How do lithium-ion smartphone batteries generate pure electrical power from chemical bonds, or why do iron bridges rust and crumble without protective paint? Oxidation and Reduction reactions are the transfer of electronic lifeblood.
Why This Chapter Matters
In Class 11 Chemistry, "Redox Reactions" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
Before You Begin (Prerequisites)
Valency from Class 9.
Redox basics from Class 10.
Displacement reactions.
What You Will Learn (Core Objectives)
Define Oxidation, Reduction, Oxidizing Agent, and Reducing Agent in terms of Electron Transfer.
Assign Oxidation Numbers to elements in compounds using formal valence rules.
Balance complex Redox Reactions using the Oxidation Number Method and Ion-Electron (Half-Reaction) Method.
Identify Disproportionation Reactions.
Explain the working of an Electrochemical Cell and Electrode Potential.
Chapter Roadmap & Progression
11. Oxidation Numbers & Electron Tra...
22. Half-Reaction Balancing Method
33. Disproportionation & Electrochem...
Complete Concept Guide (100% Curriculum Coverage)
1. Oxidation Numbers & Electron Transfer
Oxidation: Loss of electrons → Increase in Oxidation Number (LEO: Lose Electrons Oxidation).
Reduction: Gain of electrons → Decrease in Oxidation Number (GER: Gain Electrons Reduction).
Rules: Free element $= 0$; $F = -1$; $O = -2$ (except $-1$ in peroxides); $H = +1$ (except $-1$ in hydrides).
2. Half-Reaction Balancing Method
Steps in acidic medium: (1) Split into oxidation and reduction half-reactions, (2) Balance atoms other than $O$ and $H$, (3) Balance $O$ by adding $\text{H}_2\text{O}$, (4) Balance $H$ by adding $\text{H}^+$, (5) Balance charges by adding electrons ($e^-$), (6) Multiply half-reactions to equalize electrons and add!
3. Disproportionation & Electrochemical Cells
A Disproportionation Reaction is a redox process where the same element is simultaneously oxidized and reduced (e.g. $2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2$, where oxygen goes from $-1$ to $-2$ and $0$!). Electrochemical cells convert redox chemical energy into electrical voltage.
Redox Reactions - Key Conceptual Architecture & Molecular Model
Chapter Summary & 10 Key Takeaways
Takeaway 1
Oxidation State: Formal bookkeeping charge assigned assuming all bonds are fully ionic.
Takeaway 2
LEO says GER: Loss of Electrons is Oxidation; Gain of Electrons is Reduction.
Takeaway 3
Ion-Electron Method: Systematic half-reaction technique balancing mass and ionic charge.
Takeaway 4
Disproportionation: Simultaneous self-oxidation and self-reduction of a single chemical species.
Takeaway 5
Electrode Potential: Voltage differential between metal electrode and its surrounding ions.
Check Your Understanding (Diagnostic Practice Questions)
Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.
1
Calculate the oxidation number of: (i) Cr in $\text{K}_2\text{Cr}_2\text{O}_7$, (ii) Mn in $\text{KMnO}_4$.
Reveal Answer & Explanation
Answer: (i) In $\text{K}_2\text{Cr}_2\text{O}_7$: $2(+1) + 2x + 7(-2) = 0 \implies 2 + 2x - 14 = 0 \implies 2x = 12 \implies x = +6$. (ii) In $\text{KMnO}_4$: $1(+1) + x + 4(-2) = 0 \implies 1 + x - 8 = 0 \implies x = +7$. Cr is +6, Mn is +7.
2
What is a Disproportionation Reaction? Give a balanced chemical example.
Reveal Answer & Explanation
Answer: A redox reaction in which the same element in a given reacting substance undergoes both oxidation (increase in oxidation state) and reduction (decrease in oxidation state) simultaneously. Example: $2\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2$ (Oxygen goes from -1 in $\text{H}_2\text{O}_2$ to -2 in $\text{H}_2\text{O}$ and 0 in $\text{O}_2$). Same element simultaneously oxidized and reduced.
Why does fluorine never exhibit positive oxidation states in any of its compounds?
Reveal Answer & Explanation
Answer: Because fluorine is the most electronegative element in the entire periodic table ($4.0$ on Pauling scale) and has no vacant $d$-orbitals in its valence shell, strictly exhibiting an oxidation state of $-1$. Most electronegative element with no vacant d-orbitals.
5
Balance the ionic equation in acidic medium: $\text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} \to \text{Fe}^{3+} + \text{Cr}^{3+}$.
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