Oxidation states and redoxAQA A-Level Chemistry: Revision notes
Section 1
Oxidation and reduction as electron transfer
Oxidation is the process of electron loss. Reduction is the process of electron gain. (OIL RIG: Oxidation Is Loss, Reduction Is Gain.)
An oxidising agent is an electron acceptor: it oxidises another species and is itself reduced. A reducing agent is an electron donor: it reduces another species and is itself oxidised.
Example: in Zn + Cu²⁺ → Zn²⁺ + Cu, zinc loses electrons (oxidised; reducing agent) and Cu²⁺ gains electrons (reduced; oxidising agent).
The oxidising agent is the species that is reduced, and the reducing agent is the species that is oxidised. Name the right one.
Section 2
Rules for assigning oxidation states
The oxidation state is the charge an atom would have if all its bonds were ionic. Use these rules in order:
- Uncombined elements (Na, O₂, S₈): 0
- The oxidation states in a neutral compound add up to 0; in an ion they add up to the ion charge
- Group 1: +1; Group 2: +2; Al: +3
- Fluorine: −1; other halogens usually −1
- Hydrogen: +1 (but −1 in metal hydrides such as NaH)
- Oxygen: −2 (but −1 in peroxides such as H₂O₂, and +2 in OF₂)
Write the sign. The oxidation state is +6, not 6, and the charge on an ion is written the other way round (2+).
Section 3
Working out oxidation states
Set up an equation using the rules.
Worked example 1. Mn in MnO₄⁻: Mn + 4(−2) = −1, so Mn = +7.
Worked example 2. Cr in Cr₂O₇²⁻: 2x + 7(−2) = −2, so 2x = +12 and Cr = +6.
Worked example 3. S in Na₂S₂O₃: 2(+1) + 2x + 3(−2) = 0, so S = +2.
The oxidation state of an element in an ion such as Fe³⁺ equals its charge.
Section 4
Using oxidation states to identify redox
In a redox reaction:
- Oxidation = an increase in oxidation state (electron loss)
- Reduction = a decrease in oxidation state (electron gain)
Example: 8I⁻ + 10H⁺ + SO₄²⁻ → 4I₂ + H₂S + 4H₂O. Iodine goes from −1 to 0 (oxidised). Sulfur goes from +6 to −2 (reduced), a gain of 8 electrons per sulfur atom, matching the 8 iodide ions that each lose 1.
Check that the number of electrons lost equals the number gained. This confirms that the oxidation states are right.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Oxidation states and redox
- In a titration, a solution of iron(II) ions is oxidised by acidified potassium manganate(VII). The manganate(VII) ion, MnO₄⁻, is reduced to Mn²⁺, and the iron(II) ions are oxidised to iron(III) ions, Fe³⁺.State the changes in oxidation state of manganese and of iron, and identify the reducing agent.2 marks
- A technician is checking the oxidation states of an element in a range of compounds: sodium hydride, NaH; hydrogen peroxide, H₂O₂; potassium dichromate(VI), K₂Cr₂O₇; and sodium thiosulfate, Na₂S₂O₃.Deduce the oxidation state of chromium in K₂Cr₂O₇ and of sulfur in Na₂S₂O₃. Show your working.2 marks
- Solid sodium iodide reacts with concentrated sulfuric acid. Iodide ions are oxidised to iodine, and sulfuric acid is reduced to hydrogen sulfide. The ionic equation is 8I⁻ + 10H⁺ + SO₄²⁻ → 4I₂ + H₂S + 4H₂O.Explain, in terms of electrons and oxidation states, why iodide ions are oxidised in this reaction and identify the oxidising agent.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).