Chemistry Class 11 NCERT Solutions: Chapter 8 Redox Reactions Part 6

Chemistry Class 11 NCERT Solutions Chapter 8 Redox Reactions Part 6

Q: 7. Suggest a list of the substances where carbon can exhibit oxidation states from Equation and nitrogen from Equation .

Answer:

The substances where carbon can exhibit oxidation states from Equation are listed in the following table.

Q_7_2_Table of Carbon can Exhibit Oxidation States
Q_7_2_Table of Carbon can Exhibit Oxidation States

Substance

O.N. of Carbon

Equation

0

Equation

+1

Equation

-1

Equation

+2

Equation

-2

Equation

+3

Equation

-3

Equation

+4

Equation

-4

The substances where nitrogen can exhibit oxidation states from –3 to +5 are listed in the following table.

Q_7_2_Table of Nitrogen can Exhibit Oxidation States
Q_7_2_Table of Nitrogen can Exhibit Oxidation States

Substance

O.N. of Nitrogen

Equation

0

Equation

+1

Equation

-1

Equation

+2

Equation

-2

Equation

+3

Equation

-3

Equation

+4

Equation

+5

Q: 8. while sulphur dioxide and hydrogen peroxide can act as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

Q 8 Structure of Sulphur Dioxide

Q 8 Structure of Sulphur Dioxide

Q 8 Structure of Sulphur Dioxide

Q 8 Structure of Hydrogen Peroxide

Q 8 Structure of Hydrogen Peroxide

Q 8 Structure of Hydrogen Peroxide

Answer:

In sulphur dioxide, Equation the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to –2.

Therefore, Equation can act as an oxidising as well as a reducing agent.

In hydrogen peroxide Equation , the O.N. of O is –1 and the range of the O.N. that O can have is from 0 to –2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, Equation can act as an oxidising as well as a reducing agent.

In ozone ( Equation ), the O.N. of O is zero and the range of the O.N. that O can have is from 0 to –2. Therefore, the O.N. of O can only decrease in this case. Hence, Equation acts only as an oxidant.

In nitric acid, Equation the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to –3. Therefore, the O.N. of N can only decrease in this case. Hence, Equation acts only as an oxidant.

Q: 9. Consider the reactions:

(A) Equation

(B) Equation

Why it is more appropriate to write these reactions as:

(A) Equation

(B) Equation

Also, suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Answer:

(A) The process of photosynthesis involves two steps.

Step 1:

Equation Decomposes to give Equation and Equation .

Equation

Step 2:

Equation

Equation

Now, the net reaction of the process is given as:

Equation

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.

The path of this reaction can be investigated by using radioactive Equation in place of Equation .

(B) Equation is produced from each of the two reactants Equation and Equation . For this reason, Equation is written twice.

The given reaction involves two steps. First, Equation decomposes to form Equation and O. In the second step, Equation reacts with the O produced in the first step, thereby producing Equation and Equation .

Equation

The path of this reaction can be investigated by using Equation

Q: 10. The compound Equation is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?

Q 10 Structure of Silver(II) Fluoride

Q 10 Structure of Silver(II) Fluoride

Q 10 Structure of Silver(II) Fluoride

Answer

The oxidation state of Ag in Equation is +2. But, +2 is an unstable oxidation state of Ag. Therefore, whenever Equation is formed, silver readily accepts an electron to form Ag+. This helps to bring the oxidation state of Ag down from +2 to a more stable state of +1. As a result, Equation acts as a very strong oxidizing agent.

Explore Solutions for Chemistry

Sign In