Question

In the reaction $2{{H}_{2}}S+S{{O}_{2}}\to 3S+2{{H}_{2}}O$ , what has been oxidized and what is the oxidizing agent?

Answer 1

Redox is a chemical process that involves changing the oxidation states of atoms. The actual or formal transfer of electrons between chemical species is defined by redox reactions, which usually include one species (the reducing agent) suffering oxidation (losing electrons) while another species (the oxidising agent) experiences reduction (gains electrons).

Complete answer:
The chemical species that loses an electron is said to have been oxidised, whereas the chemical species that gains an electron is said to have been reduced. To put it another way, oxidation is the loss of electrons or the rise in the oxidation state of an atom, an ion, or a group of atoms in a molecule. The gain of electrons or a drop in the oxidation state of an atom, an ion, or specific atoms in a molecule is referred to as reduction (a reduction in oxidation state).
Comproportionation is a chemical process in which two reactants, each having the same element but with a different oxidation number, combine to create a product in which the oxidation numbers of the elements involved are equal. It is the polar opposite of disproportion.
The Claus process, which recovers elemental sulphur from gaseous hydrogen sulphide, is the most important gas desulfurization method. The Claus method, which was first invented in 1883 by scientist Carl Friedrich Claus, has since become the industry standard.
$2{{H}_{2}}S+S{{O}_{2}}\to 3S+2{{H}_{2}}O$
This is an example of a comproportionation process, in which sulphur compounds in various oxidation states with regard to sulphur undergo electron transfer to produce elemental sulphur (which is zerovalent).
$2 \mathrm{H}_{2} \mathrm{~S} \rightarrow S+2 \mathrm{H}^{+}+2 e^{-}$
(oxidation from $S(-I I)$ to $S(0)$ )
$\mathrm{SO}_{2}+4 \mathrm{H}^{+}+4 e^{-} \rightarrow S+2 \mathrm{H}_{2} \mathrm{O}(i i)$
(reduction from $S(+I V)$ to $S(0)$ )
So $2 \times(i)+(i i)=2 \mathrm{H}_{2} \mathrm{~S}+\mathrm{SO}_{2} \rightarrow 3 \mathrm{~S}+2 \mathrm{H}_{2} \mathrm{O}$
Hence
Sulphide has been oxidised, and the oxidising agent is sulphur dioxide.
Form $\mathbf{SO2}\to \mathbf{2S}$
The state of oxidation goes from +4 to 0.
As a result, ${{H}_{2}}S$ is oxidised and $S{{O}_{2}}$ is reduced.

Note:
Many organic reactions are redox reactions because they include changes in oxidation levels but no clear electron transfer. For example, when wood is burned with molecular oxygen, the oxidation state of carbon atoms in the wood increases while the oxidation state of oxygen atoms drops, resulting in the formation of carbon dioxide and water. The oxygen atoms go through reduction and gain electrons, whereas the carbon atoms go through oxidation and lose electrons. In this reaction, oxygen is the oxidising agent and carbon is the reducing agent.