Redox Balancing Formula
Master the half-reaction method for balancing complex redox equations
Half-Reaction Method (Acidic Solution)
Split into Half-Reactions
Separate the oxidation and reduction processes
Identify which species is oxidized and which is reduced by assigning oxidation numbers
Balance Atoms (Except O and H)
Balance all elements except oxygen and hydrogen
Balance Oxygen with H₂O
Add H₂O molecules to balance oxygen atoms
Balance Hydrogen with H⁺
Add H⁺ ions to balance hydrogen atoms
Balance Charge with e⁻
Add electrons to balance the total charge
Electrons added to the more positive side
Equalize Electrons
Multiply each half-reaction to equalize electron transfer
Add Half-Reactions
Combine the two half-reactions and cancel common terms
Electrons should cancel out completely
Verify
Check that atoms and charges balance on both sides
Worked Example (Acidic Solution)
Balance: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ (acidic)
Step 1: Split into half-reactions
Reduction: Cr₂O₇²⁻ → Cr³⁺
Oxidation: Fe²⁺ → Fe³⁺
Step 2: Balance atoms except O, H
Cr₂O₇²⁻ → 2Cr³⁺ (balance Cr)
Fe²⁺ → Fe³⁺ (already balanced)
Step 3: Balance O with H₂O
Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O (7 O on left, 7 on right)
Fe²⁺ → Fe³⁺ (no O, skip)
Step 4: Balance H with H⁺
14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O (14 H on each side)
Fe²⁺ → Fe³⁺ (no H, skip)
Step 5: Balance charge with e⁻
Reduction half:
Left: 14(+1) + (-2) = +12
Right: 2(+3) = +6
Need 6e⁻ on left:
6e⁻ + 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O
Oxidation half:
Left: +2, Right: +3
Need 1e⁻ on right:
Fe²⁺ → Fe³⁺ + 1e⁻
Step 6: Equalize electrons (LCM = 6)
6e⁻ + 14H⁺ + Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O (×1)
[Fe²⁺ → Fe³⁺ + 1e⁻] × 6
6Fe²⁺ → 6Fe³⁺ + 6e⁻
Step 7: Add and cancel
6e⁻
14H⁺ + Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺ + 6e⁻
Final Balanced Equation:
14H⁺ + Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺
Verification:
Cr: 2 = 2 ✓
O: 7 = 7 ✓
H: 14 = 14 ✓
Fe: 6 = 6 ✓
Charge: 14 + (-2) + 6(2) = 12 + 6 = +24; 2(3) + 6(3) = 6 + 18 = +24 ✓
Basic Solution Method
For reactions in basic solution, follow all acidic steps, then add one final step:
Extra Step: Neutralize H⁺ with OH⁻
1. Add OH⁻ to both sides equal to the number of H⁺
2. Combine H⁺ + OH⁻ → H₂O on the side with H⁺
3. Cancel excess H₂O from both sides
Example: Convert to Basic
Acidic form:
4H⁺ + MnO₄⁻ + 3e⁻ → MnO₂ + 2H₂O
Add 4OH⁻ to each side:
4H⁺ + 4OH⁻ + MnO₄⁻ + 3e⁻ → MnO₂ + 2H₂O + 4OH⁻
Combine H⁺ + OH⁻ → H₂O:
4H₂O + MnO₄⁻ + 3e⁻ → MnO₂ + 2H₂O + 4OH⁻
Cancel 2H₂O:
2H₂O + MnO₄⁻ + 3e⁻ → MnO₂ + 4OH⁻
Another Complete Example
Balance: MnO₄⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻ (basic)
Half-reactions (acidic first):
Reduction:
MnO₄⁻ → MnO₂
4H⁺ + MnO₄⁻ → MnO₂ + 2H₂O
3e⁻ + 4H⁺ + MnO₄⁻ → MnO₂ + 2H₂O
Oxidation:
SO₃²⁻ → SO₄²⁻
H₂O + SO₃²⁻ → SO₄²⁻ + 2H⁺
H₂O + SO₃²⁻ → SO₄²⁻ + 2H⁺ + 2e⁻
Equalize electrons (LCM = 6):
[3e⁻ + 4H⁺ + MnO₄⁻ → MnO₂ + 2H₂O] × 2
[H₂O + SO₃²⁻ → SO₄²⁻ + 2H⁺ + 2e⁻] × 3
Combined (acidic):
8H⁺ + 2MnO₄⁻ + 3SO₃²⁻ → 2MnO₂ + 3SO₄²⁻ + 4H₂O + 6H⁺
Simplify (cancel 6H⁺):
2H⁺ + 2MnO₄⁻ + 3SO₃²⁻ → 2MnO₂ + 3SO₄²⁻ + H₂O
Convert to basic (add 2OH⁻):
2H⁺ + 2OH⁻ + 2MnO₄⁻ + 3SO₃²⁻ → 2MnO₂ + 3SO₄²⁻ + H₂O + 2OH⁻
2H₂O + 2MnO₄⁻ + 3SO₃²⁻ → 2MnO₂ + 3SO₄²⁻ + H₂O + 2OH⁻
Final (cancel 1H₂O):
H₂O + 2MnO₄⁻ + 3SO₃²⁻ → 2MnO₂ + 3SO₄²⁻ + 2OH⁻
Common Mistakes
Forgetting to Balance Atoms First
Always balance Cr₂O₇²⁻ → 2Cr³⁺ before adding H₂O!
Wrong Electron Count
Calculate charge on BOTH sides carefully. Many errors here!
Not Equalizing Electrons
Find LCM of electron counts before adding half-reactions!
Always Verify Final Answer
Check: atoms balanced? Charges balanced? Electrons cancelled?