By the end of this chapter you'll be able to…

  • 1Define oxidation and reduction in classical and electronic terms (OIL RIG)
  • 2Assign oxidation numbers using the standard rules
  • 3Identify oxidising agents, reducing agents, and disproportionation reactions
  • 4Balance redox equations by the oxidation-number and half-reaction methods
  • 5Use the electrochemical series to predict spontaneity and cell EMF
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Why this chapter matters
Redox reactions -- electron transfer -- power everything from rusting and respiration to batteries and bleaching. Mastering oxidation numbers, balancing redox equations, and the electrochemical series prepares you for electrochemistry and explains biological redox processes like respiration and photosynthesis.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Redox Reactions

'Oxidation is gain of oxygen, loss of hydrogen, loss of electrons. Reduction is the opposite. But redox is about electron transfer.' — Chemistry Concepts

1. Chapter Overview

REDOX reactions are reactions involving ELECTRON TRANSFER. They are among the MOST COMMON and important chemical reactions — from rusting to respiration, from batteries to bleaching. This chapter covers OXIDATION and REDUCTION in classical and electronic terms, OXIDATION NUMBER (a powerful bookkeeping tool), BALANCING redox reactions, and the ELECTROCHEMICAL SERIES.


2. Classical Concept of Oxidation and Reduction

Oxidation (Classical)

  • Addition of OXYGEN
  • Removal of HYDROGEN
  • Gain of ELECTRONEGATIVE element

Reduction (Classical)

  • Addition of HYDROGEN
  • Removal of OXYGEN
  • Gain of ELECTROPOSITIVE element

Mnemonic

  • OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons)

3. Oxidation Number (Oxidation State)

  • Definition: The CHARGE an atom would have if electrons were COMPLETELY transferred (a bookkeeping concept)
  • Rules for Assigning ON:
    1. ON of an element in FREE state = 0 (H₂, O₂, Fe, S₈)
    2. ON of a MONOATOMIC ion = its charge (Na⁺ = +1, Cl⁻ = -1)
    3. ON of HYDROGEN = +1 (except metal hydrides where it's -1: NaH, CaH₂)
    4. ON of OXYGEN = -2 (except peroxides -1: H₂O₂; superoxides -½: KO₂; OF₂ = +2)
    5. Sum of ON in a NEUTRAL compound = 0
    6. Sum of ON in a POLYATOMIC ion = charge on the ion
    7. Halogens: -1 in most compounds (except with O or higher halogens)

Worked Problem

Q: Find ON of Cr in Cr₂O₇²⁻ and S in H₂SO₄. A: Cr₂O₇²⁻: 2x + 7(-2) = -2 → 2x = +12 → x = +6. H₂SO₄: 2(+1) + x + 4(-2) = 0 → 2 + x — 8 = 0 → x = +6.


4. Types of Redox Reactions

TypeDescriptionExample
CombinationTwo substances combine2Mg + O₂ → 2MgO
DecompositionOne substance breaks down2H₂O → 2H₂ + O₂
DisplacementOne element replaces anotherZn + CuSO₄ → ZnSO₄ + Cu
DisproportionationSame element oxidised AND reduced2H₂O₂ → 2H₂O + O₂
CombustionRapid oxidation producing heat/lightCH₄ + 2O₂ → CO₂ + 2H₂O

Disproportionation Reactions

  • The ELEMENT must have at least THREE oxidation states
  • One atom of the element is OXIDISED, another REDUCED
  • Example: 2H₂O₂ → 2H₂O + O₂ (O in H₂O₂: -1 → goes to -2 and 0)

5. Balancing Redox Equations

Method 1: Oxidation Number Method

  1. Assign ON to ALL atoms
  2. Identify which atoms change ON
  3. Calculate CHANGE in ON (increase = oxidation; decrease = reduction)
  4. Make total increase = total decrease (balance electrons)
  5. Balance O with H₂O and H with H⁺ (in acidic medium)
  6. For basic medium: after balancing in acidic, add OH⁻ to NEUTRALISE H⁺

Method 2: Half-Reaction Method (Ion-Electron Method)

  1. Write OXIDATION and REDUCTION half-reactions separately
  2. Balance ATOMS other than H and O
  3. Balance O with H₂O, H with H⁺ (acidic) or OH⁻ (basic)
  4. Balance CHARGE by adding e⁻
  5. MULTIPLY half-reactions to equalise electrons lost = gained
  6. ADD half-reactions

Worked Problem (Acidic Medium)

Q: Balance Fe²⁺ + Cr₂O₇²⁻ → Fe³⁺ + Cr³⁺ in acidic medium. A:

  • Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (×6)
  • Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
  • Adding: 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O

