Numford
Chemistry

Oxidation numbers and redox

Find any oxidation number. Balance any redox equation.

Oxidation number

Type a formula or an ion: K2Cr2O7, Cr2O7 2-, MnO4-, NH4+ or K4[Fe(CN)6]. Put a known ion in brackets when two elements are unknown. Pick an element to solve for it.

Cr+6

  1. K is an alkali metal: +1 in its compounds.
  2. Oxygen is −2 in most compounds.
  3. The oxidation numbers add up to the charge, 0: 2(+1) + 7(−2) + 2x = 0.
  4. 2x = +12, so x = +6.

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.

MnO₄⁻+5Fe²⁺+8H⁺→Mn²⁺+5Fe³⁺+4H₂O

  • Oxidising agentMnO₄⁻
  • Reducing agentFe²⁺
  • Electrons5

OxidationFe +2 → +3

5Fe²⁺→5Fe³⁺+5e⁻

loses 5 e⁻

ReductionMn +7 → +2

MnO₄⁻+8H⁺+5e⁻→Mn²⁺+4H₂O

gains 5 e⁻
  1. Write the skeleton in ionic form
    MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
  2. Split into two half reactions
    Oxidation: Fe²⁺ → Fe³⁺ (Fe +2 → +3)
    Reduction: MnO₄⁻ → Mn²⁺ (Mn +7 → +2)
  3. Balance atoms other than O and H
    Oxidation: Fe²⁺ → Fe³⁺
    Reduction: MnO₄⁻ → Mn²⁺
  4. Balance O by adding H₂O
    Reduction: MnO₄⁻ → Mn²⁺ + 4H₂O
  5. Balance H by adding H⁺
    Reduction: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
  6. Balance the charge with electrons
    Oxidation: Fe²⁺ → Fe³⁺ + e⁻
    Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
  7. Make the electrons equal
    Oxidation × 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻
    Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
  8. Add the halves and cancel what appears on both sides
    MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
  9. Check
    Each side has 5 Fe, 1 Mn, 4 O, 8 H and a total charge of +17.
NCERT rules, in order

The rules on one card

  • 0An element on its own: Na, O₂, S₈, P₄
  • = chargeA monatomic ion: Na⁺ is +1, Cl⁻ is −1
  • −1Fluorine, in all its compounds
  • +1, +2, +3Alkali metals, alkaline earth metals, aluminium
  • +1Hydrogen; −1 in metal hydrides such as NaH
  • −2Oxygen; −1 in peroxides, −1/2 in superoxides, +2 in OF₂
  • Σ = chargeAll the oxidation numbers add up to the charge
Worked pages

Oxidation number of

Balanced redox equations

Same subject: organic conversions, name reactions, electron configuration, bonding and shapes, balance equations, mole calculator.

Questions students ask

Oxidation numbers and balancing

What are the rules for finding oxidation numbers?

An element on its own is 0, and a monatomic ion has its charge. Fluorine is −1 in all its compounds. Alkali metals are +1, alkaline earth metals +2 and aluminium +3. Hydrogen is +1, but −1 in metal hydrides. Oxygen is −2, but −1 in peroxides, −1/2 in superoxides and +2 in OF₂. The oxidation numbers of all the atoms add up to the charge of the species.

Why is the oxidation number of Fe in Fe₃O₄ a fraction?

Fe₃O₄ is FeO·Fe₂O₃: one iron atom is +2 and two are +3. The rules treat all three alike and give their average, 8/3. A fraction always means the atoms of that element are not all in the same state.

What are the steps of the ion-electron method?

Write the skeleton in ionic form and split it into an oxidation half and a reduction half. In each half balance the atoms other than O and H, then O with H₂O, then H with H⁺. In basic solution add as many OH⁻ as H⁺ to both sides and turn H⁺ and OH⁻ into water. Balance the charge with electrons, make the electrons equal, add the halves and cancel.

What are the steps of the oxidation number method?

Write the oxidation numbers of every atom and find the ones that change. Balance the atoms that change, work out the total rise and the total fall, and multiply so that they are equal. Then balance the charge with H⁺ (acid) or OH⁻ (base), and the H and O with water.

How do you find the oxidising agent and the reducing agent?

The species whose oxidation number falls is reduced and is the oxidising agent. The species whose oxidation number rises is oxidised and is the reducing agent. In MnO₄⁻ + Fe²⁺, Mn falls from +7 to +2, so MnO₄⁻ is the oxidising agent, and Fe²⁺ is the reducing agent.