Numford
Balanced in acidic solution

Thermite: Al and Fe₂O₃

2Al + Fe₂O₃ → Al₂O₃ + 2Fe

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.

2Al+Fe₂O₃→Al₂O₃+2Fe

  • Oxidising agentFe₂O₃
  • Reducing agentAl
  • Electrons6

OxidationAl 0 → +3

2Al+3H₂O→Al₂O₃+6H⁺+6e⁻

loses 6 e⁻

ReductionFe +3 → 0

Fe₂O₃+6H⁺+6e⁻→2Fe+3H₂O

gains 6 e⁻
  1. Write the skeleton in ionic form
    Al + Fe₂O₃ → Al₂O₃ + Fe
  2. Split into two half reactions
    Oxidation: Al → Al₂O₃ (Al 0 → +3)
    Reduction: Fe₂O₃ → Fe (Fe +3 → 0)
  3. Balance atoms other than O and H
    Oxidation: 2Al → Al₂O₃
    Reduction: Fe₂O₃ → 2Fe
  4. Balance O by adding H₂O
    Oxidation: 2Al + 3H₂O → Al₂O₃
    Reduction: Fe₂O₃ → 2Fe + 3H₂O
  5. Balance H by adding H⁺
    Oxidation: 2Al + 3H₂O → Al₂O₃ + 6H⁺
    Reduction: Fe₂O₃ + 6H⁺ → 2Fe + 3H₂O
  6. Balance the charge with electrons
    Oxidation: 2Al + 3H₂O → Al₂O₃ + 6H⁺ + 6e⁻
    Reduction: Fe₂O₃ + 6H⁺ + 6e⁻ → 2Fe + 3H₂O
  7. The electrons are already equal
    Oxidation: 2Al + 3H₂O → Al₂O₃ + 6H⁺ + 6e⁻
    Reduction: Fe₂O₃ + 6H⁺ + 6e⁻ → 2Fe + 3H₂O
  8. Add the halves and cancel what appears on both sides
    2Al + Fe₂O₃ → Al₂O₃ + 2Fe
  9. Check
    Each side has 2 Al, 2 Fe, 3 O and a total charge of 0.
Both methods

Step by step

Ion-electron method

  1. Write the skeleton in ionic form
    Al + Fe₂O₃ → Al₂O₃ + Fe
  2. Split into two half reactions
    Oxidation: Al → Al₂O₃ (Al 0 → +3)
    Reduction: Fe₂O₃ → Fe (Fe +3 → 0)
  3. Balance atoms other than O and H
    Oxidation: 2Al → Al₂O₃
    Reduction: Fe₂O₃ → 2Fe
  4. Balance O by adding H₂O
    Oxidation: 2Al + 3H₂O → Al₂O₃
    Reduction: Fe₂O₃ → 2Fe + 3H₂O
  5. Balance H by adding H⁺
    Oxidation: 2Al + 3H₂O → Al₂O₃ + 6H⁺
    Reduction: Fe₂O₃ + 6H⁺ → 2Fe + 3H₂O
  6. Balance the charge with electrons
    Oxidation: 2Al + 3H₂O → Al₂O₃ + 6H⁺ + 6e⁻
    Reduction: Fe₂O₃ + 6H⁺ + 6e⁻ → 2Fe + 3H₂O
  7. The electrons are already equal
    Oxidation: 2Al + 3H₂O → Al₂O₃ + 6H⁺ + 6e⁻
    Reduction: Fe₂O₃ + 6H⁺ + 6e⁻ → 2Fe + 3H₂O
  8. Add the halves and cancel what appears on both sides
    2Al + Fe₂O₃ → Al₂O₃ + 2Fe
  9. Check
    Each side has 2 Al, 2 Fe, 3 O and a total charge of 0.

Oxidation number method

  1. Write the oxidation numbers and find the change
    Al: 0 in Al → +3 in Al₂O₃, a rise of 6 for 2Al → Al₂O₃ (oxidation)
    Fe: +3 in Fe₂O₃ → 0 in Fe, a fall of 6 for Fe₂O₃ → 2Fe (reduction)
  2. Make the total rise equal the total fall (6 each)
    2Al + Fe₂O₃ → Al₂O₃ + 2Fe
  3. Check
    2Al + Fe₂O₃ → Al₂O₃ + 2Fe
    Each side has 2 Al, 2 Fe, 3 O and a total charge of 0.
Who gives, who takes

Oxidising and reducing agents

  • Al 0 → +3Al to Al₂O₃: oxidised
  • Fe +3 → 0Fe₂O₃ to Fe: reduced
  • Fe₂O₃oxidising agent, gains electrons
  • Alreducing agent, loses electrons
Questions

Thermite: Al and Fe₂O₃: quick answers

What is the balanced equation for aluminium and iron(III) oxide (thermite)?

2Al + Fe₂O₃ → Al₂O₃ + 2Fe.

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

Al goes from 0 to +3 and Fe goes from +3 to 0. Fe₂O₃ is the oxidising agent (it is reduced) and Al is the reducing agent (it is oxidised).

How many electrons are transferred?

6 electrons pass from Al to Fe₂O₃ in the balanced equation as written.

Balance your own equation