Chapter Introduction
Redox (reduction-oxidation) reactions are among the most fundamental reaction types in chemistry. Every combustion reaction, every battery, every corrosion process, and most biological energy-transfer reactions are redox reactions. In CBSE Class 11 Chemistry, Chapter 8 teaches you to identify oxidation and reduction, assign oxidation numbers, balance redox equations, and understand electrode processes.
This chapter also lays the foundation for electrochemistry in Class 12. Understanding redox thoroughly now will save you significant effort later.
Key Definitions
| Term | Definition |
|---|---|
| Oxidation (classical) | Addition of oxygen or removal of hydrogen |
| Reduction (classical) | Addition of hydrogen or removal of oxygen |
| Oxidation (electronic) | Loss of one or more electrons; increase in oxidation number |
| Reduction (electronic) | Gain of one or more electrons; decrease in oxidation number |
| Oxidizing agent | Substance that oxidizes another (itself gets reduced; accepts electrons) |
| Reducing agent | Substance that reduces another (itself gets oxidized; donates electrons) |
| Oxidation number | The charge an atom would have if all bonds were ionic |
| Redox reaction | A reaction in which oxidation and reduction occur simultaneously |
| Disproportionation | A reaction where the same element is both oxidized and reduced |
| Half-reaction | Either the oxidation part or the reduction part of a redox reaction |
Classical vs Electronic Concepts
| Concept | Oxidation | Reduction |
|---|---|---|
| Classical | Addition of O₂ / Removal of H₂ | Addition of H₂ / Removal of O₂ |
| Electronic | Loss of electrons (oxidation number increases) | Gain of electrons (oxidation number decreases) |
| Example | Zn → Zn²⁺ + 2e⁻ (Zn oxidized) | Cu²⁺ + 2e⁻ → Cu (Cu²⁺ reduced) |
Key insight: Oxidation and reduction always occur together. The substance oxidized is the reducing agent; the substance reduced is the oxidizing agent.
Rules for Assigning Oxidation Numbers
| Rule | Description | Example |
|---|---|---|
| 1 | Oxidation number of an element in its free state is 0 | Na = 0, O₂ = 0, Cl₂ = 0 |
| 2 | Oxidation number of a monoatomic ion equals its charge | Na⁺ = +1, Cl⁻ = −1, Al³⁺ = +3 |
| 3 | Hydrogen is +1 in compounds (−1 in metal hydrides) | H in HCl = +1; H in NaH = −1 |
| 4 | Oxygen is −2 in compounds (−1 in peroxides, +2 in OF₂) | O in H₂O = −2; O in H₂O₂ = −1 |
| 5 | Halogens are usually −1 (except when bonded to more electronegative element) | Cl in HCl = −1; Cl in HClO₃ = +5 |
| 6 | Sum of oxidation numbers of all atoms = overall charge of the species | In SO₄²⁻: S + 4(−2) = −2, so S = +6 |
| 7 | Oxidation number can be fractional | In C₃O₂: 3C + 2(−2) = 0, average C = +4/3 |
| 8 | In compounds, more electronegative atom gets the negative oxidation number | In NF₃: F = −1 (more electronegative), N = +3 |
Practice: Oxidation Numbers
| Species | Oxidation Numbers |
|---|---|
| KMnO₄ | K = +1, Mn = +7, O = −2 |
| K₂Cr₂O₇ | K = +1, Cr = +6, O = −2 |
| H₂SO₄ | H = +1, S = +6, O = −2 |
| HNO₃ | H = +1, N = +5, O = −2 |
| Fe₂O₃ | Fe = +3, O = −2 |
| Na₂O₂ | Na = +1, O = −1 (peroxide) |
| CaH₂ | Ca = +2, H = −1 (hydride) |
| OF₂ | O = +2, F = −1 |
Types of Redox Reactions
| Type | Description | Example |
|---|---|---|
| Combination | Two elements combine; one is oxidized, the other reduced | 2Na + Cl₂ → 2NaCl |
| Decomposition | A compound breaks into elements; reverse of combination | 2H₂O → 2H₂ + O₂ |