6. Electrochemical Series

Half-ReactionE° (V)
Li⁺ + e⁻ → Li-3.04
K⁺ + e⁻ → K-2.93
Ca²⁺ + 2e⁻ → Ca-2.87
Na⁺ + e⁻ → Na-2.71
Mg²⁺ + 2e⁻ → Mg-2.36
Al³⁺ + 3e⁻ → Al-1.66
Zn²⁺ + 2e⁻ → Zn-0.76
Fe²⁺ + 2e⁻ → Fe-0.44
Ni²⁺ + 2e⁻ → Ni-0.25
2H⁺ + 2e⁻ → H₂0.00 (Reference)
Cu²⁺ + 2e⁻ → Cu+0.34
Ag⁺ + e⁻ → Ag+0.80
Au³⁺ + 3e⁻ → Au+1.40
F₂ + 2e⁻ → 2F⁻+2.87

Uses of Electrochemical Series

  • Predicting spontaneity: More NEGATIVE E° → better reducing agent (tendency to get oxidised)
  • Predicting displacement reactions: A metal with MORE negative E° reduces the ion of a metal with LESS negative E°
  • Calculating cell EMF: E°_cell = E°_cathode — E°_anode

Worked Problem

Q: Zn + Cu²⁺ → Zn²⁺ + Cu. Calculate E°_cell. (E°_Zn = -0.76 V, E°_Cu = +0.34 V) A: Zn is oxidised (anode), Cu²⁺ reduced (cathode). E°_cell = E°_Cu — E°_Zn = 0.34 — (-0.76) = 1.10 V. Positive → SPONTANEOUS.


7. Common Mistakes

  1. ON is NOT the actual charge (except for monatomic ions): It is a BOOKKEEPING convention
  2. H is +1 EXCEPT in metal hydrides: Many students forget NaH has H = -1
  3. O is -2 EXCEPT in peroxides and superoxides: H₂O₂ has O = -1, not -2
  4. Disproportionation requires the element to have an INTERMEDIATE oxidation state: Not all reactions of an element are disproportionation
  5. E°_cell positive means spontaneous, but real rate may be slow: Thermodynamic favourability ≠ fast kinetics

8. CBSE Exam Focus

  1. Oxidation number calculations (3-mark)
  2. Identifying oxidising and reducing agents (1-mark)
  3. Balancing redox equations — both methods (5-mark)
  4. Electrochemical series — predicting reactions (3-mark)
  5. Disproportionation reactions — examples (3-mark)

9. Key Formulas

  • E°_cell = E°_cathode — E°_anode (standard cell potential)
  • ΔG° = -nFE°_cell (relation between free energy and cell potential)
  • F = 96500 C/mol (Faraday constant)

10. Self-Test (5+ Q&A)

Q1: Assign ON to Mn in KMnO₄ and Cr in Cr₂O₃. A: KMnO₄: K = +1, each O = -2 → +1 + x + 4(-2) = 0 → x = +7. Cr₂O₃: 2x + 3(-2) = 0 → 2x = +6 → x = +3.

Q2: Balance in basic medium: MnO₄⁻ + I⁻ → MnO₂ + I₂. A: After balancing in acidic then converting to basic: 2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻

Q3: Which is a stronger reducing agent — Zn or Cu? Why? A: Zn (E° = -0.76 V) is stronger than Cu (E° = +0.34 V). MORE negative E° means greater tendency to LOSE electrons (be oxidised).

Q4: What is a disproportionation reaction? Give an example. A: Reaction where the SAME element is BOTH oxidised and reduced. Example: 2H₂O₂ → 2H₂O + O₂ (O goes from -1 to -2 and 0).

Q5: Calculate E°_cell for Fe²⁺ + Zn → Fe + Zn²⁺. (E°_Fe = -0.44 V, E°_Zn = -0.76 V) A: Zn is oxidised (anode), Fe²⁺ is reduced (cathode). E°_cell = E°_Fe — E°_Zn = -0.44 — (-0.76) = +0.32 V → SPONTANEOUS.


11. Conclusion

Redox reactions are FUNDAMENTAL to chemistry and life. The oxidation number system provides a POWERFUL bookkeeping tool for tracking electron transfer. Balancing redox equations is an ESSENTIAL skill. The electrochemical series is a ROADMAP of reduction potentials — predicting spontaneity, displacement reactions, and cell voltages. These concepts prepare you for the in-depth study of electrochemistry in Class 12 and are VITAL for understanding biological redox processes (respiration, photosynthesis).

