CBSE Class 11 Chemistry Chapter 7 Notes Redox Reactions

Redox reactions are chemical reactions where oxidation and reduction occur simultaneously. Every redox reaction involves an oxidising agent that accepts electrons and a reducing agent that donates them. Oxidation number tracks electron shifts across combination, decomposition, displacement, and disproportionation reactions. These class 11 chemistry redox reaction notes follow the NCERT sequence from classical definitions to electrode processes.

Iron rusts, batteries discharge, and bleach removes stains because of redox reactions. These redox reaction class 11 notes cover everything in NCERT Chapter 7: classical definitions, electron transfer, oxidation number rules, four reaction types, balancing methods, and electrode processes. The notes follow the NCERT Reprint 2026-27 sequence. The official chapter PDF is available at ncert.nic.in. For exam practice, visit Important Questions Class 11 Chemistry Chapter 7.

Students who skip oxidation number rules drop marks on balancing questions. Students who master the half-reaction method balance any equation in under three minutes. Use these class 11 chemistry redox reaction notes alongside your teacher's guidance to cover every exam-ready concept in this chapter.

Key Takeaways: CBSE Class 11 Chemistry Chapter 7

This chapter table covers all chapters connected to redox reaction class 11 and beyond.

Chapter-Wise Class 11 Chemistry Notes

Classical Idea of Oxidation and Reduction

The earliest definitions of oxidation and reduction were built around oxygen and hydrogen. As chemists studied more reactions, these definitions expanded to cover all electronegative and electropositive elements.

Oxidation is the addition of oxygen or an electronegative element to a substance, or the removal of hydrogen or an electropositive element. Reduction is the reverse: removal of oxygen or an electronegative element, or the addition of hydrogen or an electropositive element.

Examples of Oxidation

2Mg(s) + O2(g) → 2MgO(s): oxygen added to magnesium 2H2S(g) + O2(g) → 2S(s) + 2H2O(l): hydrogen removed from H2S

Examples of Reduction

2HgO(s) → 2Hg(l) + O2(g): oxygen removed from HgO CH2=CH2(g) + H2(g) → H3C-CH3(g): hydrogen added to ethene

Oxidation and reduction always occur together in the same reaction. That is why the combined term redox was coined.

Redox Reactions in Terms of Electron Transfer

The electronic definition resolves cases where classical definitions fall short, particularly in covalent compounds. This section forms the foundation of class 11 redox reaction notes and connects directly to electrochemistry in Class 12.

Oxidation is the loss of electrons by any species. Reduction is the gain of electrons. The species that accepts electrons is the oxidising agent. The species that donates electrons is the reducing agent.

Formation of NaCl: Electron Transfer Illustrated

Oxidation half-reaction: 2Na(s) → 2Na+(g) + 2e- Reduction half-reaction: Cl2(g) + 2e- → 2Cl-(g) Overall: 2Na(s) + Cl2(g) → 2NaCl(s)

Sodium is oxidised and acts as the reducing agent. Chlorine is reduced and acts as the oxidising agent.

Competitive Electron Transfer Reactions

When zinc is placed in a copper sulphate solution, this reaction occurs: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Zinc loses electrons to copper ions. The electron-releasing tendency follows: Zn > Cu > Ag. This order forms the basis of the electrochemical activity series and galvanic cell design.

Oxidation Number: Rules and Application

Oxidation number is the hypothetical charge on an atom in a compound, assuming complete electron transfer from the less electronegative to the more electronegative atom. Mastering these rules is essential before attempting any balancing question in class 11 chemistry chapter redox reaction notes.

These six NCERT rules cover every case you will encounter in the 2026 CBSE exam.

Rules for Assigning Oxidation Number

Rule 1: Free elements Oxidation number = 0. H2, O2, Na, S8, and P4 all have oxidation number zero.

Rule 2: Monoatomic ions Oxidation number equals the charge on the ion. Na+ is +1, Mg2+ is +2, Cl- is -1. Alkali metals are always +1, alkaline earth metals always +2, and aluminium always +3 in compounds.

Rule 3: Oxygen in compounds Usually -2. Exceptions: peroxides (H2O2, Na2O2) = -1; superoxides (KO2, RbO2) = -1/2; OF2 = +2; O2F2 = +1.

Rule 4: Hydrogen in compounds Usually +1. Exception: metal hydrides (NaH, LiH, CaH2) where hydrogen = -1.

Rule 5: Fluorine Always -1 in all compounds. Other halogens are -1 as halide ions but show positive oxidation states in oxoacids and oxoanions.

Rule 6: Algebraic sum Sum of oxidation numbers in a neutral molecule = 0. In a polyatomic ion, sum = charge on the ion.

Worked Examples

Mn in KMnO4: +1 + x + 4(-2) = 0, so x = +7 Cr in Cr2O72-: 2x + 7(-2) = -2, so x = +6

Stock Notation in Redox Reactions

Stock notation uses Roman numerals in parentheses after the metal symbol to indicate its oxidation state. Alfred Stock introduced this system to avoid ambiguity in naming compounds with variable valency.

