Question

When the gases sulfur dioxide and hydrogen sulfide mix in the presence of water, the reaction ${{S}}{{{O}}_2} + 2{{{H}}_2}{{S}} \to 2{{{H}}_2}{{O}} + 3{{S}}$ occurs. Here, hydrogen sulfide acts as:
A. an oxidizing agent
B. a reducing agent
C. a dehydrating agent
D. a catalyst

Answer 1

Hint:Oxidizing agent is a substance which accepts electrons from another substance. While a reducing agent is a substance which gives electrons to another substance. It undergoes oxidation and it reduces other substances. While oxidizing agents undergo reduction and it oxidizes other substances.

Complete answer:
The given chemical reaction is:
${{S}}{{{O}}_2} + 2{{{H}}_2}{{S}} \to 2{{{H}}_2}{{O}} + 3{{S}}$
According to mole concept, one mole of ${{S}}{{{O}}_2}$ reacts with $2{{mol}}$ of ${{{H}}_2}{{S}}$ to give $2{{mol}}$ of ${{{H}}_2}{{O}}$ and ${{3mol}}$ of sulfur.
We know that the oxidation number or oxidation state is the total number of electrons which is exchanged by an atom. So let’s calculate the oxidation state of each compound.
Initially, let’s calculate the oxidation state of sulfur in sulfur dioxide. Oxygen has an oxidation state $ - 2$ and let the oxidation number of sulfur be ${{x}}$. The total charge of the molecule is $0$.
Therefore, ${{x}} + \left( {2 \times - 2} \right) = 0 \Leftrightarrow {{x}} = + 4$
The oxidation state of sulfur changes from $ + 4$ to $0$ since the sulfur, in elemental form, has an oxidation state $0$.
${{S}}{{{O}}_2}$ accepts electrons which are donated by ${{{H}}_2}{{S}}$. Thus ${{S}}{{{O}}_2}$ acts as an oxidizing agent and ${{{H}}_2}{{S}}$acts as a reducing agent.
Now, it is clear that the ${{{H}}_2}{{S}}$ molecule acts as a reducing agent. It is not an oxidizing agent or a dehydrating agent or a catalyst.

Hence, the correct option is B.

Additional information:
The major property of gases is that they mix with each other. Generally, it cannot be reversed, i.e. irreversible. Thus it is a spontaneous reaction.

Note:
We have not mentioned the oxidation state of sulfur in hydrogen sulfide molecules. Hydrogen has $ + 1$ oxidation state. The sulfur in hydrogen sulfide can be calculated by:
${{x}} + 2 = 0 \Leftrightarrow {{x = - 2}}$
The oxidation state of sulfur is oxidized from $ - 2$ to $0$.