How do we identify electron transfer and balance redox reactions?
Assign oxidation numbers, identify oxidants and reductants, and balance redox half-equations and overall equations.
Oxidation numbers, the meaning of oxidation and reduction as electron transfer, identifying oxidants and reductants, and balancing redox half-equations and full ionic equations.
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What this dot point is asking
You must assign oxidation numbers, identify what is oxidised and reduced, name the oxidant and reductant, and balance redox equations.
Oxidation and reduction
Redox reactions involve transfer of electrons.
- Oxidation is the loss of electrons; the oxidation number increases.
- Reduction is the gain of electrons; the oxidation number decreases.
The oxidant (oxidising agent) is the species that is reduced, because it takes electrons from something else. The reductant (reducing agent) is the species that is oxidised, because it supplies the electrons.
Assigning oxidation numbers
Use these rules in order:
- Free elements have an oxidation number of 0 (for example, , ).
- A monatomic ion has an oxidation number equal to its charge (for example, is +1).
- Oxygen is usually -2 (except in peroxides, where it is -1).
- Hydrogen is usually +1 (except in metal hydrides, where it is -1).
- The sum of oxidation numbers equals the overall charge of the species.
Writing half-equations
Split the reaction into an oxidation half and a reduction half. For each half-equation balance the atoms, then balance charge by adding electrons. In acidic aqueous conditions, balance oxygen with and hydrogen with .
State the oxidant and reductant by name or formula in your answer, since identifying them is often a separate mark from balancing the equation.
Exam-style practice questions
Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2023 TASC6 marksIdentify the oxidation number of vanadium in V3+, VO2+, V2O4 and NH4VO3. With reference to the oxidation state of vanadium, explain whether vanadium is oxidised or reduced in the reaction 2V2O4(s) + O2(g) -> 2V2O5(s).Show worked answer →
Oxidation numbers (4 marks). Oxygen is -2 and the overall charge must balance. V3+: vanadium is +3. VO2+: V + (-2) = +2, so V = +4. V2O4: 2V + 4(-2) = 0, so V = +4. NH4VO3: NH4 is +1, so V + 3(-2) = -1, giving V = +5.
Oxidised or reduced (2 marks). In V2O4 vanadium is +4; in V2O5, 2V + 5(-2) = 0 gives V = +5. The oxidation number rises from +4 to +5, so vanadium is oxidised (it loses electrons). The oxygen gas is reduced from 0 to -2.
2022 TASC3 marksWhen melted, potassium chlorate KClO3 undergoes a disproportionation reaction in which the chlorate ion (ClO3-) produces both the perchlorate ion (ClO4-) and the chloride ion. Give the oxidation state of chlorine in the chlorate and perchlorate ions, then write the redox half-equation for chlorate ions reacting to give chloride ions and identify this as oxidation or reduction.Show worked answer →
Oxidation states (2 marks). In ClO3-, Cl + 3(-2) = -1, so Cl = +5. In ClO4-, Cl + 4(-2) = -1, so Cl = +7.
Half-equation and identification (1 mark). Going from ClO3- (Cl is +5) to Cl- (Cl is -1), chlorine gains 6 electrons, so this is reduction: ClO3-(aq) + 6H+(aq) + 6e- -> Cl-(aq) + 3H2O(l). Check: charge on the left is -1 +6 -6 = -1, matching the chloride on the right; oxygen and hydrogen balance via 3 water molecules.