Question: Consider the reaction: \({{H}_{2}}S{{O}_{3}}(aq)+S{{n}^{4+}}(aq)+{{H}_{2}}O(l)\to S{{n}^{2+}}(aq)+...

Question

Consider the reaction:
${{H}_{2}}S{{O}_{3}}(aq)+S{{n}^{4+}}(aq)+{{H}_{2}}O(l)\to S{{n}^{2+}}(aq)+HSO_{4}^{-}(aq)+3{{H}^{+}}$
Which of the following statements is correct?
(A) $S{{n}^{4+}}$ is the oxidizing agent because it undergoes oxidation
(B) $S{{n}^{4+}}$ is the reducing agent because it undergoes oxidation
(C) ${{H}_{2}}S{{O}_{3}}$ is the reducing agent because it undergoes oxidation
(D) ${{H}_{2}}S{{O}_{3}}$ is the oxidizing agent because it undergoes reduction

Answer 1

Solution

The oxidizing agent and reducing agent in a reaction can be calculated by finding the oxidation state of the atoms in the reaction. The oxidized and reduced agents are not the same of oxidizing and reducing agents.

Complete step by step solution:
The charge which an atom of the element has in its ion or appears to have when present in the combined state with other atoms can be defined as the oxidation number. We can also call the oxidation number as an oxidation state.
Oxidation is the process or a chemical change in which there is an increase in the oxidation number of an atom or atoms.
The reduction is the process or a chemical change in which there is a decrease in the oxidation number of an atom or atoms.
An oxidizing agent or an oxidant is a substance in which the oxidation number or state decreases.
A reducing agent or a reluctance is a substance in which the oxidation number or state increases.
So, in the equation, we have to find the oxidation number of every atom.
In ${{H}_{2}}S{{O}_{3}}$ , the oxidation state of H is +1, O is -2 and S is +4.
The oxidation state of $S{{n}^{4+}}$ is +4
The oxidation state of $S{{n}^{2+}}$ is +2
In $HSO_{4}^{-}$ , the oxidation state of H is +1, O is -2 and S is +6.
So, ${{H}_{2}}S{{O}_{3}}$ is undergoing oxidation and acts as a reducing agent, and $S{{n}^{4+}}$ is undergoing reduction and acts as an oxidizing agent.

Therefore, the correct answer is an option (C) ${{H}_{2}}S{{O}_{3}}$ is the reducing agent because it undergoes oxidation.

Note: You may get confused that the substance which gets oxidized also acts as an oxidizing agent and the substance which gets reduced also acts as a reducing agent. But they are the opposite.