Numford
Balanced in acidic solution

Cu and dilute HNO₃

3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 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.

3Cu+8HNO₃→3Cu(NO₃)₂+2NO+4H₂O

Ionic 3Cu+2NO₃⁻+8H⁺→3Cu²⁺+2NO+4H₂O

  • Oxidising agentNO₃⁻
  • Reducing agentCu
  • Electrons6

OxidationCu 0 → +2

3Cu→3Cu²⁺+6e⁻

loses 6 e⁻

ReductionN +5 → +2

2NO₃⁻+8H⁺+6e⁻→2NO+4H₂O

gains 6 e⁻
  1. Write the skeleton in ionic form
    Cu + NO₃⁻ → Cu²⁺ + NO
    Written as ions: HNO₃ as H⁺ + NO₃⁻; Cu(NO₃)₂ as Cu²⁺ + 2NO₃⁻.
  2. Split into two half reactions
    Oxidation: Cu → Cu²⁺ (Cu 0 → +2)
    Reduction: NO₃⁻ → NO (N +5 → +2)
  3. Balance atoms other than O and H
    Oxidation: Cu → Cu²⁺
    Reduction: NO₃⁻ → NO
  4. Balance O by adding H₂O
    Reduction: NO₃⁻ → NO + 2H₂O
  5. Balance H by adding H⁺
    Reduction: NO₃⁻ + 4H⁺ → NO + 2H₂O
  6. Balance the charge with electrons
    Oxidation: Cu → Cu²⁺ + 2e⁻
    Reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O
  7. Make the electrons equal
    Oxidation × 3: 3Cu → 3Cu²⁺ + 6e⁻
    Reduction × 2: 2NO₃⁻ + 8H⁺ + 6e⁻ → 2NO + 4H₂O
  8. Add the halves and cancel what appears on both sides
    3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O
  9. Check
    Each side has 3 Cu, 2 N, 6 O, 8 H and a total charge of +6.
  10. Put back the ions that do not change
    3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O
Both methods

Step by step

Ion-electron method

  1. Write the skeleton in ionic form
    Cu + NO₃⁻ → Cu²⁺ + NO
    Written as ions: HNO₃ as H⁺ + NO₃⁻; Cu(NO₃)₂ as Cu²⁺ + 2NO₃⁻.
  2. Split into two half reactions
    Oxidation: Cu → Cu²⁺ (Cu 0 → +2)
    Reduction: NO₃⁻ → NO (N +5 → +2)
  3. Balance atoms other than O and H
    Oxidation: Cu → Cu²⁺
    Reduction: NO₃⁻ → NO
  4. Balance O by adding H₂O
    Reduction: NO₃⁻ → NO + 2H₂O
  5. Balance H by adding H⁺
    Reduction: NO₃⁻ + 4H⁺ → NO + 2H₂O
  6. Balance the charge with electrons
    Oxidation: Cu → Cu²⁺ + 2e⁻
    Reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O
  7. Make the electrons equal
    Oxidation × 3: 3Cu → 3Cu²⁺ + 6e⁻
    Reduction × 2: 2NO₃⁻ + 8H⁺ + 6e⁻ → 2NO + 4H₂O
  8. Add the halves and cancel what appears on both sides
    3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O
  9. Check
    Each side has 3 Cu, 2 N, 6 O, 8 H and a total charge of +6.
  10. Put back the ions that do not change
    3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O

Oxidation number method

  1. Write the oxidation numbers and find the change
    Cu: 0 in Cu → +2 in Cu²⁺, a rise of 2 for Cu → Cu²⁺ (oxidation)
    N: +5 in NO₃⁻ → +2 in NO, a fall of 3 for NO₃⁻ → NO (reduction)
  2. Make the total rise equal the total fall (6 each)
    3Cu + 2NO₃⁻ → 3Cu²⁺ + 2NO
  3. Balance the charge with 8 H⁺ (acidic solution)
    3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO
  4. Balance H with 4 H₂O
    3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O
  5. Check
    3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O
    Each side has 3 Cu, 2 N, 6 O, 8 H and a total charge of +6.
  6. Put back the ions that do not change
    3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O
Who gives, who takes

Oxidising and reducing agents

  • Cu 0 → +2Cu to Cu²⁺: oxidised
  • N +5 → +2NO₃⁻ to NO: reduced
  • NO₃⁻oxidising agent, gains electrons
  • Cureducing agent, loses electrons
Questions

Cu and dilute HNO₃: quick answers

What is the balanced equation for copper and dilute nitric acid?

3Cu + 8HNO₃ → 3Cu(NO₃)₂ + 2NO + 4H₂O. In ionic form: 3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O.

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

Cu goes from 0 to +2 and N goes from +5 to +2. NO₃⁻ is the oxidising agent (it is reduced) and Cu is the reducing agent (it is oxidised).

How many electrons are transferred?

6 electrons pass from Cu to NO₃⁻ in the balanced equation as written.

Balance your own equation