Q 1 :

Which of the following cannot function as an oxidising agent?               [2024]

  • N3-

     

  • SO42-

     

  • BrO3-

     

  • MnO4-

     

(1)              

In nitride ion (N3-), nitrogen is in its minimum possible oxidation state (-3) hence it cannot be further reduced. Thus N3- cannot act as an oxidizing agent.

 



Q 2 :

Chlorine undergoes disproportionation in alkaline medium as shown below:

aCl2(g)+bOH(aq)-cClO(aq)-+dCl(aq)-+eH2O(l)

The values of a, b, c and d in a balanced redox reaction are respectively:

  • 2, 2, 1 and 3

     

  • 1, 2, 1 and 1

     

  • 2, 4, 1 and 3

     

  • 3, 4, 4 and 2

     

(2)        

C0l2(g)+2OH-(aq)Cl-(aq)+Cl+1O-(aq)+H2O(l)

a = 1, b = 2, c = 1, d = 1

 



Q 3 :

In acidic medium, K2Cr2O7 shows oxidizing action as represented in the half reaction:

Cr2O72-+XH++Ye2A+ZH2O

X, Y, Z and A, respectively, are                         [2024]

  • 8, 6, 4 and Cr2O3

     

  • 14, 7, 6 and Cr3+

     

  • 8, 4, 6 and Cr2O3

     

  • 14, 6, 7 and Cr3+

     

(4)         

In acidic medium the half reaction is balanced as follows:

(A) Balance atoms undergoing reduction i.e. chromium:

Cr2O72-(aq)2Cr3+(aq)

(B) Balance oxygen using H2O

Cr2O72-(aq)2Cr3+(aq)+7H2O(l)

(C) Balance hydrogen using H+

Cr2O72-(aq)+14H+(aq)2Cr3+(aq)+7H2O(l)

(D) Balance charge using electrons:

Cr2O72-(aq)+14H+(aq)+6e-2Cr3+(aq)+7H2O(l)

On comparing with given reaction we get,

X=14, Y=6, Z=7, A=Cr3+



Q 4 :

Thiosulphate reacts differently with iodine and bromine in the reaction given below:

2S2O32-+I2S4O62-+2I-

S2O32-+5Br2+5H2O2SO42-+4Br-+10H+

Which of the following statement justifies the above dual behaviour of thiosulphate?              [2024]

  • Bromine is a weaker oxidant than iodine

     

  • Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions

     

  • Bromine undergoes oxidation and iodine undergoes reduction in these reactions

     

  • Bromine is a stronger oxidant than iodine

     

(4) 

2S2+2O32-+I2S4+2.5O62-+2I-

S2+2O32-+5Br2+5H2O2S+6O42-+4Br-+10H+

Br2 is oxidizing S of thiosulphate to a higher oxidation state than I2. This illustrates that Br2 is a stronger oxidizing agent than I2.

 



Q 5 :

2MnO4-+bI-+cH2OxI2+yMnO2+zO¯H

If the above equation is balanced with integer coefficients, the value of z is ______.             [2024]



(8)

2MnO4-(aq)+4H2O(l)+6I-(aq)2MnO2(s)+3I2(s)+8OH-(aq)

This is required balanced chemical equation.

 



Q 6 :

H2O2 acts as a reducing agent in                    [2023]

  • Na2S+4H2O2Na2SO4+4H2O

     

  • Mn2++2H2O2MnO2+2H2O

     

  • 2Fe2++2H++H2O22Fe3++2H2O

     

  • 2NaOCl+H2O22NaCl+H2O+O2

     

(4)

H2O2 acts as a reducing agent in the following reaction

NaOCl+H2O22NaCl+H2O+O2



Q 7 :

2IO3-+xI-+12H+6I2+6H2O

What is the value of x ?                 [2023]

  • 10

     

  • 2

     

  • 6

     

  • 12

     

(1)

2IO3-+10I-+12H+6I2+6H2O



Q 8 :

Sum of oxidation states of bromine in bromic acid and perbromic acid is ________.              [2023]



(12)

HBrO4Perbromic acid

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

x=+7

HBrO3Bromic acid

(+1)+x+3(-2)=0

x=+5

Sum of oxidation numbers of Bromine=7+5=12



Q 9 :

See the following chemical reaction:

Cr2O72-+XH++6Fe2+YCr3++6Fe3++ZH2O

The sum of X, Y and Z is _______ .                    [2023]



(23)

Balance reaction is:

Cr2O72-+14H++6Fe+26Fe+3+2Cr+3+7H2O

                   X=14                                       Y=2        Z=7

Hence (X+Y+Z)=14+2+7=23