_{2}}$= $Number\ \text{of}\ \text{moles}\times \text{molecular}\ \text{mass}$
$0.043\times 32=1.376g$
Mass of ${{O}_{3}}$= $Number\ \text{of}\ \text{moles}\times \text{molecular}\ \text{mass}$
$0.002\times 48=0.96g$

- Now we have to calculate percentage of ${{O}_{3}}$ this can be given by:
$\dfrac{Mass\ \text{of}\ {{\text{O}}_{\text{3}}}}{Mass\ \text{of}\ {{\text{O}}_{\text{3}}}+Mass\ \text{of}\ {{\text{O}}_{\text{2}}}}\times 100$
$\dfrac{0.096}{0.096+1.376}\times 100=6.52%$
Hence the % of ozone (by volume) in the 2.24 L mixture is 6.52%.

Note: Oxidation is defined as the loss of electrons or we can say that increase in the oxidation state of an atom, ion or molecule while reduction can be defined as the gain of electrons or decreases in the oxidation state of any given atom, ion or molecule.