Redox Reactions Class 11 NCERT Chemistry Guide

Core Concepts and Oxidation Number Rules

A redox (reduction-oxidation) reaction involves a simultaneous transfer of electrons between chemical species. While classical chemistry defined oxidation as the addition of oxygen or removal of hydrogen, the modern electronic concept defines oxidation as the loss of electrons (de-electronation) and reduction as the gain of electrons (electronation). To simplify tracking, chemists use the concept of an Oxidation Number (or oxidation state), which represents the formal charge an atom would carry if all its bonds were completely ionic.

To calculate oxidation numbers accurately, you must apply the following NCERT guidelines:

1. In free or elemental states, every atom has an oxidation number of zero (e.g., O2, P4, Na, S8).
2. For monoatomic ions, the oxidation number equals the charge of the ion (e.g., Na+ is +1, S2- is -2).
3. Oxygen typically exhibits an oxidation state of -2. Important exceptions include peroxides (like H2O2, where oxygen is -1), superoxides (like KO2, where oxygen is -1/2), and oxygen difluoride (OF2, where oxygen is +2).
4. Hydrogen is +1 when bonded to non-metals, but is -1 in metallic hydrides (like LiH, CaH2).
5. Fluorine, being the most electronegative element, always has an oxidation state of -1 in all its compounds.
6. The algebraic sum of oxidation numbers of all atoms in a neutral molecule must equal zero, while in a polyatomic ion, it must equal the net charge of the ion.

Step-by-Step Guide: Balancing by Ion-Electron Method

1. Identify and Split — Write the net ionic equation and calculate oxidation numbers for all elements. Identify the species being oxidized and reduced, then split the skeletal equation into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
2. Balance Key Atoms — Balance all atoms in each half-reaction except for Oxygen (O) and Hydrogen (H).
3. Balance Oxygen and Hydrogen — In an acidic medium: Balance oxygen atoms by adding H2O molecules to the deficient side. Then, balance hydrogen atoms by adding H+ ions to the opposite side. (In basic medium, neutralise H+ with equal OH- on both sides to form water).
4. Balance Charge and Equate — Balance the charges by adding electrons (e-) to the side with the higher positive charge. Multiply both half-reactions by appropriate integers so that the number of electrons lost equals the number of electrons gained.
5. Combine and Verify — Add the two balanced half-reactions together and cancel terms appearing on both sides. Perform a final verification to ensure both atomic mass and electrical charges are balanced on both sides of the equation.

Step-by-Step Worked Examples

• Example 1: Determine the oxidation number of Chromium (Cr) in Potassium Dichromate (K2Cr2O7). Step 1: Let the oxidation number of Cr be x. The oxidation state of K is +1 and O is -2. Step 2: Set up the algebraic equation: 2(+1) + 2(x) + 7(-2) = 0. Step 3: Solve for x: 2 + 2x - 14 = 0 => 2x = 12 => x = +6. Final Answer: The oxidation state of Cr in K2Cr2O7 is +6.
• Example 2: Calculate the average oxidation state of Iron (Fe) in Magnetite (Fe3O4). Step 1: Let the average oxidation number of Fe be x. Oxygen is -2. Step 2: Set up the algebraic equation: 3(x) + 4(-2) = 0. Step 3: Solve for x: 3x - 8 = 0 => 3x = 8 => x = +8/3. Note: This is a fractional oxidation state because Fe3O4 is a mixed oxide consisting of FeO (where Fe is +2) and Fe2O3 (where Fe is +3).

CBSE Exam Pitfalls & High-Yield Tips

• Fractional Oxidation States: Elements like Sulfur in Na2S4O6 or Bromine in Br3O8 display fractional oxidation numbers when solved algebraically. Remember that this represents an average value. In reality, individual atoms within the molecular structure have different integer oxidation states.
• Disproportionation Reactions: A reaction where a single reactant species is simultaneously oxidized and reduced is called a disproportionation reaction (e.g., decomposition of H2O2). Keep an eye out for these in exams!
• Watch the Medium: Students often forget to convert H+ ions to OH- when balancing in basic media. Always neutralize H+ with equivalent OH- on both sides of the reaction.

Practice Questions with Solutions

• Q: Calculate the oxidation number of Phosphorus in NaH2PO4. A: Step 1: Let the oxidation number of P be x. The oxidation states of Na, H, and O are +1, +1, and -2, respectively. Step 2: Set up the equation: (+1) + 2(+1) + x + 4(-2) = 0. Step 3: Simplify: 1 + 2 + x - 8 = 0 => x - 5 = 0 => x = +5. Final answer: The oxidation number of Phosphorus in NaH2PO4 is +5.
• Q: Balance the following ionic equation in acidic medium: Fe2+ + Cr2O7^2- -> Fe3+ + Cr3+ A: Step 1: Split into half-reactions: Oxidation: Fe2+ -> Fe3+ ; Reduction: Cr2O7^2- -> Cr3+. Step 2: Balance elements other than O & H: Fe2+ -> Fe3+ ; Cr2O7^2- -> 2Cr3+. Step 3: Balance O by adding H2O: Cr2O7^2- -> 2Cr3+ + 7H2O. Step 4: Balance H by adding H+: Cr2O7^2- + 14H+ -> 2Cr3+ + 7H2O. Step 5: Balance charge with electrons: Oxidation: Fe2+ -> Fe3+ + 1e- (multiply by 6) ; Reduction: Cr2O7^2- + 14H+ + 6e- -> 2Cr3+ + 7H2O. Step 6: Combine: 6Fe2+ + Cr2O7^2- + 14H+ -> 6Fe3+ + 2Cr3+ + 7H2O. Final answer: 6Fe^2+ + Cr2O7^2- + 14H^+ -> 6Fe^3+ + 2Cr^3+ + 7H2O
• Q: Identify the oxidizing agent and reducing agent in the reaction: H2S + Cl2 -> 2HCl + S. A: Step 1: Determine oxidation numbers: In H2S, S is -2. In Cl2, Cl is 0. In HCl, Cl is -1. In S, S is 0. Step 2: Identify oxidation: S changes from -2 to 0 (loss of electrons). H2S is oxidized, making it the reducing agent. Step 3: Identify reduction: Cl changes from 0 to -1 (gain of electrons). Cl2 is reduced, making it the oxidizing agent. Final answer: Reducing agent is H2S; Oxidizing agent is Cl2.
• Q: Balance the disproportionation of Phosphorus in basic medium: P4 + OH- -> PH3 + H2PO2- A: Step 1: Write half-reactions: Oxidation: P4 -> H2PO2- ; Reduction: P4 -> PH3. Step 2: Balance P atoms: Oxidation: P4 -> 4H2PO2- ; Reduction: P4 -> 4PH3. Step 3: Balance oxygen and hydrogen under acidic rules first, then neutralize. For Oxidation half: P4 + 8H2O -> 4H2PO2- + 8H+ + 4e-. For Reduction half: P4 + 12H+ + 12e- -> 4PH3. Step 4: Combine after multiplying oxidation half by 3: 3P4 + 24H2O + P4 + 12H+ -> 12H2PO2- + 24H+ + 4PH3. This simplifies to: 4P4 + 24H2O -> 12H2PO2- + 12H+ + 4PH3. Divide by 4: P4 + 6H2O -> 3H2PO2- + 3H+ + PH3. Step 5: Add 3OH- to both sides to convert to basic medium: P4 + 3H2O + 3OH- -> 3H2PO2- + PH3. Final answer: P4 + 3H2O + 3OH^- -> 3H2PO2^- + PH3

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