How do we balance complex redox reactions?
Construct and combine oxidation and reduction half-equations to balance redox reactions.
Writing half-equations, balancing atoms and charge in acidic solution, and combining oxidation and reduction halves into a full ionic equation.
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What this dot point is asking
You must construct balanced half-equations and combine them into a balanced full redox equation.
Why half-equations help
A redox reaction is electron transfer: one species is oxidised (loses electrons) and another is reduced (gains electrons). Splitting the reaction into two halves lets you track the electrons explicitly and balance even complicated reactions reliably.
Balancing a half-equation in acidic solution
Use this order for each half:
- Balance the atoms being oxidised or reduced.
- Balance oxygen by adding .
- Balance hydrogen by adding .
- Balance charge by adding electrons .
Electrons appear on the left of a reduction half-equation (gained) and on the right of an oxidation half-equation (lost).
Checking your answer
A correctly balanced redox equation has the same number of each atom on both sides and the same total charge on both sides. Always verify the charge balance; it is the step most often skipped.
In the exam, write each half-equation clearly, follow the atom-then-charge order, scale for equal electrons, cancel them on combining, and finish by checking both atom and charge balance.
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 TASC4 marksVanadium ions, V3+, react with nitrate, NO3-, in acidic solution to form aqueous vanadyl ions, VO2+, and nitric oxide gas, NO. Determine the balanced redox reaction using the half-equation method.Show worked answer β
Oxidation half-equation. V goes from +3 (V3+) to +4 (VO2+). Balance O with water, H with H+, charge with electrons: V3+ + H2O -> VO2+ + 2H+ + e-. (Charge: left +3, right +2 +2 -1 = +3.)
Reduction half-equation. N goes from +5 (NO3-) to +2 (NO): NO3- + 4H+ + 3e- -> NO + 2H2O. (Charge: left -1 +4 -3 = 0, right 0.)
Combine. Multiply the oxidation half by 3 to supply 3 electrons: 3V3+ + 3H2O -> 3VO2+ + 6H+ + 3e-. Add to the reduction half and cancel 3e-, then cancel common H+ and H2O:
3V3+(aq) + NO3-(aq) + 2H+(aq) -> 3VO2+(aq) + NO(g) + H2O(l). (4 marks for correct half-equations, electron balance and cancellation.)
2022 TASC2 marksWhen molten, the chlorate ion (ClO3-) disproportionates to perchlorate (ClO4-) and chloride. Write the redox half-equation for chlorate ions reacting to give perchlorate ions, then write the balanced net ionic equation for the disproportionation of the chlorate ion.Show worked answer β
Oxidation half (chlorate to perchlorate, Cl +5 to +7, loses 2 electrons): ClO3- + H2O -> ClO4- + 2H+ + 2e-.
Reduction half (chlorate to chloride, Cl +5 to -1, gains 6 electrons): ClO3- + 6H+ + 6e- -> Cl- + 3H2O.
Balance electrons by multiplying the oxidation half by 3, add to the reduction half and cancel 6e-, 6H+ and 3H2O:
4ClO3- -> 3ClO4- + Cl-. (1 mark each for a correct half-equation and the cancelled overall equation. Check: 4 Cl and 12 O on each side, total charge -4 on each side.)