Hg2(I)Cl2 is the reduced form of Hg(II)Cl2. For formula references, visit Chemistry Full Forms.

Types of Redox Reactions

NCERT classifies redox reactions into four types. Identifying each type is a standard task in redox reaction class 11 questions.

Each type has a specific structural rule. Learn the rule, not just the example.

1. Combination Reactions

At least one reactant must be in elemental form for a combination reaction to qualify as a redox reaction.

C(s) + O2(g) → CO2(g): carbon goes from 0 to +4; oxygen goes from 0 to -2 3Mg(s) + N2(g) → Mg3N2(s): Mg oxidised, N reduced

All combustion reactions involving elemental O2 are redox reactions.

2. Decomposition Reactions

A compound breaks down into two or more components. At least one product must be in elemental form.

2H2O(l) → 2H2(g) + O2(g): H goes from +1 to 0; O goes from -2 to 0 2KClO3(s) → 2KCl(s) + 3O2(g): Cl goes from +5 to -1; O goes from -2 to 0

Not all decomposition reactions are redox. CaCO3 → CaO + CO2 has no change in oxidation number and is not a redox reaction.

3. Displacement Reactions

An atom or ion in a compound is replaced by another element. Two sub-types appear in class 11 redox reaction questions.

Metal displacement: A more reactive metal displaces a less reactive one. CuSO4(aq) + Zn(s) → Cu(s) + ZnSO4(aq) V2O5(s) + 5Ca(s) → 2V(s) + 5CaO(s): industrial application

Non-metal displacement: Covers hydrogen and halogen displacement. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g): lab preparation of hydrogen Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(l): Layer Test for Br- identification

Halogen oxidising power decreases down Group 17: F2 > Cl2 > Br2 > I2.

4. Disproportionation Reactions

The same element in a single compound is simultaneously oxidised and reduced. The element must exist in at least three oxidation states for this to be possible.

2H2O2(aq) → 2H2O(l) + O2(g): oxygen at -1 goes to -2 (reduced) and 0 (oxidised) Cl2(g) + 2OH-(aq) → ClO-(aq) + Cl-(aq) + H2O(l): Cl goes from 0 to +1 and -1

Fluorine cannot disproportionate. It is the most electronegative element and cannot take any positive oxidation state.

A note on fractional oxidation numbers: in C3O2, carbon appears to be at +4/3. In reality, the two terminal carbons are at +2 and the middle carbon is at 0. Fractional oxidation number is always an average across different whole-number states within the same compound.

Balancing Redox Reactions: Oxidation Number Method

The oxidation number method equalises the total increase and decrease in oxidation numbers using stoichiometric multipliers. It works well for molecular equations and is often faster for straightforward reactions.

These steps apply to all redox reaction class 11 questions that ask for balancing by this method.

Steps

1. Write correct formulas for all reactants and products.
2. Assign oxidation numbers to all elements.
3. Identify which atoms are oxidised and which are reduced.
4. Calculate total increase and decrease in oxidation number.
5. Multiply by suitable numbers to equalise them.
6. Add H+ (acidic medium) or OH- (basic medium) to balance ionic charges.
7. Add H2O to balance hydrogen atoms. Verify oxygen count.

Worked Example: Cr2O72- with SO32- in Acidic Medium

Cr: +6 to +3 (decrease of 3 per Cr atom, total decrease of 6 for 2 Cr atoms) S: +4 to +6 (increase of 2 per S atom, need 3 S atoms to equalise)

After adding 8H+ and balancing: Cr2O72-(aq) + 3SO32-(aq) + 8H+(aq) → 2Cr3+(aq) + 3SO42-(aq) + 4H2O(l)

Balancing Redox Reactions: Half-Reaction Method

The half-reaction method, also called the ion-electron method, splits the equation into separate oxidation and reduction half-reactions. Each is balanced independently for atoms and charge, then combined. This method is preferred for ionic equations in aqueous solution.

Balancing in Acidic Medium

Worked Example: Fe2+ oxidised by Cr2O72-

Step 1: Fe2+(aq) + Cr2O72-(aq) → Fe3+(aq) + Cr3+(aq)

Step 2: Split into half-reactions. Oxidation: Fe2+(aq) → Fe3+(aq) Reduction: Cr2O72-(aq) → 2Cr3+(aq)

Step 3: Balance O with H2O and H with H+. Cr2O72-(aq) + 14H+(aq) → 2Cr3+(aq) + 7H2O(l)

Step 4: Balance charge with electrons. Fe2+(aq) → Fe3+(aq) + e- Cr2O72-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l)

Step 5: Equalise electrons, multiply oxidation half by 6 and add. 6Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)

Balancing in Basic Medium

Follow all acidic medium steps first. Then for every H+ present, add equal OH- ions to both sides. Where H+ and OH- appear on the same side, combine them to give H2O.

Worked Example: MnO4- with I- in basic medium

After balancing and equalising electrons: 6I-(aq) + 2MnO4-(aq) + 4H2O(l) → 3I2(s) + 2MnO2(s) + 8OH-(aq)

Redox Reactions as the Basis for Titrations

In redox titrations, the strength of a reductant or oxidant is determined using a redox-sensitive indicator. Knowing which indicator works for which reaction is a common 1-mark and 2-mark question in CBSE 2026 papers.

