Question

Identify the oxidizing agent and the reducing agent in the following reaction:
$Mn{{O}_{4}}^{-}+S{{O}_{3}}^{2-}\to M{{n}^{2+}}+S{{O}_{4}}^{2-}$
(A) Oxidizing agent$Mn{{O}_{4}}^{-}$; reducing agent $S{{O}_{3}}^{2-}$
(B) Oxidizing agent$S{{O}_{3}}^{2-}$; reducing agent$Mn{{O}_{4}}^{-}$
(C) Can’t be predicted
(D) None of the above

Answer 1

Find the oxidation states of each element present in the reaction. Do it for both the reactant and product side. Then compare the oxidation states.

Complete step-by-step answer:
A redox reaction always involves exchange of electrons. If something is oxidised then some other thing has to be reduced or vice-versa. If we go by the definitions, oxidation is characterized as a loss of electrons from an atom of a particular element. Similarly we can define reduction is gain of electrons by any entity belonging to one of the elements of the periodic table.
The reaction given above is:
$Mn{{O}_{4}}^{-}+S{{O}_{3}}^{2-}\to M{{n}^{2+}}+S{{O}_{4}}^{2-}$
Let’s start with calculating the oxidation states of the atoms on both sides of the reaction.
It is known that the oxidation state of oxygen is always “$-2$” provided it is bonded to any element which is less electronegative than itself. We will now go from molecule to molecule starting at the left hand side.
- $Mn{{O}_{4}}^{-}$
If we take the oxidation state of manganese to be “$x$” then the equation becomes-
$x+4(-2)=-1$
because the overall charge on the molecule is “$-1$ ”.
Solving the above we get $x=+7$, which is the oxidation state of manganese.
- Taking similar methods we get the oxidation state of sulphur in $S{{O}_{3}}^{2-}$ as $+4$; that of manganese in $M{{n}^{2+}}$as $+2$; and oxidation of sulphur in $S{{O}_{4}}^{2-}$is $+6$.
It is clear that the oxidation state of manganese has decreased from $+7$to$+2$ and that of sulphur has increased from $+4$to$+6$. Therefore manganese has reduced and sulphur has oxidized. That gives manganese as the oxidizing agent and sulphur as reducing agent.
The correct answer is option is (A) Oxidizing agent $Mn{{O}_{4}}^{-}$; reducing agent $S{{O}_{3}}^{2-}$

Note: The one which oxidises itself reduces another and that makes it a reducing agent. The one which reduces itself oxidises another and that makes it an oxidizing agent. Which compound or element oxidised or reduces is decided by their respective electrode potential.
As in the above question, as long as we have the complete reaction we can figure out the oxidizing and reducing agents.