Redox Reactions Class 11 Chemistry Chapter Notes

Key Terminology in Redox Reactions

Oxidation

A process involving the loss of electrons, an increase in oxidation number, addition of oxygen, or removal of hydrogen.

Reduction

A process involving the gain of electrons, a decrease in oxidation number, removal of oxygen, or addition of hydrogen.

Redox Reaction

A chemical reaction in which oxidation and reduction occur simultaneously.

Oxidizing Agent (Oxidant)

A substance that causes oxidation by accepting electrons and gets reduced itself.

Reducing Agent (Reductant)

A substance that causes reduction by donating electrons and gets oxidized itself.

Oxidation Number (or Oxidation State)

A hypothetical charge that an atom would have if all bonds to atoms of different elements were 100% ionic. It's a tool to track electron shifts.

Disproportionation Reaction

A redox reaction where an element in one oxidation state is simultaneously oxidized and reduced.

Half-Reaction

Either the oxidation or reduction part of a redox reaction, shown with electrons gained or lost.

The Concept and Rules of Oxidation Number

The oxidation number is a central concept for understanding redox reactions. It provides a way to track which species loses electrons (is oxidized) and which gains electrons (is reduced). It's a hypothetical charge assigned to an atom based on a set of rules, assuming all bonds are ionic. An increase in oxidation number signifies oxidation, while a decrease signifies reduction. To correctly identify and balance redox reactions, you must master these rules:

1. For an element in its free or uncombined state, the oxidation number is always zero. (e.g., O₂, H₂, Na, Fe)
2. For a monatomic ion, the oxidation number is equal to the charge on the ion. (e.g., Na⁺ is +1, Cl⁻ is -1, Mg²⁺ is +2)
3. The oxidation number of oxygen is usually -2. Exceptions: in peroxides (like H₂O₂), it is -1; in superoxides (like KO₂), it is -1/2; and in compounds with fluorine (like OF₂), it is +2.
4. The oxidation number of hydrogen is usually +1. Exception: in metal hydrides (like NaH, CaH₂), it is -1.
5. The oxidation number of halogens (F, Cl, Br, I) is usually -1. Exceptions occur when they are bonded to oxygen or a more electronegative halogen.
6. The algebraic sum of oxidation numbers of all atoms in a neutral compound is zero.
7. The algebraic sum of oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion.

Must Remember for Redox Reactions

• Memory Aid: LEO says GER (Loss of Electrons is Oxidation, Gain of Electrons is Reduction). Alternatively, OIL RIG (Oxidation Is Loss, Reduction Is Gain).
• An oxidizing agent is an electron acceptor and gets reduced.
• A reducing agent is an electron donor and gets oxidized.
• The sum of oxidation numbers in a neutral molecule is always zero.
• The sum of oxidation numbers in a polyatomic ion equals the ion's charge.
• Fluorine, being the most electronegative element, always has an oxidation state of -1 in its compounds.
• Alkali metals (Group 1) always have +1, and Alkaline earth metals (Group 2) always have +2 oxidation states in their compounds.
• Balancing can be done by two main methods: the Oxidation Number Method and the Ion-Electron (Half-Reaction) Method.
• In a disproportionation reaction, a single species undergoes both oxidation and reduction.
• Fractional oxidation numbers are possible and represent an average charge over several atoms of the same element in a molecule (e.g., S in S₄O₆²⁻).

Oxidation vs. Reduction at a Glance

Balancing Redox Equations: Ion-Electron Method (Acidic Medium)

1. —
2. —
3. —
4. —
5. —
6. —
7. —
8. —

Quick Worked Examples

• {"title":"Assigning Oxidation Number","content":"Find the oxidation number of Mn in KMnO₄.\nSolution: Let the oxidation number of Mn be x. We know K is +1 and O is -2. The overall charge is 0.\n(+1) + (x) + 4(-2) = 0\n1 + x - 8 = 0\nx = +7. So, the oxidation state of Mn is +7."}
• {"title":"Identifying Oxidizing and Reducing Agents","content":"In the reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)\nSolution:\n- Zinc (Zn) goes from oxidation state 0 to +2. It loses electrons and is oxidized. Therefore, Zn is the reducing agent.\n- Copper ion (Cu²⁺) goes from oxidation state +2 to 0. It gains electrons and is reduced. Therefore, Cu²⁺ is the oxidizing agent."}
• {"title":"Disproportionation Reaction Example","content":"Reaction: 2H₂O₂(aq) → 2H₂O(l) + O₂(g)\nSolution: In hydrogen peroxide (H₂O₂), oxygen has an oxidation state of -1. In the products, it is -2 in water (H₂O) and 0 in oxygen gas (O₂). Since the same element (Oxygen) is both reduced (-1 to -2) and oxidized (-1 to 0), this is a disproportionation reaction."}

Board Exam Traps and Scoring Tips

In balancing questions, marks are awarded for each step. Never write only the final balanced equation. Clearly show the steps for both half-reactions. For the Ion-Electron method, remember to adjust for basic medium by adding OH⁻ ions to neutralize H⁺ if required. Also, be very careful with exceptions to oxidation number rules, especially for oxygen in peroxides/superoxides and hydrogen in metal hydrides, as these are common sources for MCQs. When asked to identify the oxidant/reductant, always specify the entire reactant species (e.g., MnO₄⁻), not just the element (Mn).

Quick Revision Check

• What is the oxidation state of Cr in the dichromate ion, Cr₂O₇²⁻? +6. (2x + 7(-2) = -2 => 2x - 14 = -2 => 2x = 12 => x = +6)
• Identify the reducing agent in the reaction: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. This is a disproportionation reaction. Cl₂ acts as both the oxidizing and reducing agent. Chlorine is reduced to Cl⁻ (0 to -1) and oxidized to ClO⁻ (0 to +1).
• What happens to the oxidation number of a substance during reduction? The oxidation number decreases.
• Name a type of redox reaction where the same element is both oxidized and reduced. Disproportionation reaction.

Frequently Asked Questions on Redox Reactions

Frequently Asked Questions