Question

Identify oxidant and reductant in the following redox reaction.
${{\text{P}}_4} + 5{{\text{O}}_2} \to {{\text{P}}_4}{{\text{O}}_{10}}$
${\text{A}}{\text{.}}$ Oxidant:${{\text{O}}_2}$ ; Reductant: ${{\text{P}}_4}{{\text{O}}_{10}}$
${\text{B}}{\text{.}}$ Oxidant: ${{\text{P}}_4}$ ; Reductant: ${{\text{O}}_2}$
${\text{C}}{\text{.}}$ Oxidant: ${{\text{O}}_2}$ ; Reductant: ${{\text{P}}_4}$
${\text{D}}{\text{.}}$ None of these

Answer 1

Hint- Here, we will proceed by defining an oxidation-reduction (redox) reaction. Then, we will discuss the rules for assigning oxidation states. Then, we will write the oxidation and reduction reactions corresponding to the given chemical reaction.

Complete answer:
An oxidation-reduction (redox) reaction is a form of chemical reaction which involves electrons being transferred between two species. An oxidation-reduction reaction is any chemical reaction, in which a molecule, atom, or ion's oxidation number changes by gaining or losing an electron. Redox reactions are normal and essential to some of life's basic functions, including photosynthesis , respiration, combustion, and rusting or corrosion.

Rules for Assigning Oxidation States:
The oxidation state of an element corresponds to the number of electrons that an atom loses, gains, or appears to use when joining with other atoms in compounds. The oxidation state of an atom is determined through the below mentioned seven guidelines:
$1.$ The oxidation state of an individual atom is 0.
$2.$ The total oxidation state of all atoms in a neutral species is 0 and that in an ion is equal to the ion charge.
$3.$ Group 1 metals have an oxidation state of +1 and Group 2 an oxidation state of +2
$4.$ The oxidation state of fluorine is -1 in compounds
$5.$ Hydrogen generally has an oxidation state of +1 in compounds
$6.$ Oxygen generally has an oxidation state of -2 in compounds
$7.$ Group 17 elements in binary metal compounds have an oxidizing state of -1, group 16 elements of -2, and group 15 elements of -3.
Also, it is important to note that the sum of the oxidation states is equal to zero for neutral compounds and equal to the charge for polyatomic ion species.
The given chemical reaction is an oxidation-reduction (redox) reaction. The individual oxidation and reduction reactions involved are mentioned under.
Oxidation chemical reaction, $4{{\text{P}}^0} - 20{{\text{e}}^ - } \to 4{{\text{P}}^{5 + }}$
Reduction chemical reaction, $10{{\text{O}}^0} + 20{{\text{e}}^ - } \to 10{{\text{O}}^{2 - }}$
Adding the above mentioned oxidation and reduction reactions, we have
$4{{\text{P}}^0} - 20{{\text{e}}^ - } + 10{{\text{O}}^0} + 20{{\text{e}}^ - } \to 4{{\text{P}}^{5 + }} + 10{{\text{O}}^{2 - }} \\\ 4{{\text{P}}^0} + 10{{\text{O}}^0} \to 4{{\text{P}}^{5 + }} + 10{{\text{O}}^{2 - }} \\\ {{\text{P}}_4} + 5{{\text{O}}_2} \to {{\text{P}}_4}{{\text{O}}_{10}} \\\$
Clearly in the given chemical reaction, ${{\text{P}}_4}$ is losing its electrons and ${{\text{O}}_2}$ is accepting those electrons. That’s why ${{\text{P}}_4}$ is a reducing agent (reductant) and ${{\text{O}}_2}$ is an oxidizing agent (oxidant).

Hence, option C is correct.

Note- In redox processes, the reductant or reducing agent transfers electrons to the oxidant or oxidizing agent. The reducing or reducing agent thus loses electrons and is oxidized in the reaction, and the oxidizing or oxidizing agent retains electrons, and is reduced.