H₂O₂ and MnO₄⁻ in acid
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O
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⁻+5H₂O₂+6H⁺→2Mn²⁺+5O₂+8H₂O
- Oxidising agentMnO₄⁻
- Reducing agentH₂O₂
- Electrons10
OxidationO −1 → 0
5H₂O₂→5O₂+10H⁺+10e⁻
loses 10 e⁻ReductionMn +7 → +2
2MnO₄⁻+16H⁺+10e⁻→2Mn²⁺+8H₂O
gains 10 e⁻- Write the skeleton in ionic form
MnO₄⁻ + H₂O₂ → Mn²⁺ + O₂ - Split into two half reactions
Oxidation: H₂O₂ → O₂ (O −1 → 0)
Reduction: MnO₄⁻ → Mn²⁺ (Mn +7 → +2) - Balance atoms other than O and H
Oxidation: H₂O₂ → O₂
Reduction: MnO₄⁻ → Mn²⁺ - Balance O by adding H₂O
Reduction: MnO₄⁻ → Mn²⁺ + 4H₂O - Balance H by adding H⁺
Oxidation: H₂O₂ → O₂ + 2H⁺
Reduction: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O - Balance the charge with electrons
Oxidation: H₂O₂ → O₂ + 2H⁺ + 2e⁻
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O - Make the electrons equal
Oxidation × 5: 5H₂O₂ → 5O₂ + 10H⁺ + 10e⁻
Reduction × 2: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O - Add the halves and cancel what appears on both sides
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O - Check
Each side has 16 H, 18 O, 2 Mn and a total charge of +4.
- Write the oxidation numbers and find the change
O: −1 in H₂O₂ → 0 in O₂, a rise of 2 for H₂O₂ → O₂ (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⁻ + 5H₂O₂ → 2Mn²⁺ + 5O₂ - Balance the charge with 6 H⁺ (acidic solution)
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ - Balance H with 8 H₂O
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O - Check
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O
Each side has 16 H, 18 O, 2 Mn and a total charge of +4.
Step by step
Ion-electron method
- Write the skeleton in ionic form
MnO₄⁻ + H₂O₂ → Mn²⁺ + O₂ - Split into two half reactions
Oxidation: H₂O₂ → O₂ (O −1 → 0)
Reduction: MnO₄⁻ → Mn²⁺ (Mn +7 → +2) - Balance atoms other than O and H
Oxidation: H₂O₂ → O₂
Reduction: MnO₄⁻ → Mn²⁺ - Balance O by adding H₂O
Reduction: MnO₄⁻ → Mn²⁺ + 4H₂O - Balance H by adding H⁺
Oxidation: H₂O₂ → O₂ + 2H⁺
Reduction: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O - Balance the charge with electrons
Oxidation: H₂O₂ → O₂ + 2H⁺ + 2e⁻
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O - Make the electrons equal
Oxidation × 5: 5H₂O₂ → 5O₂ + 10H⁺ + 10e⁻
Reduction × 2: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O - Add the halves and cancel what appears on both sides
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O - Check
Each side has 16 H, 18 O, 2 Mn and a total charge of +4.
Oxidation number method
- Write the oxidation numbers and find the change
O: −1 in H₂O₂ → 0 in O₂, a rise of 2 for H₂O₂ → O₂ (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⁻ + 5H₂O₂ → 2Mn²⁺ + 5O₂ - Balance the charge with 6 H⁺ (acidic solution)
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ - Balance H with 8 H₂O
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O - Check
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O
Each side has 16 H, 18 O, 2 Mn and a total charge of +4.
Oxidising and reducing agents
- O −1 → 0H₂O₂ to O₂: oxidised
- Mn +7 → +2MnO₄⁻ to Mn²⁺: reduced
- MnO₄⁻oxidising agent, gains electrons
- H₂O₂reducing agent, loses electrons
H₂O₂ and MnO₄⁻ in acid: quick answers
What is the balanced equation for hydrogen peroxide and permanganate in acid?
2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O.
Which is the oxidising agent and which is the reducing agent?
O goes from −1 to 0 and Mn goes from +7 to +2. MnO₄⁻ is the oxidising agent (it is reduced) and H₂O₂ is the reducing agent (it is oxidised).
How many electrons are transferred?
10 electrons pass from H₂O₂ to MnO₄⁻ in the balanced equation as written.