MnO₄⁻ and I⁻ in base
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻
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⁻+6I⁻+4H₂O→2MnO₂+3I₂+8OH⁻
- Oxidising agentMnO₄⁻
- Reducing agentI⁻
- Electrons6
OxidationI −1 → 0
6I⁻→3I₂+6e⁻
loses 6 e⁻ReductionMn +7 → +4
2MnO₄⁻+4H₂O+6e⁻→2MnO₂+8OH⁻
gains 6 e⁻- Write the skeleton in ionic form
MnO₄⁻ + I⁻ → MnO₂ + I₂ - Split into two half reactions
Oxidation: I⁻ → I₂ (I −1 → 0)
Reduction: MnO₄⁻ → MnO₂ (Mn +7 → +4) - Balance atoms other than O and H
Oxidation: 2I⁻ → I₂
Reduction: MnO₄⁻ → MnO₂ - Balance O by adding H₂O
Reduction: MnO₄⁻ → MnO₂ + 2H₂O - Balance H by adding H⁺
Reduction: MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O - Basic solution: add OH⁻ to both sides for each H⁺, make water and cancel
Reduction: MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻ - Balance the charge with electrons
Oxidation: 2I⁻ → I₂ + 2e⁻
Reduction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ - Make the electrons equal
Oxidation × 3: 6I⁻ → 3I₂ + 6e⁻
Reduction × 2: 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻ - Add the halves and cancel what appears on both sides
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻ - Check
Each side has 6 I, 2 Mn, 12 O, 8 H and a total charge of −8.
- Write the oxidation numbers and find the change
I: −1 in I⁻ → 0 in I₂, a rise of 2 for 2I⁻ → I₂ (oxidation)
Mn: +7 in MnO₄⁻ → +4 in MnO₂, a fall of 3 for MnO₄⁻ → MnO₂ (reduction) - Make the total rise equal the total fall (6 each)
2MnO₄⁻ + 6I⁻ → 2MnO₂ + 3I₂ - Balance the charge with 8 OH⁻ (basic solution)
2MnO₄⁻ + 6I⁻ → 2MnO₂ + 3I₂ + 8OH⁻ - Balance H with 4 H₂O
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻ - Check
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻
Each side has 6 I, 2 Mn, 12 O, 8 H and a total charge of −8.
Step by step
Ion-electron method
- Write the skeleton in ionic form
MnO₄⁻ + I⁻ → MnO₂ + I₂ - Split into two half reactions
Oxidation: I⁻ → I₂ (I −1 → 0)
Reduction: MnO₄⁻ → MnO₂ (Mn +7 → +4) - Balance atoms other than O and H
Oxidation: 2I⁻ → I₂
Reduction: MnO₄⁻ → MnO₂ - Balance O by adding H₂O
Reduction: MnO₄⁻ → MnO₂ + 2H₂O - Balance H by adding H⁺
Reduction: MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O - Basic solution: add OH⁻ to both sides for each H⁺, make water and cancel
Reduction: MnO₄⁻ + 2H₂O → MnO₂ + 4OH⁻ - Balance the charge with electrons
Oxidation: 2I⁻ → I₂ + 2e⁻
Reduction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ - Make the electrons equal
Oxidation × 3: 6I⁻ → 3I₂ + 6e⁻
Reduction × 2: 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻ - Add the halves and cancel what appears on both sides
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻ - Check
Each side has 6 I, 2 Mn, 12 O, 8 H and a total charge of −8.
Oxidation number method
- Write the oxidation numbers and find the change
I: −1 in I⁻ → 0 in I₂, a rise of 2 for 2I⁻ → I₂ (oxidation)
Mn: +7 in MnO₄⁻ → +4 in MnO₂, a fall of 3 for MnO₄⁻ → MnO₂ (reduction) - Make the total rise equal the total fall (6 each)
2MnO₄⁻ + 6I⁻ → 2MnO₂ + 3I₂ - Balance the charge with 8 OH⁻ (basic solution)
2MnO₄⁻ + 6I⁻ → 2MnO₂ + 3I₂ + 8OH⁻ - Balance H with 4 H₂O
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻ - Check
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻
Each side has 6 I, 2 Mn, 12 O, 8 H and a total charge of −8.
Oxidising and reducing agents
- I −1 → 0I⁻ to I₂: oxidised
- Mn +7 → +4MnO₄⁻ to MnO₂: reduced
- MnO₄⁻oxidising agent, gains electrons
- I⁻reducing agent, loses electrons
MnO₄⁻ and I⁻ in base: quick answers
What is the balanced equation for permanganate and iodide in basic solution?
2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻.
Which is the oxidising agent and which is the reducing agent?
I goes from −1 to 0 and Mn goes from +7 to +4. MnO₄⁻ is the oxidising agent (it is reduced) and I⁻ is the reducing agent (it is oxidised).
How many electrons are transferred?
6 electrons pass from I⁻ to MnO₄⁻ in the balanced equation as written.