Question

Consider the reaction of acidified potassium permanganate with sodium sulphite as follows:

$mn {o}^{ - } \frac{}{4} + {h}^{ + } + so \frac{2 - }{3} > m {n}^{ + 2} + so \frac{2 - }{4} + h2o$

It is a redox reaction in which one species is getting oxidized by losing electron/s and other is getting
reduced by gaining electron/s.
The number of electrons which the atom of an element loses or gains in going from its free elemental
state (which is 0) to its new state in that particular compound is called the Oxidation Number (O.N.) of the
element in that compound.
The oxidation number of an element in a compound is determined by the algebraic sum of oxidation
numbers of the individual atoms, each multiplied by the number of atoms of the element in the molecule
which is = 0 (in case of a compound) and for an ion is equal to the charge present. For example the
algebraic sum of O.N. of atoms in H2O is shown as below :
(oxidation number of oxygen is generally -2, O.N. of hydrogen = +1)

H2O
(+1) × 2 + (-2) = 0
The substance which undergoes reduction is called an oxidant and that which undergoes oxidation is
called a reductant.
Answer the following questions:
4. Which species is getting reduced in the above reaction?
(A) MnO^- /4


(B) H^+


(C) SO^2-/3


(D) Mn ^2+

5. Which species is acting as a reductant?
A) MnO^- /4


(B) H^+


(C) SO^2-/3


(D) Mn ^2+

6. If equivalent weight =Molecular weight/
Valence factor

where valence factor = no. of electrons lost or gained by one molecule of reductant or oxidant in a
reaction and is calculated by taking the difference of O. N. of that element in the reactant & the
product. It is always positive.
What will be the equivalent weight of KMnO4 in the above reaction?
(At. Mass K = 39; Mn = 55; O =16)
(A) 158 (B) 79
(C) 39.5 (D) 31.6
​

Answer 1

Answer:

Consider the reaction of acidified potassium permanganate with sodium sulphite as follows:

mn {o}^{ - } \frac{}{4} + {h}^{ + } + so \frac{2 - }{3} > m {n}^{ + 2} + so \frac{2 - }{4} + h2omno

−

4

+h

+

+so

3

2−

>mn

+2

+so

4

2−

+h2o

It is a redox reaction in which one species is getting oxidized by losing electron/s and other is getting

reduced by gaining electron/s.

The number of electrons which the atom of an element loses or gains in going from its free elemental

state (which is 0) to its new state in that particular compound is called the Oxidation Number (O.N.) of the

element in that compound.

The oxidation number of an element in a compound is determined by the algebraic sum of oxidation

numbers of the individual atoms, each multiplied by the number of atoms of the element in the molecule

which is = 0 (in case of a compound) and for an ion is equal to the charge present. For example the

algebraic sum of O.N. of atoms in H2O is shown as below :

(oxidation number of oxygen is generally -2, O.N. of hydrogen = +1)

H2O

(+1) × 2 + (-2) = 0

The substance which undergoes reduction is called an oxidant and that which undergoes oxidation is

called a reductant.

Answer the following questions:

4. Which species is getting reduced in the above reaction?

(A) MnO^- /4

(B) H^+

(C) SO^2-/3

(D) Mn ^2+

5. Which species is acting as a reductant?

A) MnO^- /4

(B) H^+

(C) SO^2-/3

(D) Mn ^2+

6. If equivalent weight =Molecular weight/

Valence factor

where valence factor = no. of electrons lost or gained by one molecule of reductant or oxidant in a

reaction and is calculated by taking the difference of O. N. of that element in the reactant & the

product. It is always positive.

What will be the equivalent weight of KMnO4 in the above reaction?

(At. Mass K = 39; Mn = 55; O =16)

(A) 158 (B) 79

(C) 39.5 (D) 31.6