What Are Redox Reactions?
Redox (reduction-oxidation) reactions involve the transfer of electrons between species. An element's oxidation number changes when it loses or gains electrons. Understanding oxidation numbers and the half-reaction method is essential for balancing complex redox equations.
Oxidation is loss of electrons (increase in oxidation number); reduction is gain of electrons (decrease in oxidation number). In redox reactions, the reducing agent is oxidized and the oxidizing agent is reduced. The half-reaction method separates oxidation and reduction steps, balancing atoms and then electrons.
- 1↓Step 1: Assign oxidation numbersDetermine oxidation state of each element before and after reaction.
- 2↓Step 2: Write half-reactionsSeparate into oxidation (lose e⁻) and reduction (gain e⁻) half-reactions.
- 3↓Step 3: Balance atomsBalance all atoms except O and H, then add H₂O or H⁺ to balance O and H.
- 4↓Step 4: Balance electronsMultiply each half-reaction so electrons lost = electrons gained.
- 5Step 5: Add half-reactionsCombine and cancel spectator ions to get the net ionic equation.
Step-by-step worked examples
Balance the redox reaction: Fe²⁺ + MnO₄⁻ → Fe³⁺ + Mn²⁺ in acidic solution.
Step 1: Oxidation numbers — Fe²⁺ (starts +2) → Fe³⁺ (ends +3), Mn in MnO₄⁻ (starts +7) → Mn²⁺ (ends +2) Step 2: Half-reactions: Oxidation: Fe²⁺ → Fe³⁺ + e⁻ Reduction: MnO₄⁻ → Mn²⁺ (Mn: +7 to +2, gains 5e⁻) → MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O Step 3: Balance electrons — multiply oxidation by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻ Step 4: Add half-reactions: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
Identify which element is oxidized and which is reduced: 2H₂S + SO₂ → 3S + 2H₂O
Oxidation numbers: H₂S: S is -2 SO₂: S is +4 S (product): S is 0 In H₂S: S goes -2 → 0 (loses 2e⁻, OXIDIZED) In SO₂: S goes +4 → 0 (gains 4e⁻, REDUCED) H₂S is the reducing agent (oxidized) SO₂ is the oxidizing agent (reduced)
Balance in basic solution: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺
Half-reactions (acidic first): Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (multiply by 6 for Cr) Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O Combine: 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O Convert to basic by adding OH⁻ to neutralize H⁺: 6Fe²⁺ + Cr₂O₇²⁻ + 14H⁺ + 14OH⁻ → 6Fe³⁺ + 2Cr³⁺ + 7H₂O + 14OH⁻ Simplify: 6Fe²⁺ + Cr₂O₇²⁻ + 7H₂O → 6Fe³⁺ + 2Cr³⁺ + 14OH⁻
Flashcards
Quick quiz
Q1.In KMnO₄, Mn's oxidation number is…
Q2.Which is oxidized in: 2Cu + O₂ → 2CuO?
Q3.In redox reactions, the reducing agent is…
Q4.Balancing MnO₄⁻ in acidic solution — Mn gains how many e⁻?
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Common mistakes
Oxidation and reduction always involve oxygen. — Correct: Oxidation = loss of e⁻, reduction = gain of e⁻. Oxygen is just one example of an oxidizing agent.
An element cannot be both oxidized and reduced in the same reaction. — Correct: Disproportionation reactions: one element can be both oxidized and reduced (e.g., S in H₂S + SO₂ → S).
The oxidizing agent gains mass. — Correct: The oxidizing agent gains electrons and is reduced; mass is not a factor.
In basic solution, only use OH⁻ from the start. — Correct: Balance as acidic first (using H⁺), then convert to basic by adding OH⁻ to both sides.
FAQ
What is an oxidation-reduction reaction?
A reaction where electrons are transferred from one species (oxidized, reducing agent) to another (reduced, oxidizing agent).
How do you find oxidation numbers?
Use the rules: element = 0, ion charge = oxidation number, O = -2 (usually), H = +1 (usually), sum = total charge.
What is a disproportionation reaction?
One element is both oxidized and reduced in the same reaction (e.g., Cl₂ + base → Cl⁻ + ClO⁻).
Why balance electrons in redox reactions?
Electrons lost in oxidation must equal electrons gained in reduction; this is conservation of charge and mass.