Numford
Balanced in basic solution

MnO₄⁻ and Br⁻ in base

2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻

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⁻+Br⁻+H₂O→2MnO₂+BrO₃⁻+2OH⁻

  • Oxidising agentMnO₄⁻
  • Reducing agentBr⁻
  • Electrons6

OxidationBr −1 → +5

Br⁻+6OH⁻→BrO₃⁻+3H₂O+6e⁻

loses 6 e⁻

ReductionMn +7 → +4

2MnO₄⁻+4H₂O+6e⁻→2MnO₂+8OH⁻

gains 6 e⁻
  1. Write the skeleton in ionic form
    MnO₄⁻ + Br⁻ → MnO₂ + BrO₃⁻
  2. Split into two half reactions
    Oxidation: Br⁻ → BrO₃⁻ (Br −1 → +5)
    Reduction: MnO₄⁻ → MnO₂ (Mn +7 → +4)
  3. Balance atoms other than O and H
    Oxidation: Br⁻ → BrO₃⁻
    Reduction: MnO₄⁻ → MnO₂
  4. Balance O by adding H₂O
    Oxidation: Br⁻ + 3H₂O → BrO₃⁻
    Reduction: MnO₄⁻ → MnO₂ + 2H₂O
  5. Balance H by adding H⁺
    Oxidation: Br⁻ + 3H₂O → BrO₃⁻ + 6H⁺
    Reduction: MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O
  6. Basic solution: add OH⁻ to both sides for each H⁺, make water and cancel
    Oxidation: Br⁻ + 6OH⁻ → BrO₃⁻ + 3H₂O
    Reduction: MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻
  7. Balance the charge with electrons
    Oxidation: Br⁻ + 6OH⁻ → BrO₃⁻ + 3H₂O + 6e⁻
    Reduction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
  8. Make the electrons equal
    Oxidation: Br⁻ + 6OH⁻ → BrO₃⁻ + 3H₂O + 6e⁻
    Reduction × 2: 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻
  9. Add the halves and cancel what appears on both sides
    2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻
  10. Check
    Each side has 1 Br, 2 H, 9 O, 2 Mn and a total charge of −3.
Both methods

Step by step

Ion-electron method

  1. Write the skeleton in ionic form
    MnO₄⁻ + Br⁻ → MnO₂ + BrO₃⁻
  2. Split into two half reactions
    Oxidation: Br⁻ → BrO₃⁻ (Br −1 → +5)
    Reduction: MnO₄⁻ → MnO₂ (Mn +7 → +4)
  3. Balance atoms other than O and H
    Oxidation: Br⁻ → BrO₃⁻
    Reduction: MnO₄⁻ → MnO₂
  4. Balance O by adding H₂O
    Oxidation: Br⁻ + 3H₂O → BrO₃⁻
    Reduction: MnO₄⁻ → MnO₂ + 2H₂O
  5. Balance H by adding H⁺
    Oxidation: Br⁻ + 3H₂O → BrO₃⁻ + 6H⁺
    Reduction: MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O
  6. Basic solution: add OH⁻ to both sides for each H⁺, make water and cancel
    Oxidation: Br⁻ + 6OH⁻ → BrO₃⁻ + 3H₂O
    Reduction: MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻
  7. Balance the charge with electrons
    Oxidation: Br⁻ + 6OH⁻ → BrO₃⁻ + 3H₂O + 6e⁻
    Reduction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
  8. Make the electrons equal
    Oxidation: Br⁻ + 6OH⁻ → BrO₃⁻ + 3H₂O + 6e⁻
    Reduction × 2: 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻
  9. Add the halves and cancel what appears on both sides
    2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻
  10. Check
    Each side has 1 Br, 2 H, 9 O, 2 Mn and a total charge of −3.

Oxidation number method

  1. Write the oxidation numbers and find the change
    Br: −1 in Br⁻ → +5 in BrO₃⁻, a rise of 6 for Br⁻ → BrO₃⁻ (oxidation)
    Mn: +7 in MnO₄⁻ → +4 in MnO₂, a fall of 3 for MnO₄⁻ → MnO₂ (reduction)
  2. Make the total rise equal the total fall (6 each)
    2MnO₄⁻ + Br⁻ → 2MnO₂ + BrO₃⁻
  3. Balance the charge with 2 OH⁻ (basic solution)
    2MnO₄⁻ + Br⁻ → 2MnO₂ + BrO₃⁻ + 2OH⁻
  4. Balance H with 1 H₂O
    2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻
  5. Check
    2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻
    Each side has 1 Br, 2 Mn, 9 O, 2 H and a total charge of −3.
Who gives, who takes

Oxidising and reducing agents

  • Br −1 → +5Br⁻ to BrO₃⁻: oxidised
  • Mn +7 → +4MnO₄⁻ to MnO₂: reduced
  • MnO₄⁻oxidising agent, gains electrons
  • Br⁻reducing agent, loses electrons
Questions

MnO₄⁻ and Br⁻ in base: quick answers

What is the balanced equation for permanganate and bromide in basic solution?

2MnO₄⁻ + Br⁻ + H₂O → 2MnO₂ + BrO₃⁻ + 2OH⁻.

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

Br goes from −1 to +5 and Mn goes from +7 to +4. MnO₄⁻ is the oxidising agent (it is reduced) and Br⁻ is the reducing agent (it is oxidised).

How many electrons are transferred?

6 electrons pass from Br⁻ to MnO₄⁻ in the balanced equation as written.

Balance your own equation