| Displacement | More reactive element displaces less reactive from compound | Zn + CuSO₄ → ZnSO₄ + Cu |
| Disproportionation | Same element is both oxidized and reduced | 2Cu⁺ → Cu + Cu²⁺ |
| Competitive electron transfer | Two species compete for electrons from a third | Cl₂ + 2KBr → 2KCl + Br₂ |
Common Oxidizing and Reducing Agents
Oxidizing Agents (get reduced)
| Agent | Change in Oxidation Number |
|---|---|
| KMnO₄ (in acid) | Mn: +7 → +2 |
| K₂Cr₂O₇ (in acid) | Cr: +6 → +3 |
| HNO₃ (conc.) | N: +5 → +4 or +2 |
| H₂SO₄ (conc.) | S: +6 → +4 |
| Cl₂, Br₂, F₂, O₃ | Halogens: 0 → −1; O₃: 0 → −2 |
| H₂O₂ | O: −1 → −2 (acts as oxidizing agent) |
Reducing Agents (get oxidized)
| Agent | Change in Oxidation Number |
|---|---|
| Metals (Na, Zn, Fe, Al) | Metal: 0 → positive oxidation number |
| H₂ | H: 0 → +1 |
| C, CO | C: 0 → +4; C: +2 → +4 |
| SO₂ | S: +4 → +6 |
| H₂O₂ | O: −1 → 0 (acts as reducing agent) |
Note: H₂O₂ can act as both an oxidizing and reducing agent depending on the reaction.
Balancing Redox Reactions
Method 1: Oxidation Number Method
Steps: 1. Write the skeleton equation and identify atoms whose oxidation numbers change. 2. Calculate the total increase and total decrease in oxidation numbers. 3. Find the LCM of the increase and decrease, and assign coefficients so that total increase = total decrease. 4. Balance all other atoms (except O and H) by inspection. 5. Balance oxygen atoms by adding H₂O to the side deficient in O. 6. Balance hydrogen atoms by adding H⁺ (acidic medium) or H₂O/OH⁻ (basic medium).
Method 2: Half-Reaction (Ion-Electron) Method
In Acidic Medium:
1. Write the skeleton equation in ionic form. 2. Split into two half-reactions (oxidation and reduction). 3. Balance each half-reaction: a. Balance all atoms except O and H. b. Balance O by adding H₂O. c. Balance H by adding H⁺. d. Balance charge by adding electrons. 4. Multiply half-reactions by appropriate integers so electrons cancel. 5. Add the two half-reactions and simplify.
Example (Acidic Medium): Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
Reduction half-reaction:
- MnO₄⁻ → Mn²⁺
- Balance O: MnO₄⁻ → Mn²⁺ + 4H₂O
- Balance H: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
- Balance charge: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Oxidation half-reaction:
- Fe²⁺ → Fe³⁺
- Balance charge: Fe²⁺ → Fe³⁺ + e⁻
Multiply and add:
- Multiply oxidation by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻
- Add: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
In Basic Medium:
Same steps as acidic, but after balancing with H⁺, add OH⁻ to both sides to neutralize H⁺ (converting H⁺ + OH⁻ → H₂O). Simplify by cancelling H₂O.
Example (Basic Medium): Balance MnO₄⁻ + I⁻ → MnO₂ + I₂
Reduction half-reaction:
- MnO₄⁻ → MnO₂
- Balance O (acidic first): MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O
- Balance charge: MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O
- Convert to basic: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
Oxidation half-reaction:
- 2I⁻ → I₂ + 2e⁻
Multiply and add:
- Multiply reduction by 2, oxidation by 3: 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻
- 6I⁻ → 3I₂ + 6e⁻
- Add: 2MnO₄⁻ + 4H₂O + 6I⁻ → 2MnO₂ + 8OH⁻ + 3I₂
Disproportionation Reactions
A disproportionation reaction occurs when an element in one oxidation state splits into two different oxidation states — one higher (oxidized) and one lower (reduced).
