Question

Identify oxidant and reductant in the following redox reaction.
$2Mn{O_2} + 4KOH + {O_2} \to 2{K_2}Mn{O_4} + 2{H_2}O$
A.Oxidant: ${O_2}$; Reductant : $Mn{O_2}$
B. Oxidant: ${O_2}$; Reductant : $KOH$
C. Oxidant: $Mn{O_2}$; Reductant : ${O_2}$
D. None of these

Answer 1

An element having higher oxidation number is considered to be a good oxidizing agent. On the other hand, an element having lower oxidation number is considered to be a good reducing agent.

Complete step by step answer:
Given reaction is,
$2Mn{O_2} + 4KOH + {O_2} \to 2{K_2}Mn{O_4} + 2{H_2}O$
In the reaction, we see that the oxidation number of ${O_2}$ molecules decreases from 0 to -1. So we can say that oxygen is reduced in the reaction. Therefore, ${O_2}$​ acts as an oxidant.
In the above reaction, we see that the oxidation number of $Mn$ increases from +4 to +6. Therefore, $Mn$ is oxidized in the reaction. So, $Mn{O_2}$ acts as a reductant.

So, the correct answer is Option A .

Note:
One important example of redox reaction is, during the combustion of wood with the molecular oxygen, the oxidation state of carbon atoms in the wood increases and the oxidation state of the oxygen atoms decrease and carbon dioxide and water are formed. The oxygen atoms in this combustion reaction undergo reduction, by gaining electrons, and the carbon atoms during the combustion undergo oxidation, by losing electrons. Thus, oxygen acts as the oxidizing agent and carbon acts as the reducing agent in this reaction.
In a redox reaction, there exists one reducing agent and one oxidizing agent. The reducing agent is the one that loses an electron and the oxidizing agent is marked by the gaining of an electron.