Redox Reaction: Chemistry

• Answers are in BOLD Option name.

 Q.1.  The oxidation state of nitrogen in   is 

A) 

B) +3

C) -1

D)  

Explanation-

 

Q.2. A,B and C are three elements forming a part of compound in oxidation states of +2,+5 and -2 respectively. What could be the compound if molecule is neutral ?

A) A(BC)

B) A(BC)

C) A(BC)

D) ABC

Explanation-

As the molecule is neutral , the sum of oxidation numbers must be zero .

Q.3. If three electrons are lost by a metal ion  , its final oxidation number would be 

A) Zero

B) +6

C) +2

D) +3

Explanation-

 

Q.4.  The oxidation number of chromium in  is 

A) -6

B) +6

C) +2

D) -2

Explanation-

 

Q.5.  The colour of  changes from red orange to lemon yellow on treatment with aqueous KOH because of 

A) Reduction of Cr(VI) to (III)

B) Formation of chromium ion and chromate ion 

C) Conversion of dichromate ion to chromate ion 

D) Oxidation of potassium hydroxide to potassium peroxide 

Explanation-

IN basic medium , dichromate ion changes to chromate ion 

 

Q.6.  Which pf the following species do not show disproportionation reaction ?

A) CIO

B) CIO 

C) CIO

D) CIO 

Explanation-

  does not disproportionate because in this oxoanion chlorine is present in its highest oxidation state that is , + 7. The disproportionation reactions for  the other three oxoanions of chlorine are as follows :

 

Q.7.  Which of the following is an example of displacement reaction ?

A) X+YZ

B) X+YZXZ+Y

C) ZX+Y

D) None of these 

Explanation-

The reaction in which an atom or ion in a compound is displaced by another atom or ion is known as displacement reaction .

Q.8. In Ostwald's process for the manufacture of nitric acid , the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam .What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g of ammonia and 20.00 g of oxygen ?

A) 10 g

B) 120 g

C) 15 g

D) 20 g

Explanation-

The reaction involved in the manufacturing process is : 

68 of NH reacts with 160 g O to produce 120 g of NO  . 

  10 g of ammonia will require 23.52 g of O react completely . 

but oxygen which is actually available is 20.0 g . Therefore ,

oxygen is the limiting reagent .

Now 160 g of O will form NO = 120 g

 20 g of O will form NO =  

Q.9.  In  the oxidation numbers of Cr and O are 

A) O = + 2 , Cr = - 6

B) O = + 3 , Cr = +5

C) O = + 6 , Cr = - 6

D) O = - 2 , Cr = + 6

Explanation-

Oxidation state of oxygen is - 2 and that Cr can be calculate as 

 

Q.10.  The number of electrons involved in the reduction of one nitrate ion to hydrazine is 

A) 0

B) 2

C) 6

D) 5

Explanation-

Oxidation state of N in  - is + 5 .Oxidation state of N in  is - 2 .

So number of electrons involved in the reduction of one nitrate ion to hydrogen is 6 .