MnO₄⁻ and oxalate in acid
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 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⁻+5C₂O₄²⁻+16H⁺→2Mn²⁺+10CO₂+8H₂O
- Oxidising agentMnO₄⁻
- Reducing agentC₂O₄²⁻
- Electrons10
OxidationC +3 → +4
5C₂O₄²⁻→10CO₂+10e⁻
loses 10 e⁻ReductionMn +7 → +2
2MnO₄⁻+16H⁺+10e⁻→2Mn²⁺+8H₂O
gains 10 e⁻- Write the skeleton in ionic form
MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ - Split into two half reactions
Oxidation: C₂O₄²⁻ → CO₂ (C +3 → +4)
Reduction: MnO₄⁻ → Mn²⁺ (Mn +7 → +2) - Balance atoms other than O and H
Oxidation: C₂O₄²⁻ → 2CO₂
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: C₂O₄²⁻ → 2CO₂ + 2e⁻
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O - Make the electrons equal
Oxidation × 5: 5C₂O₄²⁻ → 10CO₂ + 10e⁻
Reduction × 2: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O - Add the halves and cancel what appears on both sides
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O - Check
Each side has 10 C, 28 O, 2 Mn, 16 H and a total charge of +4.
- Write the oxidation numbers and find the change
C: +3 in C₂O₄²⁻ → +4 in CO₂, a rise of 2 for C₂O₄²⁻ → 2CO₂ (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⁻ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ - Balance the charge with 16 H⁺ (acidic solution)
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ - Balance H with 8 H₂O
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O - Check
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
Each side has 10 C, 28 O, 2 Mn, 16 H and a total charge of +4.
Step by step
Ion-electron method
- Write the skeleton in ionic form
MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂ - Split into two half reactions
Oxidation: C₂O₄²⁻ → CO₂ (C +3 → +4)
Reduction: MnO₄⁻ → Mn²⁺ (Mn +7 → +2) - Balance atoms other than O and H
Oxidation: C₂O₄²⁻ → 2CO₂
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: C₂O₄²⁻ → 2CO₂ + 2e⁻
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O - Make the electrons equal
Oxidation × 5: 5C₂O₄²⁻ → 10CO₂ + 10e⁻
Reduction × 2: 2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O - Add the halves and cancel what appears on both sides
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O - Check
Each side has 10 C, 28 O, 2 Mn, 16 H and a total charge of +4.
Oxidation number method
- Write the oxidation numbers and find the change
C: +3 in C₂O₄²⁻ → +4 in CO₂, a rise of 2 for C₂O₄²⁻ → 2CO₂ (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⁻ + 5C₂O₄²⁻ → 2Mn²⁺ + 10CO₂ - Balance the charge with 16 H⁺ (acidic solution)
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ - Balance H with 8 H₂O
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O - Check
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
Each side has 10 C, 28 O, 2 Mn, 16 H and a total charge of +4.
Oxidising and reducing agents
- C +3 → +4C₂O₄²⁻ to CO₂: oxidised
- Mn +7 → +2MnO₄⁻ to Mn²⁺: reduced
- MnO₄⁻oxidising agent, gains electrons
- C₂O₄²⁻reducing agent, loses electrons
MnO₄⁻ and oxalate in acid: quick answers
What is the balanced equation for permanganate and oxalate in acid?
2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.
Which is the oxidising agent and which is the reducing agent?
C goes from +3 to +4 and Mn goes from +7 to +2. MnO₄⁻ is the oxidising agent (it is reduced) and C₂O₄²⁻ is the reducing agent (it is oxidised).
How many electrons are transferred?
10 electrons pass from C₂O₄²⁻ to MnO₄⁻ in the balanced equation as written.