| Reaction | Element | Oxidized to | Reduced to |
|---|---|---|---|
| 2Cu⁺ → Cu²⁺ + Cu | Cu (+1) | Cu²⁺ (+2) | Cu (0) |
| 2H₂O₂ → 2H₂O + O₂ | O (−1) | O₂ (0) | H₂O (−2) |
| 3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O | Cl (0) | ClO₃⁻ (+5) | Cl⁻ (−1) |
| 3MnO₄²⁻ + 2H₂O → MnO₂ + 2MnO₄⁻ + 4OH⁻ | Mn (+6) | MnO₄⁻ (+7) | MnO₂ (+4) |
How to identify: If the same element appears on both sides with different oxidation numbers AND the element on the reactant side has a single intermediate oxidation state, check if it disproportionates.
Redox Reactions and Electrode Processes
Electrochemical Cells
- Galvanic/Voltaic cell: Converts chemical energy to electrical energy (spontaneous redox reaction). Example: Daniel cell (Zn | Zn²⁺ || Cu²⁺ | Cu).
- Electrolytic cell: Converts electrical energy to chemical energy (non-spontaneous, driven by external power).
Key Terms
| Term | Definition |
|---|---|
| Anode | Electrode where oxidation occurs (loss of electrons) |
| Cathode | Electrode where reduction occurs (gain of electrons) |
| Oxidation half-cell | Where the reducing agent is oxidized |
| Reduction half-cell | Where the oxidizing agent is reduced |
| Salt bridge | Maintains electrical neutrality by allowing ion flow between half-cells |
Mnemonic: "AN OX, RED CAT" — Anode = Oxidation, Reduction = Cathode
Daniel Cell Reaction
- At anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻
- At cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)
- Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Applications of Redox Reactions
| Application | Description |
|---|---|
| Corrosion | Oxidation of metals by atmospheric oxygen and moisture (e.g., rusting of iron: 4Fe + 3O₂ + 6H₂O → 4Fe(OH)₃). Prevention: painting, galvanization, alloying, sacrificial anode |
| Rancidity | Oxidation of fats and oils in food producing unpleasant smell/taste. Prevention: antioxidants (BHA, BHT), nitrogen flushing, refrigeration |
| Batteries | Convert chemical energy to electrical energy via redox reactions (dry cell, lead-acid battery, Ni-Cd battery) |
| Photosynthesis | 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ (C is reduced, O in H₂O is oxidized) |
| Respiration | C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O (reverse of photosynthesis, C is oxidized) |
| Combustion | Fuel + O₂ → CO₂ + H₂O (exothermic redox) |
| Water treatment | Chlorination: Cl₂ + H₂O → HCl + HOCl (oxidizes bacteria) |
Common Mistakes
1. Confusing oxidizing and reducing agents: The oxidizing agent is REDUCED (gains electrons). The reducing agent is OXIDIZED (loses electrons). Many students reverse these.
2. Wrong oxidation number assignments: Remember that H is −1 only in metal hydrides (NaH, CaH₂), and O is −1 only in peroxides (H₂O₂, Na₂O₂). In all other compounds, H = +1 and O = −2.
3. Forgetting to balance charge: After balancing atoms, always verify that total charge is equal on both sides. This catches many errors.
4. Disproportionation identification: Check whether the same element appears with different oxidation numbers on both sides. If an element goes from a single intermediate state to both a higher and lower state, it is disproportionation.
5. Half-reaction method in basic medium: Students forget to convert H⁺ to H₂O by adding OH⁻ to both sides. This step is essential for basic medium problems.
6. Counting electrons incorrectly: The total electrons lost in oxidation must equal the total electrons gained in reduction. Use LCM to find the multiplication factors.
7. Oxidation number vs valency: Valency has no sign and is always a whole number. Oxidation number can be negative, zero, or fractional (e.g., Fe in Fe₃O₄ has an average oxidation number of +8/3).
