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NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reactions covers the classical and electronic concepts of oxidation-reduction, oxidation number rules, balancing redox equations, and redox reactions as the basis of titrations.

Overview of Topics & Subtopics (Click to Jump)

1. Classical and Electronic Concepts of Redox Reactions

Short Note:

• Oxidation: loss of electrons (or addition of oxygen/removal of hydrogen); Reduction: gain of electrons (or addition of hydrogen/removal of oxygen).

• A redox reaction involves both oxidation and reduction occurring simultaneously, with electrons transferred from the reducing agent to the oxidising agent.

2. Oxidation Number

Short Note:

• The oxidation number of an atom is the charge it would have if the compound were composed of ions, assigned according to a set of rules (e.g., oxidation number of an element in its free state is 0; of oxygen is usually −2; of hydrogen is usually +1).

• Oxidation is characterised by an increase in oxidation number; reduction by a decrease in oxidation number.

3. Balancing Redox Reactions

Short Note:

• Oxidation Number Method and Half-Reaction (Ion-Electron) Method are the two standard techniques used to balance redox equations, especially in acidic or basic medium.

4. Redox Reactions in Titrations

Short Note:

• Redox titrations, such as those involving KMnO4 or K2Cr2O7 as oxidising agents, are used to estimate the concentration of reducing agents in a solution based on the stoichiometry of electron transfer.

NCERT Textbook Exercise Questions

Calculate the oxidation number of sulphur in H2SO4, SO2 and S8.
Answer: In H2SO4: 2(+1) + S + 4(−2) = 0 ⇒ S = +6. In SO2: S + 2(−2) = 0 ⇒ S = +4. In S8 (free element): oxidation number = 0.
Justify that the reaction 2Cu2O(s) + Cu2S(s) → 6Cu(s) + SO2(g) is a redox reaction. Identify the species oxidised and reduced.
Answer: In Cu2O, Cu is +1; in the product Cu, oxidation number is 0 — so Cu is reduced. In Cu2S, S is −2; in the product SO2, S is +4 — so sulphur is oxidised. Since both oxidation and reduction occur, it is a redox reaction.
Balance the equation Fe2+ + Cr2O7^2- → Fe3+ + Cr3+ (in acidic medium) using the ion-electron method.
Answer: Oxidation half: Fe²+ → Fe³+ + e− (×6). Reduction half: Cr2O7²− + 14H+ + 6e− → 2Cr³+ + 7H2O. Combining: 6Fe²+ + Cr2O7²− + 14H+ → 6Fe³+ + 2Cr³+ + 7H2O.
Identify the oxidising and reducing agent in the reaction: MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O.
Answer: Mn goes from +4 (in MnO2) to +2 (in MnCl2), so it is reduced — MnO2 is the oxidising agent. Cl goes from −1 (in HCl) to 0 (in Cl2), so it is oxidised — HCl is the reducing agent.
Why can hydrogen peroxide (H2O2) act as both an oxidising agent and a reducing agent?
Answer: In H2O2, oxygen has an oxidation state of −1, which is intermediate between 0 (in O2) and −2 (in H2O). Since it can be either oxidised to O2 (oxidation number increases to 0) or reduced to H2O (decreases to −2), H2O2 can act as both an oxidising and a reducing agent depending on the reaction.
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Class 11 › Chemistry › Redox Reactions
Chapter notesRead the chapterPractice MCQsCheck what you rememberChapter test10 questions on one pageOnline testTest 1, 2 and 3, one question at a timePYQPrevious year questionsStudy materialPDF notes
Topics in this chapter: Classical & Electron Transfer Concepts of Redox ReactionsOxidation Number Concept & Stock NotationTypes of Redox Reactions & DisproportionationBalancing Redox Reactions & Electrochemical Cells