NCERT Solutions for Class 11-science Chemistry Chapter 8 - Redox Reactions

Page / Exercise

Chapter 8 - Redox Reactions Exercise 272

Solution 1

a. NaH2PO4

Let the oxidation number of P be x.

We know that,

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = –2

begin mathsize 14px style stack rightwards double arrow Na with plus 1 on top stack straight H subscript 2 with plus 1 on top straight P with x on top straight O with negative 2 on top subscript 4 end style

Then, we have

1(+1) +2(+1)+1(x)+4(-2) =0

rightwards double arrow1+2+x-8 = 0

begin mathsize 14px style rightwards double arrow end stylex = +5

Hence, the oxidation number of P is +5.

 

b. NaHSO4

Hence, the oxidation number of S is + 6.

 

c. H4P2O7

Hence, the oxidation number of P is + 5.

 

d. K2MnO4

Hence, the oxidation number of Mn is + 6.


e. CaO2

Hence, the oxidation number of O is – 1.

 

f. NaBH4

Hence, the oxidation number of B is + 3.


g. H2S2O7

Hence, the oxidation number of S is + 6.

 

h. KAl(SO4)2.12H2O

Or,

We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have

Hence, the oxidation number of S is + 6.


Solution 2

a. KI3

In KI3, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is begin mathsize 12px style negative 1 third end style. However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3 to find the oxidation states.
In a KI3 molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.
begin mathsize 12px style stack straight K to the power of plus with plus 1 on top open square brackets straight I with 0 on top minus straight I with 0 on top leftwards arrow straight I with negative 1 on top close square brackets to the power of minus end style
Hence, in a KI3 molecule, the O.N. of the two I atoms forming the I2 molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.


 b. H2S4O6

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

 The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.


 c. Fe3O4

On taking the O.N. of O as –2, the O.N. of Fe is found to be begin mathsize 12px style plus 2 2 over 3 end style. However, O.N. cannot be fractional.

Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3.

  d. CH3CH2OH

 Hence, the O.N. of C is –2.


   e. CH3COOH

However, 0 is average O.N. of C. The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and –2 in CH3COOH.

Solution 3

Solution 4
Solution 5


Solution 6
Solution 7

Solution 8
Solution 9

Chapter 8 - Redox Reactions Exercise 273

Solution 1
Solution 2

Solution 3
Solution 4

Solution 5
Solution 6
Solution 7
Solution 8

Chapter 8 - Redox Reactions Exercise 274

Solution 1



Solution 2







Solution 3
Solution 4
Solution 5
Solution 6

Solution 7
Solution 8
Solution 9

Chapter 8 - Redox Reactions Exercise 275

Solution 1

Solution 2
Solution 3
Solution 4

Loading...