Balancing Redox Half reactions

Step 1 of half-reaction method

Divide the skeleton reaction into two half-reactions (oxidation and reduction), each with the oxidized and reduced forms of one species

Step 2 of half-reaction method

Balance atoms and charge in each half-reaction

Order for balancing atoms in a half-reaction

1) Atoms other than O and H, 2) O, 3) H

Where do electrons go in the reduction half-reaction?

On the LEFT (reactant) side

Where do electrons go in the oxidation half-reaction?

On the RIGHT (product) side

Step 3 of half-reaction method

Multiply half-reactions by integers so electrons gained = electrons lost

Step 4 of half-reaction method

Add the half-reactions, cancel electrons (and any common species), and include states

Acidic solution: how to balance O

Add H₂O to the side that needs O

Acidic solution: how to balance H

Add H⁺ to the side that needs H

Basic solution: method

Balance as if acidic first, then add OH⁻ to both sides equal to the number of H⁺, combine H⁺ + OH⁻ → H₂O, and cancel extra water

Final check for a balanced redox equation

Atoms balanced AND total charge equal on both sides

Balance: Cu²⁺(aq) + Cr(s) → Cr³⁺(aq) + Cu(s) – oxidation half

Cr(s) → Cr³⁺(aq) + 3e⁻

Balance: Cu²⁺(aq) + Cr(s) → Cr³⁺(aq) + Cu(s) – reduction half

Cu²⁺(aq) + 2e⁻ → Cu(s)

Multipliers for Cu²⁺/Cr reaction

Ox × 2, Red × 3 (6 electrons each)

Final: Cu²⁺ + Cr → Cr³⁺ + Cu

3Cu²⁺(aq) + 2Cr(s) → 2Cr³⁺(aq) + 3Cu(s)

Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ (acidic): reduction half

Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l)

Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ (acidic): oxidation half

2I⁻(aq) → I₂(s) + 2e⁻

Multipliers for Cr₂O₇²⁻/I⁻ reaction

Reduction × 1, oxidation × 3 (6 electrons)

Final: Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ (acidic)

Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6I⁻(aq) → 2Cr³⁺(aq) + 3I₂(s) + 7H₂O(l)

MnO₄⁻ + C₂O₄²⁻ → MnO₂ + CO₃²⁻ (basic): reduction half

MnO₄⁻(aq) + 2H₂O(l) + 3e⁻ → MnO₂(s) + 4OH⁻(aq)

MnO₄⁻ + C₂O₄²⁻ → MnO₂ + CO₃²⁻ (basic): oxidation half

C₂O₄²⁻(aq) + 4OH⁻(aq) → 2CO₃²⁻(aq) + 2H₂O(l) + 2e⁻

Multipliers for MnO₄⁻/C₂O₄²⁻ reaction

Reduction × 2, oxidation × 3 (6 electrons)

Final: MnO₄⁻ + C₂O₄²⁻ → MnO₂ + CO₃²⁻ (basic)

2MnO₄⁻(aq) + 3C₂O₄²⁻(aq) + 4OH⁻(aq) → 2MnO₂(s) + 6CO₃²⁻(aq) + 2H₂O(l)