Redox Balancing Formula

Master the half-reaction method for balancing complex redox equations

Half-Reaction Method (Acidic Solution)

1

Split into Half-Reactions

Separate the oxidation and reduction processes

Identify which species is oxidized and which is reduced by assigning oxidation numbers

2

Balance Atoms (Except O and H)

Balance all elements except oxygen and hydrogen

3

Balance Oxygen with Hâ‚‚O

Add Hâ‚‚O molecules to balance oxygen atoms

4

Balance Hydrogen with H⁺

Add H⁺ ions to balance hydrogen atoms

5

Balance Charge with e⁻

Add electrons to balance the total charge

Electrons added to the more positive side

6

Equalize Electrons

Multiply each half-reaction to equalize electron transfer

7

Add Half-Reactions

Combine the two half-reactions and cancel common terms

Electrons should cancel out completely

8

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 Hâ‚‚O

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 Hâ‚‚O 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 2Hâ‚‚O:

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 1Hâ‚‚O):

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?

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