Numford
Balanced in acidic solution

Ag and dilute HNO₃

3Ag + 4HNO₃ → 3AgNO₃ + NO + 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.

3Ag+4HNO₃→3AgNO₃+NO+2H₂O

Ionic 3Ag+NO₃⁻+4H⁺→3Ag⁺+NO+2H₂O

  • Oxidising agentNO₃⁻
  • Reducing agentAg
  • Electrons3

OxidationAg 0 → +1

3Ag→3Ag⁺+3e⁻

loses 3 e⁻

ReductionN +5 → +2

NO₃⁻+4H⁺+3e⁻→NO+2H₂O

gains 3 e⁻
  1. Write the skeleton in ionic form
    Ag + NO₃⁻ → Ag⁺ + NO
    Written as ions: HNO₃ as H⁺ + NO₃⁻; AgNO₃ as Ag⁺ + NO₃⁻.
  2. Split into two half reactions
    Oxidation: Ag → Ag⁺ (Ag 0 → +1)
    Reduction: NO₃⁻ → NO (N +5 → +2)
  3. Balance atoms other than O and H
    Oxidation: Ag → Ag⁺
    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: Ag → Ag⁺ + e⁻
    Reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O
  7. Make the electrons equal
    Oxidation × 3: 3Ag → 3Ag⁺ + 3e⁻
    Reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O
  8. Add the halves and cancel what appears on both sides
    3Ag + NO₃⁻ + 4H⁺ → 3Ag⁺ + NO + 2H₂O
  9. Check
    Each side has 3 Ag, 1 N, 3 O, 4 H and a total charge of +3.
  10. Put back the ions that do not change
    3Ag + 4HNO₃ → 3AgNO₃ + NO + 2H₂O
Both methods

Step by step

Ion-electron method

  1. Write the skeleton in ionic form
    Ag + NO₃⁻ → Ag⁺ + NO
    Written as ions: HNO₃ as H⁺ + NO₃⁻; AgNO₃ as Ag⁺ + NO₃⁻.
  2. Split into two half reactions
    Oxidation: Ag → Ag⁺ (Ag 0 → +1)
    Reduction: NO₃⁻ → NO (N +5 → +2)
  3. Balance atoms other than O and H
    Oxidation: Ag → Ag⁺
    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: Ag → Ag⁺ + e⁻
    Reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O
  7. Make the electrons equal
    Oxidation × 3: 3Ag → 3Ag⁺ + 3e⁻
    Reduction: NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O
  8. Add the halves and cancel what appears on both sides
    3Ag + NO₃⁻ + 4H⁺ → 3Ag⁺ + NO + 2H₂O
  9. Check
    Each side has 3 Ag, 1 N, 3 O, 4 H and a total charge of +3.
  10. Put back the ions that do not change
    3Ag + 4HNO₃ → 3AgNO₃ + NO + 2H₂O

Oxidation number method

  1. Write the oxidation numbers and find the change
    Ag: 0 in Ag → +1 in Ag⁺, a rise of 1 for Ag → Ag⁺ (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 (3 each)
    3Ag + NO₃⁻ → 3Ag⁺ + NO
  3. Balance the charge with 4 H⁺ (acidic solution)
    3Ag + NO₃⁻ + 4H⁺ → 3Ag⁺ + NO
  4. Balance H with 2 H₂O
    3Ag + NO₃⁻ + 4H⁺ → 3Ag⁺ + NO + 2H₂O
  5. Check
    3Ag + NO₃⁻ + 4H⁺ → 3Ag⁺ + NO + 2H₂O
    Each side has 3 Ag, 1 N, 3 O, 4 H and a total charge of +3.
  6. Put back the ions that do not change
    3Ag + 4HNO₃ → 3AgNO₃ + NO + 2H₂O
Who gives, who takes

Oxidising and reducing agents

  • Ag 0 → +1Ag to Ag⁺: oxidised
  • N +5 → +2NO₃⁻ to NO: reduced
  • NO₃⁻oxidising agent, gains electrons
  • Agreducing agent, loses electrons
Questions

Ag and dilute HNO₃: quick answers

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

3Ag + 4HNO₃ → 3AgNO₃ + NO + 2H₂O. In ionic form: 3Ag + NO₃⁻ + 4H⁺ → 3Ag⁺ + NO + 2H₂O.

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

Ag goes from 0 to +1 and N goes from +5 to +2. NO₃⁻ is the oxidising agent (it is reduced) and Ag is the reducing agent (it is oxidised).

How many electrons are transferred?

3 electrons pass from Ag to NO₃⁻ in the balanced equation as written.

Balance your own equation