MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Get instant 20% OFF on Online Material.
coupon code: MOB20 | View Course list

Get extra R 260 off
USE CODE: omar10

Solved Examples:

Question 1:

Find the oxidation number of 

a. S in SO42- ion

b. S in HSO3- ion

c. Pt in (PtCl6)2-

Solution:

a. Let the oxidation state of S in SO4= x

We know that ox. no. of O = -2

So, x +  4(-2) = -2

or x =+6

b. Let the oxidation state of S in HSO3- = x

We know that ox. no. of O = -2

&  ox. no. of H = +1

So, +1 +x+3(-2) = -1

or x = +4

c. Let the oxidation state of Pt in (PtCl6)2- = x

We know that ox. no. of Cl = -1

So, x+ 6(-1) = -2

or x =  -2 + 6 = +4

 ______________________________________________________________

Question 2: 

What is the oxidation number of Mn is KMnO4 and of S in Na2S2O3?

Solution:

Let the Ox.no. of Mn is KMnO4 be x.

We know that

 Ox.no. of K = +1

Ox.no. of O = –2

So    
Ox.no. K + Os.No. Mn+4 (Ox.no.O) = 0

or   +1+ x + (4–2)  = 0

or     +1+ x  – 8  = 0

or      x = +8 – 1 = +7

Hence, Ox.no. of Mn in KMnO4 is +7

________________________________________________

Question 3: 

What is the oxidation number of Cr in K2Cr3O7 ?

Solution:

Let the oxidation number of Fe be x.

We know that

Ox.no. of K =+1

Ox.no. of (CN) = –1

So     4(Ox.no. K) + Os.No. Fe+6(Ox.no. CN) = 0

or      4(+1)      +     x      +      (6–1)        = 0

or         +4       +     x       –       6           = 0

or           x = +6 – 4 = +2

The oxidation number of iron in K4Fe(CN)6 is +2.

______________________________________________________

Question 4:

Which compound amongst the following has the highest oxidation number for Mn?

KMnO4, K2MnO4, MnO2 and Mn2O3

Solution:

KMnO4,     
+1+x –8 = 0             
ot x = + 7 

Ox.no. of Mn =- +7

K2MnO4,   

+2 +x –8 = 0                 

x = +6                         

MnO2      

x – 4 = 0

or  x = +4        

Mn2O3        

2x – 6 = 0

 x = + 3                            

Thus, the highest oxidation number for the Mn is in KMnO4.

________________________________________________________________________________

Question 5:

Can oxidation number of any element in a compound ever be zero? Justify the answer.

Solution

Sometimes, oxidation numbers have such values which at first sight appear strange. For example, the oxidation number of carbon in cane sugar (C12H22O11), glucose (C6H12O6), dichloromethane, etc., is zero.

Cane Sugar (C12H22O11)                         Glucose (C6H12O6)

12×x+22×1+11(–2)=0                           6×x+12×1+6(–2)=0

12x+22 – 22 = 0                                   6x+12–12=0

So x = 0                                              So x = 0

                 Dichloromethane (CH2Cl2)

                  x = 2 × 1 + 2(–1) = 0

                       x + 2 – 2 =0

                          So x = 0

_______________________________________________________________

Question 6: 

Balance the equation Cr(OH)3 + IO3 -   I - + CrO42-

Solution:

Oxidation : Cr(OH)3  CrO42-     Reduction : IO3-  I-

Balancing oxidation reaction : Balancing atoms other than H and O

Cr(OH)3  CrO42-

Balancing O;         2OH- + Cr(OH)3  CrO42- + H2O

Balancing H;         3OH- + 2OH- + Cr(OH)3 CrO42- + H2O + 3H2O

Balancing charge; 5OH- + Cr(OH)3  CrO42- + 4H2O + 3e- (1)

Balancing reduction reaction :

Balancing O;         IO3- + 3H2O 6OH- + I-

Balance the charges 6e- + IO3- + 3H2O ® 6OH- + I- (2)

Multiply (1) by 2 and adding to (2)

2Cr(OH)3 + IO3- + 4OH- 2CrO42- + 5H2O + I-.     

____________________________________________


Question 7: 

Balance the equation

Cr2O72- + C2O42- + H+ ® Cr+3 + CO2 + H2O

Solution:

Reduction : Cr2O72- Cr+3    Oxidation : C2O42-  CO2

Balancing atoms other than O and H : Cr2O72-  2Cr+3

Reaction is taking place in acidic medium.

Balancing O

Cr2O72-  2Cr+3 + 7H2O

Balancing H

14H+ + Cr2O72-  2Cr+3 + 7H2O

Balancing charges, we get reduction half-reaction.

 6e- + 14H+ + Cr2O72- →  2Cr+3 + 7H2O(1)

Balancing oxidation reaction :

Balancing C    C2O42- 2CO2

Balancing O    C2O42-  2CO2

Balancing charges, we get oxidation half-reaction

C2O42-  2CO2 + 2e-             (2)

Multiplying (2) by 3 and adding to (1) (To cancel out the electrons)

Cr2O72- + 14H+ + 3C2O42- → 2Cr+3 + 7H2O + 6CO2

_________________________________________________

Question 8:

One mole of N2H4 loses 10 mole electrons to form a new compound Y. Assuming that all the N2 appears in new compound, what is the oxidation state of Nitrogen in Y? (There is no change in the oxidation state of H)

Solution:

N2H4 →  (Y) + 10e-

( Y contains all N atoms)

i.e.  (2N)x + 10e-

2x - (-4) = 10

x = +3

______________________________________________

Related Resources

You can also have a look at balancing redox reactions

Click here to refer the past year papers of IIT JEE

You can also refer to the syllabus of chemistry for IIT JEE

To read more, Buy study materials of Redox Reactions comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.

  • Complete JEE Main/Advanced Course and Test Series
  • OFFERED PRICE: R 15,000
  • View Details
Get extra R 1,500 off
USE CODE: omar10