Question

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

Answer 1

Hint: 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.
\[2\left( { + 1} \right) + x = 0 \\
2 + x = 0 \\
x = - 2 \\\]
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.