Question

: In the reaction:
\[Mn{O_2} + 4HCl \to MnC{l_2} + 2{H_2}O + C{l_2}\]
A.Name of compound (i) oxidised (ii) reduced
B.Define oxidation of reduction on its basis.

Answer 1

Hint: Generally the compound whose oxidation number increases during the reaction is known as oxidised compound and the compound whose oxidation number decreases during the reaction is known as reduced compound.

Complete step by step answer:
Let us first talk about the oxidations number of elements in the compounds given in the reaction.
Oxidation number of an element: It is defined as the total number of electrons that an atom accepts or loses in order to make a chemical bond which results in the formation of a chemical compound.
Here we are given with the transition metal i.e. manganese. The atomic number of manganese is $25$. Hence its electronic configuration is as: $Ar3{d^5}4{s^2}$. So it can show a number of valencies by losing five d-electrons and two s-electrons. And the compound of manganese on the reactant side is \[Mn{O_2}\] and we know that the oxidation number of oxygen is $ - 2$. And let’s suppose that the oxidation number of manganese in this compound be $x$. So the equation of sum of oxidation number will be as
 $
  x + 2( - 2) = 0 \\
  x = + 4 \\
 $.
So the oxidation number of Manganese on the reactant side is $ + 4$. Now on the product side the compound of manganese is \[MnC{l_2}\] and we know that the oxidation number of chlorine is $ - 1$. And let’s suppose that the oxidation number of manganese in this compound is $y$. So the equation of sum of oxidation number will be as
$
  y + 2( - 1) = 0 \\
  y = + 2 \\
 $.
So the oxidation number of Manganese on the product side is $ + 2$. So we can say that the oxidation number of manganese decreases hence $Mn{O_2}$ is a reduced compound of the reaction.
Now on the reactant side the compound is $HCl$ in which the oxidation number of hydrogen is $ + 1$ and oxidation number of chlorine is $ - 1$. And on the product side the compound is chlorine in which the oxidation number of chlorine is zero. So we can say that the oxidation number of chlorine increases hence $HCl$ is an oxidised compound of the reaction.

Hence oxidised compound is $HCl$ and reduced compound is $Mn{O_2}$.

B.Oxidation: The compound which loses the electron during the reaction or the compound whose oxidation number increases during the reaction.
Reduction: The compound which gains the electron during the reaction or the compound whose oxidation number decreases during the reaction.

Note:
Generally the compound which is oxidised in the reaction will act as reducing reagent (which oxidises the other compounds) and the compound which is reduced in the reaction will act as oxidising reagent (which reduces the other compounds) for the reaction.