CBSE Class 11 Chemistry Redox Reactions Notes Set 07

These Class 11 CBSE Chemistry notes for Chapter 7 Redox Reactions cover oxidation and reduction, the rules for oxidation numbers, oxidising and reducing agents, disproportionation and comproportionation, and balancing redox equations by the ion-electron method in acidic and basic media. Four solved examples show the working step by step.

Important Notes and Formulas

Oxidation and Reduction

Oxidation Number

1. Oxidation number is the imaginary or apparent charge developed on an atom of an element when it goes from its elemental free state to the combined state in a molecule.
2. It is calculated on the basis of an arbitrary set of rules, and it is a relative charge in a particular bonded state.
3. It is a practical method to keep track of electron shifts in chemical reactions that form compounds.
4. It is always assumed that there is a complete transfer of electrons from the less electronegative atom to the more electronegative atom.

Rules for Finding Oxidation Number

1. Fluorine is the most electronegative element. Its oxidation number is always −1 in all its compounds.
2. Oxygen is generally −2, including in its oxides. In peroxides (H₂O₂, Na₂O₂) it is −1, in superoxides (KO₂) −1/2, in ozonides (KO₃) −1/3, in OF₂ it is +2 and in O₂F₂ it is +1.
3. Hydrogen is generally +1, but −1 in metal hydrides such as NaH and KH.
4. Halogens (Cl, Br, I) are generally −1. If a halogen is attached to a more electronegative atom, it shows a positive oxidation number, e.g. Cl is +5 in KClO₃, I is +5 in HIO₃, Cl is +7 in HClO₄ and Br is +5 in KBrO₃.
5. Metals: alkali metals (Li, Na, K, Rb) are always +1, alkaline earth metals (Be, Mg, Ca) are always +2, and aluminium is always +3. A metal may also have a negative or zero oxidation number in some compounds.
6. The oxidation number of an element in its free state or in its allotropic forms is always zero, e.g. O₂, S₈, P₄ and O₃.
7. The sum of the oxidation numbers of all atoms in a molecule is zero.
8. The sum of the oxidation numbers of all atoms in an ion is equal to the charge on the ion.
9. If the group number of a p-block element in the modern periodic table is n, its oxidation number can vary from (n − 18) to (n − 10). For example, N is in group 15, so it varies from −3 to +5 (NH₃ −3, N₂H₄ −2, NH₂OH −1, N₂ 0, N₂O +1, NO +2, N₂O₃ +3, NO₂ +4, N₂O₅ +5).
10. The maximum possible oxidation number of an element in a compound is never more than the number of electrons in its valence shell (mainly applicable to p-block elements).

Average and Fractional Oxidation Number

When an element in a compound has a fractional oxidation number, the value is the average of the oxidation states of all its atoms. The atoms are actually present in different oxidation states. For example, in C₃O₂ the average oxidation number of carbon is +4/3, but the two terminal carbon atoms are +2 and the middle one is 0. In Br₃O₈ the average for Br is +16/3 (two terminal atoms at +6 and one at +4), and in S₄O₆²⁻ the average for S is +2.5 (two atoms at +5 and two at 0). The idea of a fractional oxidation state should be used with care.

Oxidising and Reducing Agents

Oxidising agent (oxidant): a substance that oxidises others and is itself reduced. The oxidation number of one of its elements decreases, or it gains electrons. Examples: KMnO4, K2Cr2O7, HNO3 and conc. H2SO4 are powerful oxidising agents.

Reducing agent (reductant): a substance that reduces others and is itself oxidised. The oxidation number of one of its elements increases, or it loses electrons. Examples: KI and Na2S2O3 are powerful reducing agents.

Some compounds can act as both an oxidising and a reducing agent, e.g. H₂O₂ and NO₂⁻.

How to identify an oxidising or reducing agent: find the oxidation state of the central atom. If it is the maximum oxidation state of the element, the substance is an oxidising agent. If it is the minimum oxidation state, it is a reducing agent. If it is an intermediate state, it can act as both, and may also disproportionate.

Redox Reactions

A reaction in which oxidation and reduction take place simultaneously is called a redox reaction. In every redox reaction, the total increase in oxidation number equals the total decrease in oxidation number.
e.g. 10FeSO₄ + 2KMnO₄ + 8H₂SO₄ → 5Fe₂(SO₄)₃ + 2MnSO₄ + K₂SO₄ + 8H₂O

Disproportionation Reaction

A redox reaction in which the same element present in one compound, in a definite oxidation state, is oxidised as well as reduced at the same time is called a disproportionation reaction. One of the reactants always contains an element that can exist in at least three oxidation states. That element is in the intermediate state, and both a higher and a lower oxidation state of it are formed. Examples:

1. 2H₂O₂ (−1) → 2H₂O (−2) + O₂ (0)
2. S₈ + 12OH⁻ → 4S²⁻ + 2S₂O₃²⁻ + 6H₂O
3. Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O
4. 4KClO₃ → 3KClO₄ + KCl
5. Fluorine does not undergo disproportionation, because it is the most electronegative element and shows only 0 and −1 states.

Not every reaction that looks like it is disproportionation. In 2KClO₃ → 2KCl + 3O₂, Cl is reduced and O is oxidised. Different elements change, so it is not a disproportionation reaction.