Three endpoint detection methods are covered in NCERT.

Self-indicator: MnO4- is intensely purple. After all reductant is consumed, the first lasting pink from excess MnO4- marks the endpoint.

External redox indicator: Cr2O72- oxidises diphenylamine just after the equivalence point to give an intense blue colour.

Iodometric method: Reagents that oxidise I- liberate I2, which gives an intense blue colour with starch. I2 is then titrated against thiosulphate ions. I2(aq) + 2S2O32-(aq) → 2I-(aq) + S4O62-(aq)

Redox Reactions and Electrode Processes

When the Zn-CuSO4 reaction runs with electrons transferring through an external wire, it becomes a Daniell cell. This section connects class 11 chemistry chapter 7 notes directly to Class 12 electrochemistry.

Understanding electrode processes here makes the Class 12 chapter significantly easier.

Daniell Cell

Zinc rod in ZnSO4 solution = anode (oxidation) Copper rod in CuSO4 solution = cathode (reduction) Salt bridge (KCl or NH4NO3 in agar-agar) completes the internal circuit Electrons flow from anode to cathode through the external wire Current flows opposite to electron flow

At anode: Zn(s) → Zn2+(aq) + 2e- At cathode: Cu2+(aq) + 2e- → Cu(s)

The two redox couples are Zn2+/Zn and Cu2+/Cu (oxidised form written before reduced form).

Standard Electrode Potential

Standard electrode potential (E°) is measured at 298 K when all species are at unity concentration and gases are at 1 atm. The H+/H2 electrode is the reference point at 0.00 V.

More positive E° = stronger oxidising agent. More negative E° = stronger reducing agent.

Standard Electrode Potentials: Key Values at 298 K

For chemistry terminology reference, visit Chemistry Full Forms.

Redox Reaction Class 11 Short Notes

These redox reaction class 11 short notes cover every must-know concept for last-minute revision before the 2026 CBSE exam.

Oxidation: Loss of electrons / increase in oxidation number / addition of O or electronegative element / removal of H or electropositive element.

Reduction: Gain of electrons / decrease in oxidation number / removal of O / addition of H or electropositive element.

Oxidation number rules to memorise:

1. Free element = 0
2. O = -2 (except: peroxide -1, superoxide -1/2, OF2 = +2)
3. H = +1 (except metal hydrides: -1)
4. F always = -1
5. Alkali metals +1; alkaline earth metals +2; Al always +3

Four reaction types:

1. Combination: at least one elemental reactant
2. Decomposition: at least one elemental product
3. Displacement: metal or non-metal displacement
4. Disproportionation: same element, same compound, oxidised and reduced simultaneously

Balancing: Acidic medium: use H+ and H2O Basic medium: balance as acidic, then replace H+ with OH- and H2O

Key rules: Disproportionation requires element to exist in at least 3 oxidation states. Fluorine never disproportionates. Daniell cell: Zn = anode, Cu = cathode. Electrons flow from Zn to Cu. Salt bridge completes the circuit.

The complete chapter-wise notes are available through the links in the chapter table above. Use these redox reactions class 11 notes alongside full section revision for best results.

Redox Reaction Class 11 Important Questions and Answers

These redox reaction class 11 important questions come directly from NCERT exercises. Working through these builds the answer-writing speed needed for CBSE 2026 exams. For a complete set of important questions of redox reaction class 11, visit Important Questions Class 11 Chemistry Chapter 7.

Worked NCERT Questions

Q1. Assign oxidation numbers. (a) NaH2PO4: P = +5 (b) NaHSO4: S = +6 (c) K2MnO4: Mn = +6 (d) CaO2: O = -1 (peroxide) (e) NaBH4: B = -5

Q2. Justify: 2Na(s) + H2(g) → 2NaH(s) is a redox reaction. NaH is ionic: Na+H-. Two Na atoms are oxidised from 0 to +1. Two H atoms in H2 are reduced from 0 to -1. Both half-reactions occur simultaneously. This confirms it is a redox reaction.

Q3. Classify the following reactions. (a) N2 + O2 → 2NO: Combination (b) 2Pb(NO3)2 → 2PbO + 4NO2 + O2: Decomposition (c) NaH + H2O → NaOH + H2: Displacement (d) 2NO2 + 2OH- → NO2- + NO3- + H2O: Disproportionation (N at +4 goes to +3 and +5)

Q4. Which oxoanion of chlorine does not disproportionate? ClO4- does not disproportionate because chlorine is already at its highest oxidation state (+7). ClO-, ClO2-, and ClO3- all undergo disproportionation.

Q5. Predict if Fe3+(aq) and I-(aq) can react. E° for Fe3+/Fe2+ = +0.77 V; E° for I2/I- = +0.54 V. Fe3+ has higher E° and is a stronger oxidant than I2. Fe3+ can oxidise I- to I2. The reaction is feasible.