How are oxidation and reduction tracked using oxidation numbers and half-equations?
Assign oxidation numbers, identify oxidation and reduction, and balance redox equations using half-equations.
How to assign oxidation numbers using the standard rules, identify the species oxidised and reduced, and combine balanced half-equations into an overall redox equation.
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What this dot point is asking
You must assign oxidation numbers, identify the oxidant and reductant, and balance redox equations using half-equations.
Oxidation and reduction
Rules for assigning oxidation numbers
- An element in its pure form has oxidation number (e.g. , ).
- A monatomic ion's oxidation number equals its charge (e.g. is ).
- Oxygen is usually (except in peroxides, where it is ).
- Hydrogen is usually (except in metal hydrides, where it is ).
- The sum of oxidation numbers equals the overall charge: for a neutral compound, or the ion charge for a polyatomic ion.
Identifying oxidation and reduction
Compare oxidation numbers before and after. If an element's oxidation number increases, it was oxidised; if it decreases, it was reduced.
Balancing with half-equations
Split the reaction into an oxidation half and a reduction half, balance each, then combine:
- Balance the atoms other than O and H.
- Balance O by adding .
- Balance H by adding (in acidic solution).
- Balance charge by adding electrons ().
- Multiply the half-equations so the electrons are equal, then add and cancel.
Exam-style practice questions
Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2024 SACE Stage 22 marksComplete the half-equation for the conversion of S2O3 2- to S4O6 2-.Show worked answer β
Two thiosulfate ions combine to form one tetrathionate ion, losing electrons (oxidation).
2S2O3 2- -> S4O6 2- + 2e-
Check: 4 sulfur atoms and 6 oxygen atoms each side. Charge on the left is -4; on the right S4O6 2- is -2 plus 2 electrons (-2), giving -4. Balanced. One mark for the correct species/coefficients, one for balancing charge with the two electrons on the product side.
2024 SACE Stage 22 marksIn the reaction 2Fe3+ + 2I- -> I2 + 2Fe2+, identify the species oxidised and the species reduced, with reference to changes in oxidation number.Show worked answer β
Track each element's oxidation number.
Iodine: I- has oxidation number -1 and becomes 0 in I2. The oxidation number increases, so iodide (I-) is oxidised (it is the reducing agent).
Iron: Fe3+ has oxidation number +3 and becomes +2 in Fe2+. The oxidation number decreases, so iron(III) (Fe3+) is reduced (it is the oxidising agent).
One mark for identifying I- as oxidised (with the -1 to 0 change), one for identifying Fe3+ as reduced (with the +3 to +2 change).