Cu and concentrated HNO₃
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
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.
Cu+4HNO₃→Cu(NO₃)₂+2NO₂+2H₂O
Ionic Cu+2NO₃⁻+4H⁺→Cu²⁺+2NO₂+2H₂O
- Oxidising agentNO₃⁻
- Reducing agentCu
- Electrons2
OxidationCu 0 → +2
Cu→Cu²⁺+2e⁻
loses 2 e⁻ReductionN +5 → +4
2NO₃⁻+4H⁺+2e⁻→2NO₂+2H₂O
gains 2 e⁻- Write the skeleton in ionic form
Cu + NO₃⁻ → Cu²⁺ + NO₂
Written as ions: HNO₃ as H⁺ + NO₃⁻; Cu(NO₃)₂ as Cu²⁺ + 2NO₃⁻. - Split into two half reactions
Oxidation: Cu → Cu²⁺ (Cu 0 → +2)
Reduction: NO₃⁻ → NO₂ (N +5 → +4) - Balance atoms other than O and H
Oxidation: Cu → Cu²⁺
Reduction: NO₃⁻ → NO₂ - Balance O by adding H₂O
Reduction: NO₃⁻ → NO₂ + H₂O - Balance H by adding H⁺
Reduction: NO₃⁻ + 2H⁺ → NO₂ + H₂O - Balance the charge with electrons
Oxidation: Cu → Cu²⁺ + 2e⁻
Reduction: NO₃⁻ + 2H⁺ + e⁻ → NO₂ + H₂O - Make the electrons equal
Oxidation: Cu → Cu²⁺ + 2e⁻
Reduction × 2: 2NO₃⁻ + 4H⁺ + 2e⁻ → 2NO₂ + 2H₂O - Add the halves and cancel what appears on both sides
Cu + 2NO₃⁻ + 4H⁺ → Cu²⁺ + 2NO₂ + 2H₂O - Check
Each side has 1 Cu, 2 N, 6 O, 4 H and a total charge of +2. - Put back the ions that do not change
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
- 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⁻ → +4 in NO₂, a fall of 1 for NO₃⁻ → NO₂ (reduction) - Make the total rise equal the total fall (2 each)
Cu + 2NO₃⁻ → Cu²⁺ + 2NO₂ - Balance the charge with 4 H⁺ (acidic solution)
Cu + 2NO₃⁻ + 4H⁺ → Cu²⁺ + 2NO₂ - Balance H with 2 H₂O
Cu + 2NO₃⁻ + 4H⁺ → Cu²⁺ + 2NO₂ + 2H₂O - Check
Cu + 2NO₃⁻ + 4H⁺ → Cu²⁺ + 2NO₂ + 2H₂O
Each side has 1 Cu, 2 N, 6 O, 4 H and a total charge of +2. - Put back the ions that do not change
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
Step by step
Ion-electron method
- Write the skeleton in ionic form
Cu + NO₃⁻ → Cu²⁺ + NO₂
Written as ions: HNO₃ as H⁺ + NO₃⁻; Cu(NO₃)₂ as Cu²⁺ + 2NO₃⁻. - Split into two half reactions
Oxidation: Cu → Cu²⁺ (Cu 0 → +2)
Reduction: NO₃⁻ → NO₂ (N +5 → +4) - Balance atoms other than O and H
Oxidation: Cu → Cu²⁺
Reduction: NO₃⁻ → NO₂ - Balance O by adding H₂O
Reduction: NO₃⁻ → NO₂ + H₂O - Balance H by adding H⁺
Reduction: NO₃⁻ + 2H⁺ → NO₂ + H₂O - Balance the charge with electrons
Oxidation: Cu → Cu²⁺ + 2e⁻
Reduction: NO₃⁻ + 2H⁺ + e⁻ → NO₂ + H₂O - Make the electrons equal
Oxidation: Cu → Cu²⁺ + 2e⁻
Reduction × 2: 2NO₃⁻ + 4H⁺ + 2e⁻ → 2NO₂ + 2H₂O - Add the halves and cancel what appears on both sides
Cu + 2NO₃⁻ + 4H⁺ → Cu²⁺ + 2NO₂ + 2H₂O - Check
Each side has 1 Cu, 2 N, 6 O, 4 H and a total charge of +2. - Put back the ions that do not change
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
Oxidation number method
- 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⁻ → +4 in NO₂, a fall of 1 for NO₃⁻ → NO₂ (reduction) - Make the total rise equal the total fall (2 each)
Cu + 2NO₃⁻ → Cu²⁺ + 2NO₂ - Balance the charge with 4 H⁺ (acidic solution)
Cu + 2NO₃⁻ + 4H⁺ → Cu²⁺ + 2NO₂ - Balance H with 2 H₂O
Cu + 2NO₃⁻ + 4H⁺ → Cu²⁺ + 2NO₂ + 2H₂O - Check
Cu + 2NO₃⁻ + 4H⁺ → Cu²⁺ + 2NO₂ + 2H₂O
Each side has 1 Cu, 2 N, 6 O, 4 H and a total charge of +2. - Put back the ions that do not change
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O
Oxidising and reducing agents
- Cu 0 → +2Cu to Cu²⁺: oxidised
- N +5 → +4NO₃⁻ to NO₂: reduced
- NO₃⁻oxidising agent, gains electrons
- Cureducing agent, loses electrons
Cu and concentrated HNO₃: quick answers
What is the balanced equation for copper and concentrated nitric acid?
Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O. In ionic form: Cu + 2NO₃⁻ + 4H⁺ → Cu²⁺ + 2NO₂ + 2H₂O.
Which is the oxidising agent and which is the reducing agent?
Cu goes from 0 to +2 and N goes from +5 to +4. NO₃⁻ is the oxidising agent (it is reduced) and Cu is the reducing agent (it is oxidised).
How many electrons are transferred?
2 electrons pass from Cu to NO₃⁻ in the balanced equation as written.