Oxidation and reduction
The earliest definitions used oxygen and hydrogen:
Oxidation: addition of oxygen or removal of hydrogen.
Reduction: removal of oxygen or addition of hydrogen.
The modern definition is based on electron transfer:
Oxidation is the loss of electrons. Reduction is the gain of electrons.
A useful memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain.
Oxidation and reduction always happen together, so such reactions are called redox reactions.
In 2Na + Cl₂ → 2NaCl, sodium loses electrons (Na → Na⁺ + e⁻) and is oxidised, while chlorine gains electrons and is reduced.
The oxidising agent (oxidant) accepts electrons and is itself reduced. Here, Cl₂.
The reducing agent (reductant) donates electrons and is itself oxidised. Here, Na.
Oxidation numbers
The oxidation number (O.N.) is the charge an atom would carry if all its bonds were completely ionic. The rules:
An element in its free state has O.N. 0: Na, O₂, P₄.
For a monoatomic ion, O.N. = the charge on the ion: Fe³⁺ is +3.
Oxygen is usually −2. The exceptions are peroxides (H₂O₂, where it is −1), superoxides (KO₂, −½) and OF₂ (+2).
Hydrogen is +1 with non-metals and −1 in metal hydrides (NaH).
Fluorine is always −1. Alkali metals are +1 and alkaline-earth metals +2 in their compounds.
The sum of O.N.s is 0 for a neutral compound and equals the charge for an ion.
Example. Find the O.N. of Cr in K₂Cr₂O₇.
2(+1) + 2x + 7(−2) = 0, so 2x = 12 and x = +6.
Example. Find the O.N. of S in S₂O₃²⁻.
2x + 3(−2) = −2, so x = +2 (this is an average over the two sulphur atoms).
In terms of oxidation numbers:
An increase in O.N. is oxidation. A decrease is reduction.
Types of redox reactions
Type | Description | Example |
|---|---|---|
Combination | two substances combine | C + O₂ → CO₂ |
Decomposition | one compound breaks down | 2KClO₃ → 2KCl + 3O₂ |
Displacement | an element displaces another from a compound | Zn + CuSO₄ → ZnSO₄ + Cu |
Disproportionation | the same element is both oxidised and reduced | 2H₂O₂ → 2H₂O + O₂ |
In the decomposition of H₂O₂, oxygen goes from −1 to −2 (in H₂O) and to 0 (in O₂), so it is reduced and oxidised at the same time.
Balancing redox equations
Method 1: oxidation-number method
Write the skeletal equation and the O.N. of every atom.
Identify the atoms whose O.N. changes, and find the increase and decrease.
Multiply so that the total increase equals the total decrease.
Balance the remaining atoms. Balance O with H₂O, then H with H⁺ (in acid) or OH⁻ (in base).
Method 2: half-reaction (ion-electron) method
Example. Balance MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ in acidic solution.
Oxidation half: Fe²⁺ → Fe³⁺ + e⁻
Reduction half: MnO₄⁻ → Mn²⁺. Balance O with water: MnO₄⁻ → Mn²⁺ + 4H₂O. Balance H with H⁺: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O. Balance the charge with electrons: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
Multiply the oxidation half by 5 so that the electrons cancel, and add the halves:
MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Check: the charge on the left is −1 + 10 + 8 = +17, and on the right +2 + 15 = +17. ✓
This reaction is used in titrations. Permanganate is a deep purple self-indicator that becomes colourless when it is reduced to Mn²⁺.