Exam Tips
1. Oxidation number calculation (1–2 marks): Practice finding oxidation numbers of the underlined element in compounds like KMnO₄ (Mn = +7), K₂Cr₂O₇ (Cr = +6), H₂S₂O₈ (S = +6, O in peroxide linkage = −1).
2. Identifying oxidizing/reducing agent (1 mark): A very common question. The substance that is reduced is the oxidizing agent; the substance that is oxidized is the reducing agent.
3. Balancing by half-reaction method (3–5 marks): This is almost guaranteed. Practice both acidic and basic medium. Follow the steps systematically — never try to balance by inspection alone.
4. Disproportionation identification (2 marks): Given a reaction, identify whether it is a disproportionation reaction and justify using oxidation numbers.
5. Applications (1–2 marks): Corrosion prevention methods and rancidity prevention are commonly asked. Write specific methods, not vague statements.
| 6. Cell representation (2 marks): Know how to write the Daniel cell: Zn(s) | Zn²⁺(aq) | Cu²⁺(aq) | Cu(s). The anode is on the left, cathode on the right. |
|---|
Quick Revision Notes
- Oxidation = loss of electrons (oxidation number increases).
- Reduction = gain of electrons (oxidation number decreases).
- Oxidizing agent gets reduced; reducing agent gets oxidized.
- Oxidation number rules: Free element = 0; H = +1 (−1 in hydrides); O = −2 (−1 in peroxides); sum = charge.
- Half-reaction method: Split → Balance atoms → Balance O with H₂O → Balance H with H⁺/OH⁻ → Balance charge with e⁻ → Combine.
- Acidic medium: Add H⁺ to balance H. Basic medium: Add H₂O and OH⁻ after balancing in acidic form.
- Disproportionation: Same element oxidized AND reduced. 2Cu⁺ → Cu + Cu²⁺; 2H₂O₂ → 2H₂O + O₂.
- AN OX, RED CAT: Anode = Oxidation, Cathode = Reduction.
- Corrosion prevention: Painting, galvanization, alloying, sacrificial anode.
- Rancidity prevention: Antioxidants, nitrogen flushing, refrigeration.
- H₂O₂: Acts as both oxidizing and reducing agent.
- Fractional oxidation numbers: Average over all atoms of that element (e.g., C in C₃O₂ = +4/3).
NCERT Exercise Focus Areas
The following NCERT exercise question types are most likely to appear in exams. Prioritize these during revision:
| Question Type | Chapter Examples | Marks |
|---|---|---|
| Assign oxidation numbers | Q1, Q2 | 1–2 |
| Identify oxidizing/reducing agent | Q3, Q4 | 1–2 |
| Balance redox reactions (oxidation number method) | Q5, Q6 | 3 |
| Balance redox reactions (half-reaction method, acidic) | Q7, Q8 | 3–5 |
| Balance redox reactions (half-reaction method, basic) | Q9, Q10 | 3–5 |
| Identify disproportionation reactions | Q14 | 2 |
| Applications — corrosion and rancidity | Q16, Q17 | 2 |
Practice: Step-by-Step Balancing Worked Example
Balance: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ (acidic medium)
Step 1 — Split into half-reactions:
- Reduction: Cr₂O₇²⁻ → Cr³⁺
- Oxidation: Fe²⁺ → Fe³⁺
Step 2 — Balance atoms (except O and H):
- Reduction: Cr₂O₇²⁻ → 2Cr³⁺
- Oxidation: Fe²⁺ → Fe³⁺ (already balanced)
Step 3 — Balance O by adding H₂O:
- Reduction: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O
Step 4 — Balance H by adding H⁺:
- Reduction: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O
Step 5 — Balance charge by adding e⁻:
- Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (charge: −2+14−6 = +6 on left; +6 on right)
- Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (charge: +2 on left; +3−1 = +2 on right)
Step 6 — Multiply to equalize electrons:
- Multiply oxidation by 6: 6Fe²⁺ → 6Fe³⁺ + 6e⁻
Step 7 — Add half-reactions and simplify:
- Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