Numford
Balanced in acidic solution

Fe and Cu²⁺

Fe + Cu²⁺ → Fe²⁺ + Cu

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.

Fe+Cu²⁺→Fe²⁺+Cu

  • Oxidising agentCu²⁺
  • Reducing agentFe
  • Electrons2

OxidationFe 0 → +2

Fe→Fe²⁺+2e⁻

loses 2 e⁻

ReductionCu +2 → 0

Cu²⁺+2e⁻→Cu

gains 2 e⁻
  1. Write the skeleton in ionic form
    Fe + Cu²⁺ → Fe²⁺ + Cu
  2. Split into two half reactions
    Oxidation: Fe → Fe²⁺ (Fe 0 → +2)
    Reduction: Cu²⁺ → Cu (Cu +2 → 0)
  3. Balance atoms other than O and H
    Oxidation: Fe → Fe²⁺
    Reduction: Cu²⁺ → Cu
  4. Balance the charge with electrons
    Oxidation: Fe → Fe²⁺ + 2e⁻
    Reduction: Cu²⁺ + 2e⁻ → Cu
  5. The electrons are already equal
    Oxidation: Fe → Fe²⁺ + 2e⁻
    Reduction: Cu²⁺ + 2e⁻ → Cu
  6. Add the halves and cancel what appears on both sides
    Fe + Cu²⁺ → Fe²⁺ + Cu
  7. Check
    Each side has 1 Fe, 1 Cu and a total charge of +2.
Both methods

Step by step

Ion-electron method

  1. Write the skeleton in ionic form
    Fe + Cu²⁺ → Fe²⁺ + Cu
  2. Split into two half reactions
    Oxidation: Fe → Fe²⁺ (Fe 0 → +2)
    Reduction: Cu²⁺ → Cu (Cu +2 → 0)
  3. Balance atoms other than O and H
    Oxidation: Fe → Fe²⁺
    Reduction: Cu²⁺ → Cu
  4. Balance the charge with electrons
    Oxidation: Fe → Fe²⁺ + 2e⁻
    Reduction: Cu²⁺ + 2e⁻ → Cu
  5. The electrons are already equal
    Oxidation: Fe → Fe²⁺ + 2e⁻
    Reduction: Cu²⁺ + 2e⁻ → Cu
  6. Add the halves and cancel what appears on both sides
    Fe + Cu²⁺ → Fe²⁺ + Cu
  7. Check
    Each side has 1 Fe, 1 Cu and a total charge of +2.

Oxidation number method

  1. Write the oxidation numbers and find the change
    Fe: 0 in Fe → +2 in Fe²⁺, a rise of 2 for Fe → Fe²⁺ (oxidation)
    Cu: +2 in Cu²⁺ → 0 in Cu, a fall of 2 for Cu²⁺ → Cu (reduction)
  2. Make the total rise equal the total fall (2 each)
    Fe + Cu²⁺ → Fe²⁺ + Cu
  3. Check
    Fe + Cu²⁺ → Fe²⁺ + Cu
    Each side has 1 Fe, 1 Cu and a total charge of +2.
Who gives, who takes

Oxidising and reducing agents

  • Fe 0 → +2Fe to Fe²⁺: oxidised
  • Cu +2 → 0Cu²⁺ to Cu: reduced
  • Cu²⁺oxidising agent, gains electrons
  • Fereducing agent, loses electrons
Questions

Fe and Cu²⁺: quick answers

What is the balanced equation for iron and copper(II) ions?

Fe + Cu²⁺ → Fe²⁺ + Cu.

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

Fe goes from 0 to +2 and Cu goes from +2 to 0. Cu²⁺ is the oxidising agent (it is reduced) and Fe is the reducing agent (it is oxidised).

How many electrons are transferred?

2 electrons pass from Fe to Cu²⁺ in the balanced equation as written.

Balance your own equation