Zn and NO₃⁻ in base
4Zn + NO₃⁻ + 7OH⁻ → 4ZnO₂²⁻ + NH₃ + 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.
4Zn+NO₃⁻+7OH⁻→4ZnO₂²⁻+NH₃+2H₂O
- Oxidising agentNO₃⁻
- Reducing agentZn
- Electrons8
OxidationZn 0 → +2
4Zn+16OH⁻→4ZnO₂²⁻+8H₂O+8e⁻
loses 8 e⁻ReductionN +5 → −3
NO₃⁻+6H₂O+8e⁻→NH₃+9OH⁻
gains 8 e⁻- Write the skeleton in ionic form
Zn + NO₃⁻ → ZnO₂²⁻ + NH₃ - Split into two half reactions
Oxidation: Zn → ZnO₂²⁻ (Zn 0 → +2)
Reduction: NO₃⁻ → NH₃ (N +5 → −3) - Balance atoms other than O and H
Oxidation: Zn → ZnO₂²⁻
Reduction: NO₃⁻ → NH₃ - Balance O by adding H₂O
Oxidation: Zn + 2H₂O → ZnO₂²⁻
Reduction: NO₃⁻ → NH₃ + 3H₂O - Balance H by adding H⁺
Oxidation: Zn + 2H₂O → ZnO₂²⁻ + 4H⁺
Reduction: NO₃⁻ + 9H⁺ → NH₃ + 3H₂O - Basic solution: add OH⁻ to both sides for each H⁺, make water and cancel
Oxidation: Zn + 4OH⁻ → ZnO₂²⁻ + 2H₂O
Reduction: NO₃⁻ + 6H₂O → NH₃ + 9OH⁻ - Balance the charge with electrons
Oxidation: Zn + 4OH⁻ → ZnO₂²⁻ + 2H₂O + 2e⁻
Reduction: NO₃⁻ + 6H₂O + 8e⁻ → NH₃ + 9OH⁻ - Make the electrons equal
Oxidation × 4: 4Zn + 16OH⁻ → 4ZnO₂²⁻ + 8H₂O + 8e⁻
Reduction: NO₃⁻ + 6H₂O + 8e⁻ → NH₃ + 9OH⁻ - Add the halves and cancel what appears on both sides
4Zn + NO₃⁻ + 7OH⁻ → 4ZnO₂²⁻ + NH₃ + 2H₂O - Check
Each side has 4 Zn, 10 O, 7 H, 1 N and a total charge of −8.
- Write the oxidation numbers and find the change
Zn: 0 in Zn → +2 in ZnO₂²⁻, a rise of 2 for Zn → ZnO₂²⁻ (oxidation)
N: +5 in NO₃⁻ → −3 in NH₃, a fall of 8 for NO₃⁻ → NH₃ (reduction) - Make the total rise equal the total fall (8 each)
4Zn + NO₃⁻ → 4ZnO₂²⁻ + NH₃ - Balance the charge with 7 OH⁻ (basic solution)
4Zn + NO₃⁻ + 7OH⁻ → 4ZnO₂²⁻ + NH₃ - Balance H with 2 H₂O
4Zn + NO₃⁻ + 7OH⁻ → 4ZnO₂²⁻ + NH₃ + 2H₂O - Check
4Zn + NO₃⁻ + 7OH⁻ → 4ZnO₂²⁻ + NH₃ + 2H₂O
Each side has 4 Zn, 1 N, 10 O, 7 H and a total charge of −8.
Step by step
Ion-electron method
- Write the skeleton in ionic form
Zn + NO₃⁻ → ZnO₂²⁻ + NH₃ - Split into two half reactions
Oxidation: Zn → ZnO₂²⁻ (Zn 0 → +2)
Reduction: NO₃⁻ → NH₃ (N +5 → −3) - Balance atoms other than O and H
Oxidation: Zn → ZnO₂²⁻
Reduction: NO₃⁻ → NH₃ - Balance O by adding H₂O
Oxidation: Zn + 2H₂O → ZnO₂²⁻
Reduction: NO₃⁻ → NH₃ + 3H₂O - Balance H by adding H⁺
Oxidation: Zn + 2H₂O → ZnO₂²⁻ + 4H⁺
Reduction: NO₃⁻ + 9H⁺ → NH₃ + 3H₂O - Basic solution: add OH⁻ to both sides for each H⁺, make water and cancel
Oxidation: Zn + 4OH⁻ → ZnO₂²⁻ + 2H₂O
Reduction: NO₃⁻ + 6H₂O → NH₃ + 9OH⁻ - Balance the charge with electrons
Oxidation: Zn + 4OH⁻ → ZnO₂²⁻ + 2H₂O + 2e⁻
Reduction: NO₃⁻ + 6H₂O + 8e⁻ → NH₃ + 9OH⁻ - Make the electrons equal
Oxidation × 4: 4Zn + 16OH⁻ → 4ZnO₂²⁻ + 8H₂O + 8e⁻
Reduction: NO₃⁻ + 6H₂O + 8e⁻ → NH₃ + 9OH⁻ - Add the halves and cancel what appears on both sides
4Zn + NO₃⁻ + 7OH⁻ → 4ZnO₂²⁻ + NH₃ + 2H₂O - Check
Each side has 4 Zn, 10 O, 7 H, 1 N and a total charge of −8.
Oxidation number method
- Write the oxidation numbers and find the change
Zn: 0 in Zn → +2 in ZnO₂²⁻, a rise of 2 for Zn → ZnO₂²⁻ (oxidation)
N: +5 in NO₃⁻ → −3 in NH₃, a fall of 8 for NO₃⁻ → NH₃ (reduction) - Make the total rise equal the total fall (8 each)
4Zn + NO₃⁻ → 4ZnO₂²⁻ + NH₃ - Balance the charge with 7 OH⁻ (basic solution)
4Zn + NO₃⁻ + 7OH⁻ → 4ZnO₂²⁻ + NH₃ - Balance H with 2 H₂O
4Zn + NO₃⁻ + 7OH⁻ → 4ZnO₂²⁻ + NH₃ + 2H₂O - Check
4Zn + NO₃⁻ + 7OH⁻ → 4ZnO₂²⁻ + NH₃ + 2H₂O
Each side has 4 Zn, 1 N, 10 O, 7 H and a total charge of −8.
Oxidising and reducing agents
- Zn 0 → +2Zn to ZnO₂²⁻: oxidised
- N +5 → −3NO₃⁻ to NH₃: reduced
- NO₃⁻oxidising agent, gains electrons
- Znreducing agent, loses electrons
Product written as the zincate ion ZnO2 2-; some books write [Zn(OH)4]2-, which gives 4Zn + NO3- + 7OH- + 6H2O -> 4[Zn(OH)4]2- + NH3.
Zn and NO₃⁻ in base: quick answers
What is the balanced equation for zinc and nitrate in basic solution?
4Zn + NO₃⁻ + 7OH⁻ → 4ZnO₂²⁻ + NH₃ + 2H₂O.
Which is the oxidising agent and which is the reducing agent?
Zn goes from 0 to +2 and N goes from +5 to −3. NO₃⁻ is the oxidising agent (it is reduced) and Zn is the reducing agent (it is oxidised).
How many electrons are transferred?
8 electrons pass from Zn to NO₃⁻ in the balanced equation as written.