Comproportionation Reaction

The reverse of disproportionation is called comproportionation: an element from two different oxidation states is converted into a single oxidation state. Example: NH₄NO₂ → N₂ + 2H₂O (nitrogen at −3 and +3 both become 0), and 5I⁻ + IO₃⁻ + 6H⁺ → 3I₂ + 3H₂O.

Balancing of Redox Reactions

A balanced equation must satisfy two conditions: (1) atom (mass) balance, the same number of atoms of each kind on both sides, and (2) charge balance, the same total charge on both sides.

Redox equations are balanced by the oxidation-number change method or the ion-electron (half-reaction) method. The ion-electron method is described below for acidic and basic media.

Ion-Electron Method in Acidic Medium

1. Assign oxidation numbers to every element in the reaction.
2. Convert the reaction into ionic form by leaving out the species that undergo neither oxidation nor reduction.
3. Identify which species is oxidised and which is reduced.
4. Split the ionic reaction into two half-reactions, one for oxidation and one for reduction.
5. Balance the atoms other than O and H in each half-reaction.
6. Balance O and H: for each excess oxygen atom add one H₂O to the other side and 2H⁺ to the same side.
7. Balance the charge of each half-reaction by adding electrons to the electrically positive side.
8. Multiply the half-reactions by suitable factors so that the electrons lost equal the electrons gained, then add them to get the balanced ionic equation.
9. Convert the ionic equation back to molecular form by adding the species left out in step 2.

Ion-Electron Method in Basic Medium

All steps are the same except step 6. Balance O and H as in acidic medium, then add as many OH⁻ ions to both sides as there are H⁺ ions. Combine H⁺ and OH⁻ to form H₂O and cancel the water that appears on both sides.

Solved Examples

Example 1. Calculate the oxidation number of sulphur in (a) Na₂S₂O₃ and (b) Na₂S₄O₆.
Answer:
(a) Let the oxidation number of S be x. (+1) × 2 + (x) × 2 + (−2) × 3 = 0, so x = +2.
(b) (+1) × 2 + (x) × 4 + (−2) × 6 = 0, so x = +2.5.
These are average values; the sulphur atoms in these compounds are not all in the same state.

Example 2. Find the average and individual oxidation numbers of Fe in Fe₃O₄ and of Pb in Pb₃O₄, which are mixed oxides.
Answer:
(i) Fe₃O₄ is a mixture of FeO and Fe₂O₃ in the ratio 1 : 1, so the individual oxidation numbers are +2 and +3. Average = [1(+2) + 2(+3)] ÷ 3 = +8/3.
(ii) Pb₃O₄ is a mixture of PbO and PbO₂ in the ratio 2 : 1, so the individual oxidation numbers are +2 and +4. Average = [2(+2) + 1(+4)] ÷ 3 = +8/3.

Example 3. Balance the following redox reaction (acidic medium): FeSO₄ + KMnO₄ + H₂SO₄ → Fe₂(SO₄)₃ + MnSO₄ + H₂O + K₂SO₄
Answer:
Step I. Fe in FeSO₄ is +2 and Mn in KMnO₄ is +7. In the products Fe is +3 and Mn is +2.
Step II (ionic form). Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺
Steps III–IV. Fe²⁺ is oxidised and MnO₄⁻ is reduced: Fe²⁺ → Fe³⁺ and MnO₄⁻ → Mn²⁺
Steps V–VII. Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (1). Reduction: 5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O (2)
Step VIII. Multiply (1) by 5 and (2) by 1 and add: 5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O
Step IX (molecular form). 5FeSO₄ + KMnO₄ + 4H₂SO₄ → 5/2 Fe₂(SO₄)₃ + MnSO₄ + 4H₂O + 1/2 K₂SO₄, or 10FeSO₄ + 2KMnO₄ + 8H₂SO₄ → 5Fe₂(SO₄)₃ + 2MnSO₄ + 8H₂O + K₂SO₄

Example 4. Balance the following redox reaction in basic medium: ClO⁻ + CrO₂⁻ + OH⁻ → Cl⁻ + CrO₄²⁻ + H₂O
Answer:
Half-reactions. Reduction: ClO⁻ (+1) → Cl⁻ (−1). Oxidation: CrO₂⁻ (+3) → CrO₄²⁻ (+6).
Balance O and H as in acid. 2H⁺ + ClO⁻ → Cl⁻ + H₂O and 2H₂O + CrO₂⁻ → CrO₄²⁻ + 4H⁺
Convert to basic medium. Add OH⁻ on both sides to cancel H⁺: H₂O + ClO⁻ → Cl⁻ + 2OH⁻ and 4OH⁻ + CrO₂⁻ → CrO₄²⁻ + 2H₂O
Add electrons. 2e⁻ + H₂O + ClO⁻ → Cl⁻ + 2OH⁻ (×3) and 4OH⁻ + CrO₂⁻ → CrO₄²⁻ + 2H₂O + 3e⁻ (×2)
Add. 3ClO⁻ + 2CrO₂⁻ + 2OH⁻ → 3Cl⁻ + 2CrO₄²⁻ + H₂O