Class XI · Chapter 7 · NCERT Chemistry

Chapter 07

Redox Reactions

The Dance of Electrons

Every oxidation demands a reduction. In chemistry, electrons are the currency of transformation.

\(\text{OIL RIG: Oxidation Is Loss, Reduction Is Gain}\)
8 CBSE Marks
Difficulty
9 Topics
Medium JEE / NEET Weight

🧬 Topics Covered

9 key topics in this chapter — each linked to NCERT exercises and exam questions.

Classical Concept of Oxidation & Reduction
Electronic Concept of Redox Reactions
Oxidation Number (State) Rules
Types of Redox Reactions
Balancing Redox Reactions: Oxidation Number Method
Balancing Redox Reactions: Half-Reaction Method
Redox Reactions & Electrode Processes
Disproportionation Reactions
Comproportionation Reactions

📚 Study Resources

Everything you need to master this chapter — from concept notes to previous year questions.

𝑓 Key Formulae

Essential expressions — understand derivations, not just results.

Oxidation Number of O
\[\text{O} = -2 \text{ (usually)},\; -1 \text{ in peroxides},\; 0 \text{ in O}_2\]
📌 Exception: OF₂ gives O = +2
Oxidation Number of H
\[\text{H} = +1 \text{ (non-metals)},\; -1 \text{ (metal hydrides)}\]
📌 H in NaH, CaH₂ = −1
Sum Rule (neutral)
\[\sum (\text{ON} \times \text{subscript}) = 0\]
📌 Algebraic sum of all oxidation numbers = 0
Sum Rule (ion)
\[\sum (\text{ON} \times \text{subscript}) = \text{charge of ion}\]
📌 e.g. Cr₂O₇²⁻: sum = −2
Electron Transfer
\[\text{Oxidation: } M \to M^{n+} + ne^-\]
📌 Reduction: M^n+ + ne⁻ → M
Change in ON
\[\Delta\text{ON}_{\text{ox}} = \Delta\text{ON}_{\text{red}} \text{ (electrons balanced)}\]
📌 Increase in ON = oxidation

🎯 Exam-Ready Insights

Important points to remember — curated from CBSE Board and entrance exam question patterns.

01

CBSE always asks "calculate the oxidation state of X in compound Y" — apply the sum rule systematically.

02

Know all exceptions to oxidation number rules: O in Na₂O₂ = −1; O in OF₂ = +2; H in NaH = −1.

03

Balancing by half-reaction method: separate into oxidation and reduction half-reactions, balance atoms and charge separately, then combine.

04

Disproportionation = same element is both oxidised and reduced. Common example: Cl₂ → Cl⁻ + ClO⁻ in basic medium.

05

For CBSE 5-mark balancing questions: always state the change in oxidation number before balancing electron transfer.

06

Electrode processes: oxidation occurs at anode (−ve in electrolytic cell); reduction at cathode.

🏆 Competitive Exam Strategy

Targeted tips for JEE Main, JEE Advanced, NEET, and BITSAT.

JEE Main

JEE tests complex oxidation state assignments in polyoxoanions like Cr₂O₇²⁻, MnO₄⁻, S₂O₃²⁻ — work methodically using the sum rule.

JEE Main

Disproportionation vs comproportionation identification is a quick JEE MCQ — one element, two different oxidation states in products (or reactants).

NEET

NEET tests biological redox: NAD⁺/NADH, FAD/FADH₂ in respiration. Know that NADH is the reduced form (gains electrons).

BITSAT

BITSAT gives rapid oxidation state questions — memorise Mn (+7 in KMnO₄, +4 in MnO₂, +2 in MnSO₄) and Cr (+6 in K₂Cr₂O₇, +3 in Cr₂O₃).

⚠️ Common Mistakes to Avoid

Students consistently lose marks on these — know them before your exam.

Confusing the oxidising agent (gets reduced) with the reducing agent (gets oxidised).

Forgetting the oxidation number of H = −1 in metal hydrides like NaH, KH, CaH₂.

Ignoring the sign of the ion when applying the sum rule (e.g., MnO₄⁻: sum = −1, not 0).

Not multiplying half-reactions by the LCM to equalise electrons before adding them.

💡 Key Takeaways

The non-negotiable concepts every student must carry out of this chapter.

Oxidation = loss of electrons = increase in oxidation number.

Reduction = gain of electrons = decrease in oxidation number.

The oxidising agent is reduced; the reducing agent is oxidised.

Oxidation number of a free element is always zero.

The algebraic sum of oxidation numbers in a neutral compound is zero; in an ion it equals the charge.

Both the oxidation number method and the half-reaction (ion-electron) method must be mastered.

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