Solved Examples

 

Question 1:

Find the oxidation number of 

a. S in SO₄^2- ion

b. S in HSO^3- ion

c. Pt in (PtCl₆)^2-

Solution:

a. Let the oxidation state of S in SO₄² = 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

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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

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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.

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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.

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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

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Question 6: 

Balance the equation Cr(OH)₃ + IO₃ ^- →  I ^- + CrO₄^2-

Solution:

Oxidation : Cr(OH)₃ → CrO₄^2-     Reduction : IO₃^- → I^-

Balancing oxidation reaction : Balancing atoms other than H and O

Cr(OH)₃ → CrO₄^2-

Balancing O;         2OH^- + Cr(OH)₃ → CrO₄^2- + H₂O

Balancing H;         3OH^- + 2OH^- + Cr(OH)₃ → CrO₄^2- + H₂O + 3H₂O

Balancing charge; 5OH^- + Cr(OH)₃ → CrO₄^2- + 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^- → 2CrO₄^2- + 5H₂O + I^-.     

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Question 7: 

Balance the equation

Cr₂O₇^2- + C₂O₄^2- + H⁺ ® Cr⁺³ + CO₂ + H₂O

Solution:

Reduction : Cr₂O₇^2- →Cr⁺³    Oxidation : C₂O₄^2- → CO₂

Balancing atoms other than O and H : Cr₂O₇^2- → 2Cr⁺³

Reaction is taking place in acidic medium.

Balancing O

Cr₂O₇^2- → 2Cr⁺³ + 7H₂O

Balancing H

14H⁺ + Cr₂O₇^2- → 2Cr⁺³ + 7H₂O

Balancing charges, we get reduction half-reaction.

 6e^- + 14H⁺ + Cr₂O₇^2- →  2Cr⁺³ + 7H₂O(1)

Balancing oxidation reaction :

Balancing C    C₂O₄^2- →2CO₂

Balancing O    C₂O₄^2- → 2CO₂

Balancing charges, we get oxidation half-reaction

C₂O₄^2- → 2CO₂ + 2e^-             (2)

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

Cr₂O₇^2- + 14H⁺ + 3C₂O₄^2- → 2Cr⁺³ + 7H₂O + 6CO₂

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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

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Related Resources

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