Question

n factor for ${{\text{H}}_2}{\text{S}}$ during its oxidation to ${\text{S}}{{\text{O}}_2}$ is:

Answer 1

In a redox reaction, the n factor gives the number of moles of electrons gained or lost by one mole of the species undergoing reduction or oxidation. n factor can also be defined as the change in the oxidation number of the species undergoing reduction or oxidation.

Complete step by step answer:
First calculate the oxidation number of sulphur in ${{\text{H}}_2}{\text{S}}$ and ${\text{S}}{{\text{O}}_2}$. Then calculate the change in the oxidation number of Sulphur. This change in the oxidation number of Sulphur is the n factor.
Let x be the oxidation number of sulphur in ${{\text{H}}_2}{\text{S}}$. The oxidation number of hydrogen is +1. The sum of the oxidation numbers of all the elements in a neutral molecule is zero.

2 + x = 0 \\\ x = - 2 \\\\$$ Hence, the oxidation number of sulphur in $${{\text{H}}_2}{\text{S}}$$ is -2. Let y be the oxidation number of sulphur in $${\text{S}}{{\text{O}}_2}$$. The oxidation number of oxygen is -2. The sum of the oxidation numbers of all the elements in a neutral molecule is zero. $$y + 2\left( { - 2} \right) = 0 \\\ y - 4 = 0 \\\ y = + 4 \\\\$$ Hence, the oxidation number of sulphur in $${\text{S}}{{\text{O}}_2}$$ is +4. Thus, during oxidation of $${{\text{H}}_2}{\text{S}}$$ to $${\text{S}}{{\text{O}}_2}$$ the oxidation number of sulphur increases from -2 to +4. The increase in the oxidation number of sulphur is $$4 - \left( { - 2} \right) = 6$$. The N factor is the increase in the oxidation number. Hence, n factor for $${{\text{H}}_2}{\text{S}}$$ during its oxidation to $${\text{S}}{{\text{O}}_2}$$ is 6. **Note:** n factor is used to obtain equivalent weight of an oxidizing or reducing agent. For an oxidizing or reducing agent, the n factor is the ratio of the molecular weight of the oxidizing or reducing agent to the n factor.