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Standard cell potential
E_cell = E_cathode - E_anode
A positive E_cell means the cell reaction is spontaneous.
Free energy and cell potential
delta-G = -nF E_cell
F = 96500 C/mol; n = moles of electrons transferred.
Oxidation number rules
Free element = 0; sum in neutral compound = 0; sum in ion = charge
H = +1 (hydrides -1); O = -2 (peroxides -1, OF2 +2).
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Treating oxidation number as the real charge
Oxidation number is a bookkeeping convention; only monatomic ions carry the actual charge.
WATCH OUT
Always assigning H = +1 and O = -2
In metal hydrides H = -1 (NaH); in peroxides O = -1 (H2O2) and in OF2 O = +2.
WATCH OUT
Calling any single-element reaction disproportionation
Disproportionation needs an element in an intermediate oxidation state that is both oxidised and reduced.
WATCH OUT
Assuming a positive E_cell means a fast reaction
A positive E_cell means thermodynamically spontaneous, not necessarily fast -- kinetics may be slow.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1MEDIUM· Oxidation Number
Assign the oxidation number of Mn in KMnO4 and Cr in Cr2O3.
Show solution
KMnO4: +1 + x + 4(-2) = 0, so x = +7. Cr2O3: 2x + 3(-2) = 0, so x = +3.
Q2HARD· Balancing
Balance Fe2+ + Cr2O7 2- -> Fe3+ + Cr3+ in acidic medium.
Show solution
Oxidation: Fe2+ -> Fe3+ + e- (x6). Reduction: Cr2O7 2- + 14H+ + 6e- -> 2Cr3+ + 7H2O. Adding: 6Fe2+ + Cr2O7 2- + 14H+ -> 6Fe3+ + 2Cr3+ + 7H2O.
Q3HARD· Balancing
Balance MnO4- + I- -> MnO2 + I2 in basic medium.
Show solution
2MnO4- + 6I- + 4H2O -> 2MnO2 + 3I2 + 8OH-.
Q4MEDIUM· Electrochemical Series
Calculate E_cell for Zn + Cu2+ -> Zn2+ + Cu (E_Zn = -0.76 V, E_Cu = +0.34 V).
Show solution
Zn is oxidised (anode), Cu2+ reduced (cathode). E_cell = E_cathode - E_anode = 0.34 - (-0.76) = +1.10 V, so the reaction is spontaneous.
Q5EASY· Concept
What is a disproportionation reaction? Give an example.
Show solution
A reaction in which the same element is simultaneously oxidised and reduced. Example: 2H2O2 -> 2H2O + O2, where oxygen (-1 in H2O2) goes to -2 in water and 0 in O2.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • OIL RIG: Oxidation Is Loss, Reduction Is Gain of electrons.
  • Oxidation number rules: free element 0, H = +1 (hydrides -1), O = -2 (peroxides -1).
  • Types: combination, decomposition, displacement, disproportionation, combustion.
  • Disproportionation needs an intermediate oxidation state.
  • Balance redox by oxidation-number or half-reaction (ion-electron) method.
  • Electrochemical series: more negative E means stronger reducing agent.
  • E_cell = E_cathode - E_anode; positive means spontaneous; delta-G = -nF E_cell.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 5-7 marks across the chapter

Question typeMarks eachTypical countWhat it tests
Balancing redox equations3-51Half-reaction and oxidation-number methods
Oxidation number / agents1-31-2Assigning oxidation states and identifying agents
Electrochemical series2-31Predicting reactions and cell potential
Prep strategy
  • Memorise the oxidation-number rules and exceptions
  • Practise both balancing methods in acidic and basic media
  • Learn the electrochemical series order
  • Use E_cell = E_cathode - E_anode to judge spontaneity

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Batteries and fuel cells

Redox reactions convert chemical energy into electrical energy in cells from torch batteries to electric vehicles.

Corrosion control

Understanding the electrochemical series guides methods like galvanising and cathodic protection to prevent rusting.

Metabolism

Respiration and photosynthesis are large-scale biological redox processes that power living organisms.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Assign oxidation numbers first to spot what is oxidised and reduced
  2. Balance charge with electrons and atoms with H2O/H+ or OH-
  3. Use the electrochemical series to identify anode and cathode
  4. State spontaneity from the sign of E_cell

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Use the Nernst equation to find cell potential under non-standard conditions.
  • Analyse redox titrations (permanganate, dichromate, iodometry) and their stoichiometry.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Chemistry examHigh
JEE Main and Advanced (Redox)High
NEET ChemistryHigh

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

The oxidising agent is the species that gets reduced -- its oxidation number decreases because it gains electrons. For example, in the reaction of zinc with copper sulphate, Cu2+ is reduced to Cu, so Cu2+ (copper sulphate) is the oxidising agent, while Zn is the reducing agent because it is oxidised.

Absolute electrode potentials cannot be measured directly, only differences. The standard hydrogen electrode (2H+ + 2e- -> H2) is chosen as the universal reference and arbitrarily assigned E = 0.00 V, so all other electrode potentials are measured relative to it.
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