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.
यह अध्याय क्यों महत्वपूर्ण है
In Class 11 Chemistry, "Redox Reactions" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus. Use the worked examples and examiner traps below to convert definitions into reliable board-exam problem-solving steps.
अध्ययन से पूर्व (आवश्यक ज्ञान)
Valency from Class 9.
Redox basics from Class 10.
Displacement reactions.
इस अध्याय के लक्ष्य
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.
अध्याय रूपरेखा एवं प्रगति
11. Oxidation Numbers & Electron Tra...
22. Half-Reaction Balancing Method
33. Disproportionation & Electrochem...
सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन
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.
महत्वपूर्ण रासायनिक सूत्र, अभिक्रियाएँ एवं समीकरण
Oxidation number
$\text{oxidation number} = \text{formal charge assigned by rules}$
The $n$-factor depends on the reaction, not only the compound.
Nernst form
$E = E^\circ - \frac{0.0591}{n}\log Q$
This form is valid at $298\text{ K}$.
अवधारणात्मक हल उदाहरण एवं अनुप्रयोग (Solved Examples)
उदाहरण 1
Balance $MnO_4^-+Fe^{2+}\to Mn^{2+}+Fe^{3+}$ in acidic medium.
विस्तृत समाधान / उत्तर:
Half-reactions give $MnO_4^-+8H^++5Fe^{2+}\to Mn^{2+}+4H_2O+5Fe^{3+}$.
उदाहरण 2
Find the oxidation number of chromium in $K_2Cr_2O_7$.
विस्तृत समाधान / उत्तर:
$2(+1)+2x+7(-2)=0$, so $x=+6$ for each chromium atom.
सामान्य गलतियाँ एवं परीक्षक के जाल (Examiner Traps)
सामान्य भ्रम / गलत उत्तर
Changing subscripts while balancing a redox equation.
सही वैज्ञानिक तथ्य
Balance only coefficients; changing a formula changes the chemical species.
सामान्य भ्रम / गलत उत्तर
Assuming oxidation always means adding oxygen.
सही वैज्ञानिक तथ्य
Oxidation is loss of electrons or increase in oxidation number; oxygen transfer is only one special case.
Redox Reactions - Key Conceptual Architecture & Molecular Model
अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष
मुख्य बिंदु 1
Oxidation State: Formal bookkeeping charge assigned assuming all bonds are fully ionic.
मुख्य बिंदु 2
LEO says GER: Loss of Electrons is Oxidation; Gain of Electrons is Reduction.
मुख्य बिंदु 3
Ion-Electron Method: Systematic half-reaction technique balancing mass and ionic charge.
मुख्य बिंदु 4
Disproportionation: Simultaneous self-oxidation and self-reduction of a single chemical species.
मुख्य बिंदु 5
Electrode Potential: Voltage differential between metal electrode and its surrounding ions.
स्व-मूल्यांकन अभ्यास (Check Your Understanding)
मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।
1
Calculate the oxidation number of: (i) Cr in $\text{K}_2\text{Cr}_2\text{O}_7$, (ii) Mn in $\text{KMnO}_4$.
उत्तर एवं व्याख्या देखें
उत्तर: (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.
उत्तर एवं व्याख्या देखें
उत्तर: 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?
उत्तर एवं व्याख्या देखें
उत्तर: 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: $\text{Fe}^{2+} + \text{Cr}_2\text{O}_7^{2-} \to \text{Fe}^{3+} + \text{Cr}^{3+}$.
