This category explains two key methods for balancing redox reactions: the oxidation number method and the half-reaction method. Practice balancing equations with step-by-step guidance and examples.
S (-2 in 5FeS) becomes +6 in 2Fe₂(SO₄)3 (6 S atoms). Total change = 6 × (+6 - (-2)) = 48; per FeS = 48/5 = 9.6, but reaction context gives +8.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
F (0 in 2F₂) to -1 in 4HF: 4 F atoms gain 4 electrons total. Per F₂ (2 F atoms) = 4/2 = 2 electrons.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
Cu (+2 to +1): 2 Cu atoms gain 2 electrons. I (-1 in 4KI) to 0 in I₂: 2 I atoms lose 2 electrons. Total = 2.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
Hg (+1 in Hg₂Cl₂) becomes +2 in HgCl₂. Change = +2 - (+1) = +1.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
PbO₂ (Pb: +4 to +2) accepts electrons from Pb (0 to +2), acting as the oxidizing agent.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
Let Re = x. O = -2. Equation: x + 3(-2) = 0, x - 6 = 0, x = +6.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
Oxidation involves electron loss. Br (-1 in KBr) becomes 0 in Br₂, losing electrons.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
In 2PbO + PbS → 3Pb + SO₂, Pb (+2 in PbO) is reduced to 0, and S (-2 in PbS) is oxidized to +4.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
Let N = x. H = +1, O = -2. Equation: +1 + x + 2(-2) = 0, 1 + x - 4 = 0, x = +3.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
N (+5 in HNO₃) to +2 in NO: 2 N gain 6 electrons. S (-2 to 0): 3 S lose 6 electrons. Per HNO₃ = 6/2 = 3.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
In 2HI + Cl₂ → 2HCl + I₂, I (-1) loses H and electrons to become 0 in I₂, with no oxygen present.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
S (-2 in H₂S) becomes 0 in S. Change = 0 - (-2) = +2.
Ref: NCERT Class 11 Chemistry > Chapter 7: Redox Reactions > Topic: Balancing Redox Reactions - Oxidation Number Method and Half Reaction
