Oxidation numbers and redoxEdexcel International A Level Chemistry: Revision notes
Section 1
Oxidation numbers and the rules
The oxidation number of an atom is the charge it would have if all its bonds were treated as fully ionic. It is written with a sign, for example +2 or −1.
Rules, in order of priority:
- Atoms in an uncombined element have oxidation number 0 (Na, O₂, S₈, Cl₂).
- The oxidation numbers in a neutral compound sum to 0. In an ion they sum to the charge on the ion.
- Group 1 is +1, Group 2 is +2, aluminium is +3.
- Fluorine is always −1. Hydrogen is +1, except in metal hydrides (NaH, CaH₂) where it is −1.
- Oxygen is −2, except in peroxides (H₂O₂, Na₂O₂) where it is −1.
- Chlorine is −1 except when combined with oxygen or fluorine.
Write the sign before the number (+2, not 2+). The sign is part of an oxidation number, the charge on an ion is written the other way.
Section 2
Calculating oxidation numbers
Use the rules and solve for the unknown.
Worked example 1. Sulfur in SO₄²⁻: S + 4(−2) = −2, so S = +6.
Worked example 2. Manganese in MnO₄⁻: Mn + 4(−2) = −1, so Mn = +7.
Worked example 3. Oxygen in H₂O₂: 2(+1) + 2(O) = 0, so O = −1 (a peroxide).
Worked example 4. Hydrogen in NaH: (+1) + H = 0, so H = −1 (a metal hydride).
In a Roman numeral name, the oxidation number of the element is shown in brackets: iron(III) oxide is Fe₂O₃, and sodium sulfate(IV) is Na₂SO₃. To write a formula from a name, balance the charges: vanadium(V) oxide has V⁵⁺ and O²⁻, so V₂O₅.
For an ion, set the sum of the oxidation numbers equal to the ion's charge, not to zero.
Section 3
Redox as electron transfer
Oxidation is loss of electrons, so the oxidation number increases. Reduction is gain of electrons, so the oxidation number decreases.
- Metals form positive ions: their oxidation number increases (Mg 0 to +2). Non-metals form negative ions: their oxidation number decreases (Cl 0 to −1).
- An oxidising agent gains electrons (it is reduced). A reducing agent loses electrons (it is oxidised).
This applies to s-block metals such as sodium (0 to +1) and p-block elements such as chlorine and sulfur.
Example. 2Na + Cl₂ → 2NaCl. Na: 0 to +1, oxidised, reducing agent. Cl: 0 to −1, reduced, oxidising agent.
The oxidising agent is the one that is reduced. State the species, not the change.
Section 4
Disproportionation and classifying reactions
Disproportionation is when an element in a single species is simultaneously oxidised and reduced.
Example: chlorine with cold dilute sodium hydroxide: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. Chlorine goes 0 to −1 (reduced) and 0 to +1 (oxidised). Another example: 2Cu⁺ → Cu + Cu²⁺.
To classify a reaction, write oxidation numbers on both sides:
- No change: not redox (for example neutralisation).
- Change in two different elements: redox.
- The same element goes both up and down, from one species: disproportionation.
A disproportionation is a special case of redox.
Section 5
Half-equations and full ionic equations
A half-equation shows one species gaining or losing electrons. Balance atoms, then add H₂O and H⁺ for oxygen and hydrogen (in acid), then add electrons to balance charge.
Reduction of manganate(VII): MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O Oxidation of iron(II): Fe²⁺ → Fe³⁺ + e⁻
To combine, multiply so the electrons are equal, add, and cancel the electrons:
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
Check the charges: left −1 + 8 + 10 = +17; right +2 + 15 = +17.
Always check that atoms and charges balance on both sides. Electrons must not appear in the final equation.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Oxidation numbers and redox
- Sulfur forms compounds and ions in which its oxidation number ranges from −2 to +6. Examples include hydrogen sulfide, H₂S, sulfur dioxide, SO₂, the sulfate ion, SO₄²⁻, and the sulfite ion, SO₃²⁻. In the contact process sulfur dioxide is oxidised to sulfur trioxide by oxygen.In the reaction 2SO₂ + O₂ → 2SO₃, state which element is oxidised and which is reduced. Give the oxidation numbers to support your answer.2 marks
- Hydrides and peroxides contain elements in unusual oxidation states. Sodium hydride, NaH, reacts with water: NaH + H₂O → NaOH + H₂. Hydrogen peroxide, H₂O₂, decomposes when a catalyst is added: 2H₂O₂ → 2H₂O + O₂.Deduce the changes in oxidation number of hydrogen in the reaction NaH + H₂O → NaOH + H₂, and state which species is the reducing agent.2 marks
- Acidified potassium manganate(VII) solution is purple. When it is added to a solution containing iron(II) ions, the purple colour disappears and iron(III) ions and manganese(II) ions form.Write ionic half-equations for (i) the reduction of MnO₄⁻ to Mn²⁺ in acid solution and (ii) the oxidation of Fe²⁺ to Fe³⁺.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).