Redox Reactions • Topic 2 of 3

Balancing Redox Reactions

A balanced redox equation must conserve both mass (every atom) and charge (total electrons lost = total electrons gained). Two systematic methods are used.

Oxidation-number method

This works directly on the molecular equation.

Assign oxidation numbers and identify the atoms that change.
Find the increase (oxidation) and decrease (reduction) per atom.
Multiply each half by suitable factors so the total increase equals the total decrease.
Balance the remaining atoms by inspection, then balance O with H₂O and H with H⁺ (acidic) or OH^- (basic).

Example: in MnO₄^- + Fe²⁺, Mn falls +7 → +2 (gain of 5 e^-) and Fe rises +2 → +3 (loss of 1 e^-); so five Fe²⁺ are needed per MnO₄^-.

Half-reaction (ion–electron) method

This splits the reaction into an oxidation half and a reduction half, balances each fully, then recombines them.

Write the two half-reactions (ionic form).
Balance all atoms except O and H.
Balance O by adding H₂O.
Balance H by adding H⁺.
Balance charge by adding electrons.
Multiply the halves so electrons cancel, then add.
For basic medium: add OH^- to both sides equal to the H⁺ present; H⁺ + OH^- combine to H₂O, and excess water is cancelled.

Acidic medium — worked outline

For MnO₄^- + Fe²⁺ in acid: reduction half MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O; oxidation half Fe²⁺ → Fe³⁺ + e^-. Multiplying the second by 5 and adding gives the balanced ionic equation.

Basic medium — the OH^- trick

Balance first as if acidic, then for every H⁺ add one OH^- to both sides; combine H⁺ + OH^- into H₂O and cancel any duplicated water. This converts the acidic equation cleanly to the basic form.

Half-Reaction (Ion–Electron) Method Steps
Step	Action
1	Split into oxidation and reduction halves
2	Balance atoms other than O and H
3	Balance O by adding H₂O
4	Balance H by adding H⁺
5	Balance charge with electrons
6	Equalise e⁻, add halves, cancel
7	For base: add OH⁻ for each H⁺, form H₂O

Solved Examples

Worked Example

Balance the reduction half-reaction of MnO₄^- to Mn²⁺ in acidic medium.

Solution

Write the skeleton: MnO₄^- → Mn²⁺.
Balance O with water: MnO₄^- → Mn²⁺ + 4H₂O.
Balance H with H⁺: MnO₄^- + 8H⁺ → Mn²⁺ + 4H₂O.
Balance charge: left = −1+8 = +7, right = +2; add 5e^- to the left.

Answer: MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O.

Worked Example

Combine the halves to balance MnO₄^- + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic medium.

Solution

Reduction: MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O.
Oxidation: Fe²⁺ → Fe³⁺ + e^-; multiply by 5.
Add the two halves; the 5 electrons cancel.

Answer: MnO₄^- + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.

Worked Example

Balance Cr₂O₇^2- + Fe²⁺ → Cr³⁺ + Fe³⁺ in acidic medium.

Solution

Reduction: Cr₂O₇^2- + 14H⁺ + 6e^- → 2Cr³⁺ + 7H₂O.
Oxidation: Fe²⁺ → Fe³⁺ + e^-; multiply by 6.
Add the halves; the 6 electrons cancel.

Answer: Cr₂O₇^2- + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.

Worked Example

Balance the disproportionation Cl₂ → Cl^- + ClO^- in basic medium.

Solution

Reduction: Cl₂ + 2e^- → 2Cl^-.
Oxidation (basic): Cl₂ + 4OH^- → 2ClO^- + 2H₂O + 2e^-.
Electrons are already equal (2 each); add the two halves.
Combine: 2Cl₂ + 4OH^- → 2Cl^- + 2ClO^- + 2H₂O; divide by 2.

Answer: Cl₂ + 2OH^- → Cl^- + ClO^- + H₂O.

Worked Example

Use the oxidation-number method to balance Cu + HNO₃ → Cu(NO₃)₂ + NO + H₂O.

Solution

Cu: 0 → +2 (loss of 2e^-); N in HNO₃: +5 → +2 in NO (gain of 3e^-).
LCM of 2 and 3 is 6: take 3 Cu and 2 reduced N atoms (3×2 = 2×3 = 6 e^-).
Place 3Cu and 2NO; the Cu(NO₃)₂ needs 6 NO₃^-, so total HNO₃ = 6 + 2 = 8.
Balance H and O with 4H₂O.

Answer: 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O.

Worked Example

Balance MnO₄^- + C₂O₄^2- → Mn²⁺ + CO₂ in acidic medium.

Solution

Reduction: MnO₄^- + 8H⁺ + 5e^- → Mn²⁺ + 4H₂O; multiply by 2.
Oxidation: C₂O₄^2- → 2CO₂ + 2e^-; multiply by 5 (10 e^- each side).
Add: 2MnO₄^- + 16H⁺ + 5C₂O₄^2- → 2Mn²⁺ + 8H₂O + 10CO₂.

Answer: 2MnO₄^- + 16H⁺ + 5C₂O₄^2- → 2Mn²⁺ + 10CO₂ + 8H₂O.

Key Points

A balanced redox equation conserves both atoms (mass) and total charge (electrons lost = electrons gained).
Oxidation-number method: find the increase and decrease per atom and cross-multiply so the totals are equal.
Half-reaction method: balance O with H₂O, H with H⁺, charge with electrons, then equalise electrons and add.
In acidic medium use H⁺ and H₂O; permanganate gives MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
For basic medium, balance as acidic then add one OH⁻ per H⁺ to both sides and combine H⁺ + OH⁻ into H₂O.

Quick Check — 4 MCQs

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Q1.In a balanced redox reaction, the total electrons lost must:

Explanation: Charge conservation requires electrons lost in oxidation to equal electrons gained in reduction.

Q2.The number of electrons in MnO₄^- + 8H⁺ + ne^- → Mn²⁺ + 4H₂O is:

Explanation: Mn goes from +7 to +2, a gain of 5 electrons.

Q3.While balancing in basic medium, after balancing as acidic you add:

Explanation: Add OH⁻ equal to the H⁺ present on both sides; H⁺ + OH⁻ combine into H₂O.

Q4.In 3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O, the change in oxidation number of N (in NO) is:

Explanation: Nitrogen is +5 in HNO₃ and +2 in NO, a gain of 3 electrons per N.

Practice more MCQs → 📝 Assignment · 30 marks

Oxidation-number method

Half-reaction (ion–electron) method

Acidic medium — worked outline

Basic medium — the OH- trick

Key Points

Basic medium — the OH^- trick