Numford
Redox balancing

The ion-electron method

Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O

Try the method

Type the reactants and products with an arrow, without numbers: MnO4- + Fe2+ -> Mn2+ + Fe3+. Leave out H+, OH- and water; they are added for you. Salts and acids such as KMnO4 or HNO3 are fine.

Cr₂O₇²⁻+3SO₃²⁻+8H⁺→2Cr³⁺+3SO₄²⁻+4H₂O

  • Oxidising agentCr₂O₇²⁻
  • Reducing agentSO₃²⁻
  • Electrons6

OxidationS +4 → +6

3SO₃²⁻+3H₂O→3SO₄²⁻+6H⁺+6e⁻

loses 6 e⁻

ReductionCr +6 → +3

Cr₂O₇²⁻+14H⁺+6e⁻→2Cr³⁺+7H₂O

gains 6 e⁻
  1. Write the skeleton in ionic form
    Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻
  2. Split into two half reactions
    Oxidation: SO₃²⁻ → SO₄²⁻ (S +4 → +6)
    Reduction: Cr₂O₇²⁻ → Cr³⁺ (Cr +6 → +3)
  3. Balance atoms other than O and H
    Oxidation: SO₃²⁻ → SO₄²⁻
    Reduction: Cr₂O₇²⁻ → 2Cr³⁺
  4. Balance O by adding H₂O
    Oxidation: SO₃²⁻ + H₂O → SO₄²⁻
    Reduction: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O
  5. Balance H by adding H⁺
    Oxidation: SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺
    Reduction: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O
  6. Balance the charge with electrons
    Oxidation: SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻
    Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
  7. Make the electrons equal
    Oxidation × 3: 3SO₃²⁻ + 3H₂O → 3SO₄²⁻ + 6H⁺ + 6e⁻
    Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
  8. Add the halves and cancel what appears on both sides
    Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O
  9. Check
    Each side has 3 S, 16 O, 8 H, 2 Cr and a total charge of 0.
Both methods

Step by step

Ion-electron method

  1. Write the skeleton in ionic form
    Cr₂O₇²⁻ + SO₃²⁻ → Cr³⁺ + SO₄²⁻
  2. Split into two half reactions
    Oxidation: SO₃²⁻ → SO₄²⁻ (S +4 → +6)
    Reduction: Cr₂O₇²⁻ → Cr³⁺ (Cr +6 → +3)
  3. Balance atoms other than O and H
    Oxidation: SO₃²⁻ → SO₄²⁻
    Reduction: Cr₂O₇²⁻ → 2Cr³⁺
  4. Balance O by adding H₂O
    Oxidation: SO₃²⁻ + H₂O → SO₄²⁻
    Reduction: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O
  5. Balance H by adding H⁺
    Oxidation: SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺
    Reduction: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O
  6. Balance the charge with electrons
    Oxidation: SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻
    Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
  7. Make the electrons equal
    Oxidation × 3: 3SO₃²⁻ + 3H₂O → 3SO₄²⁻ + 6H⁺ + 6e⁻
    Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
  8. Add the halves and cancel what appears on both sides
    Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O
  9. Check
    Each side has 3 S, 16 O, 8 H, 2 Cr and a total charge of 0.

Oxidation number method

  1. Write the oxidation numbers and find the change
    S: +4 in SO₃²⁻ → +6 in SO₄²⁻, a rise of 2 for SO₃²⁻ → SO₄²⁻ (oxidation)
    Cr: +6 in Cr₂O₇²⁻ → +3 in Cr³⁺, a fall of 6 for Cr₂O₇²⁻ → 2Cr³⁺ (reduction)
  2. Make the total rise equal the total fall (6 each)
    Cr₂O₇²⁻ + 3SO₃²⁻ → 2Cr³⁺ + 3SO₄²⁻
  3. Balance the charge with 8 H⁺ (acidic solution)
    Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻
  4. Balance H with 4 H₂O
    Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O
  5. Check
    Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O
    Each side has 3 S, 16 O, 2 Cr, 8 H and a total charge of 0.
Who gives, who takes

Oxidising and reducing agents

  • S +4 → +6SO₃²⁻ to SO₄²⁻: oxidised
  • Cr +6 → +3Cr₂O₇²⁻ to Cr³⁺: reduced
  • Cr₂O₇²⁻oxidising agent, gains electrons
  • SO₃²⁻reducing agent, loses electrons
Questions

The ion-electron method: quick answers

What is the balanced equation for the ion-electron method, worked on dichromate and sulphite?

Cr₂O₇²⁻ + 3SO₃²⁻ + 8H⁺ → 2Cr³⁺ + 3SO₄²⁻ + 4H₂O.

Which is the oxidising agent and which is the reducing agent?

S goes from +4 to +6 and Cr goes from +6 to +3. Cr₂O₇²⁻ is the oxidising agent (it is reduced) and SO₃²⁻ is the reducing agent (it is oxidised).

How many electrons are transferred?

6 electrons pass from SO₃²⁻ to Cr₂O₇²⁻ in the balanced equation as written.

Balance your own equation