MnO₄⁻ and Fe²⁺ in acid
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
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.
MnO₄⁻+5Fe²⁺+8H⁺→Mn²⁺+5Fe³⁺+4H₂O
- Oxidising agentMnO₄⁻
- Reducing agentFe²⁺
- Electrons5
OxidationFe +2 → +3
5Fe²⁺→5Fe³⁺+5e⁻
loses 5 e⁻ReductionMn +7 → +2
MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O
gains 5 e⁻- Write the skeleton in ionic form
MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ - Split into two half reactions
Oxidation: Fe²⁺ → Fe³⁺ (Fe +2 → +3)
Reduction: MnO₄⁻ → Mn²⁺ (Mn +7 → +2) - Balance atoms other than O and H
Oxidation: Fe²⁺ → Fe³⁺
Reduction: MnO₄⁻ → Mn²⁺ - Balance O by adding H₂O
Reduction: MnO₄⁻ → Mn²⁺ + 4H₂O - Balance H by adding H⁺
Reduction: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O - Balance the charge with electrons
Oxidation: Fe²⁺ → Fe³⁺ + e⁻
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O - Make the electrons equal
Oxidation × 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O - Add the halves and cancel what appears on both sides
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O - Check
Each side has 5 Fe, 1 Mn, 4 O, 8 H and a total charge of +17.
- Write the oxidation numbers and find the change
Fe: +2 in Fe²⁺ → +3 in Fe³⁺, a rise of 1 for Fe²⁺ → Fe³⁺ (oxidation)
Mn: +7 in MnO₄⁻ → +2 in Mn²⁺, a fall of 5 for MnO₄⁻ → Mn²⁺ (reduction) - Make the total rise equal the total fall (5 each)
MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ - Balance the charge with 8 H⁺ (acidic solution)
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ - Balance H with 4 H₂O
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O - Check
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Each side has 5 Fe, 1 Mn, 4 O, 8 H and a total charge of +17.
Step by step
Ion-electron method
- Write the skeleton in ionic form
MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ - Split into two half reactions
Oxidation: Fe²⁺ → Fe³⁺ (Fe +2 → +3)
Reduction: MnO₄⁻ → Mn²⁺ (Mn +7 → +2) - Balance atoms other than O and H
Oxidation: Fe²⁺ → Fe³⁺
Reduction: MnO₄⁻ → Mn²⁺ - Balance O by adding H₂O
Reduction: MnO₄⁻ → Mn²⁺ + 4H₂O - Balance H by adding H⁺
Reduction: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O - Balance the charge with electrons
Oxidation: Fe²⁺ → Fe³⁺ + e⁻
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O - Make the electrons equal
Oxidation × 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O - Add the halves and cancel what appears on both sides
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O - Check
Each side has 5 Fe, 1 Mn, 4 O, 8 H and a total charge of +17.
Oxidation number method
- Write the oxidation numbers and find the change
Fe: +2 in Fe²⁺ → +3 in Fe³⁺, a rise of 1 for Fe²⁺ → Fe³⁺ (oxidation)
Mn: +7 in MnO₄⁻ → +2 in Mn²⁺, a fall of 5 for MnO₄⁻ → Mn²⁺ (reduction) - Make the total rise equal the total fall (5 each)
MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ - Balance the charge with 8 H⁺ (acidic solution)
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ - Balance H with 4 H₂O
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O - Check
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Each side has 5 Fe, 1 Mn, 4 O, 8 H and a total charge of +17.
Oxidising and reducing agents
- Fe +2 → +3Fe²⁺ to Fe³⁺: oxidised
- Mn +7 → +2MnO₄⁻ to Mn²⁺: reduced
- MnO₄⁻oxidising agent, gains electrons
- Fe²⁺reducing agent, loses electrons
MnO₄⁻ and Fe²⁺ in acid: quick answers
What is the balanced equation for permanganate and iron(II) in acid?
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O.
Which is the oxidising agent and which is the reducing agent?
Fe goes from +2 to +3 and Mn goes from +7 to +2. MnO₄⁻ is the oxidising agent (it is reduced) and Fe²⁺ is the reducing agent (it is oxidised).
How many electrons are transferred?
5 electrons pass from Fe²⁺ to MnO₄⁻ in the balanced equation as written.