MnO₄⁻ and SO₂
2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻
Redox balancer
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.
2MnO₄⁻+5SO₂+2H₂O+H⁺→2Mn²⁺+5HSO₄⁻
- Oxidising agentMnO₄⁻
- Reducing agentSO₂
- Electrons10
OxidationS +4 → +6
5SO₂+10H₂O→5HSO₄⁻+15H⁺+10e⁻
loses 10 e⁻ReductionMn +7 → +2
2MnO₄⁻+16H⁺+10e⁻→2Mn²⁺+8H₂O
gains 10 e⁻- Write the skeleton in ionic form
MnO₄⁻ + SO₂ → Mn²⁺ + HSO₄⁻ - Split into two half reactions
Oxidation: SO₂ → HSO₄⁻ (S +4 → +6)
Reduction: MnO₄⁻ → Mn²⁺ (Mn +7 → +2) - Balance atoms other than O and H
Oxidation: SO₂ → HSO₄⁻
Reduction: MnO₄⁻ → Mn²⁺ - Balance O by adding H₂O
Oxidation: SO₂ + 2H₂O → HSO₄⁻
Reduction: MnO₄⁻ → Mn²⁺ + 4H₂O - Balance H by adding H⁺
Oxidation: SO₂ + 2H₂O → HSO₄⁻ + 3H⁺
Reduction: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O - Balance the charge with electrons
Oxidation: SO₂ + 2H₂O → HSO₄⁻ + 3H⁺ + 2e⁻
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O - Make the electrons equal
Oxidation × 5: 5SO₂ + 10H₂O → 5HSO₄⁻ + 15H⁺ + 10e⁻
Reduction × 2: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O - Add the halves and cancel what appears on both sides
2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻ - Check
Each side has 5 S, 20 O, 5 H, 2 Mn and a total charge of −1.
- Write the oxidation numbers and find the change
S: +4 in SO₂ → +6 in HSO₄⁻, a rise of 2 for SO₂ → HSO₄⁻ (oxidation)
Mn: +7 in MnO₄⁻ → +2 in Mn²⁺, a fall of 5 for MnO₄⁻ → Mn²⁺ (reduction) - Make the total rise equal the total fall (10 each)
2MnO₄⁻ + 5SO₂ → 2Mn²⁺ + 5HSO₄⁻ - Balance the charge with 1 H⁺ (acidic solution)
2MnO₄⁻ + 5SO₂ + H⁺ → 2Mn²⁺ + 5HSO₄⁻ - Balance H with 2 H₂O
2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻ - Check
2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻
Each side has 5 S, 20 O, 2 Mn, 5 H and a total charge of −1.
Step by step
Ion-electron method
- Write the skeleton in ionic form
MnO₄⁻ + SO₂ → Mn²⁺ + HSO₄⁻ - Split into two half reactions
Oxidation: SO₂ → HSO₄⁻ (S +4 → +6)
Reduction: MnO₄⁻ → Mn²⁺ (Mn +7 → +2) - Balance atoms other than O and H
Oxidation: SO₂ → HSO₄⁻
Reduction: MnO₄⁻ → Mn²⁺ - Balance O by adding H₂O
Oxidation: SO₂ + 2H₂O → HSO₄⁻
Reduction: MnO₄⁻ → Mn²⁺ + 4H₂O - Balance H by adding H⁺
Oxidation: SO₂ + 2H₂O → HSO₄⁻ + 3H⁺
Reduction: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O - Balance the charge with electrons
Oxidation: SO₂ + 2H₂O → HSO₄⁻ + 3H⁺ + 2e⁻
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O - Make the electrons equal
Oxidation × 5: 5SO₂ + 10H₂O → 5HSO₄⁻ + 15H⁺ + 10e⁻
Reduction × 2: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O - Add the halves and cancel what appears on both sides
2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻ - Check
Each side has 5 S, 20 O, 5 H, 2 Mn and a total charge of −1.
Oxidation number method
- Write the oxidation numbers and find the change
S: +4 in SO₂ → +6 in HSO₄⁻, a rise of 2 for SO₂ → HSO₄⁻ (oxidation)
Mn: +7 in MnO₄⁻ → +2 in Mn²⁺, a fall of 5 for MnO₄⁻ → Mn²⁺ (reduction) - Make the total rise equal the total fall (10 each)
2MnO₄⁻ + 5SO₂ → 2Mn²⁺ + 5HSO₄⁻ - Balance the charge with 1 H⁺ (acidic solution)
2MnO₄⁻ + 5SO₂ + H⁺ → 2Mn²⁺ + 5HSO₄⁻ - Balance H with 2 H₂O
2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻ - Check
2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻
Each side has 5 S, 20 O, 2 Mn, 5 H and a total charge of −1.
Oxidising and reducing agents
- S +4 → +6SO₂ to HSO₄⁻: oxidised
- Mn +7 → +2MnO₄⁻ to Mn²⁺: reduced
- MnO₄⁻oxidising agent, gains electrons
- SO₂reducing agent, loses electrons
MnO₄⁻ and SO₂: quick answers
What is the balanced equation for permanganate and sulphur dioxide in acid?
2MnO₄⁻ + 5SO₂ + 2H₂O + H⁺ → 2Mn²⁺ + 5HSO₄⁻.
Which is the oxidising agent and which is the reducing agent?
S goes from +4 to +6 and Mn goes from +7 to +2. MnO₄⁻ is the oxidising agent (it is reduced) and SO₂ is the reducing agent (it is oxidised).
How many electrons are transferred?
10 electrons pass from SO₂ to MnO₄⁻ in the balanced equation as written.