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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, balancing disproportionation, and combining oxidation and reduction halves into a full ionic equation, with worked TASC-style examples.

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What this dot point is asking

TASC expects you to construct balanced half-equations in acidic solution and combine them into a balanced full ionic equation, including disproportionation reactions.

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. It also makes the link to electrochemical cells direct, since each electrode reaction is a half-equation.

Balancing a half-equation in acidic solution

Use this order for each half-equation:

  1. Balance the atoms being oxidised or reduced.
  2. Balance oxygen by adding H2O\text{H}_2\text{O}.
  3. Balance hydrogen by adding H+\text{H}^+.
  4. Balance charge by adding electrons eβˆ’e^-.

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. For a disproportionation, the same element appears in both halves, so check the atom totals of that element carefully.

Balancing in basic solution

Many TASC redox reactions occur in acidic solution, but some are in basic conditions. The simplest approach is to balance the equation first as if it were acidic (using H+\text{H}^+ and H2O\text{H}_2\text{O}), then neutralise the H+\text{H}^+ by adding the same number of OHβˆ’\text{OH}^- to both sides. The H+\text{H}^+ and OHβˆ’\text{OH}^- on one side combine to form water, and you cancel any water that then appears on both sides. The result is a half-equation containing OHβˆ’\text{OH}^- and H2O\text{H}_2\text{O} but no H+\text{H}^+, appropriate for a basic medium.

Linking half-equations to cells

Each balanced half-equation is exactly the reaction that occurs at one electrode of an electrochemical cell: the oxidation half is the anode reaction and the reduction half is the cathode reaction. The number of electrons in the half-equation is the same number used in Faraday's law calculations and in finding standard cell potentials. Building the skill of writing clean half-equations therefore pays off directly in the electrochemistry and electrolysis questions later in Unit 3.

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.

TCE 20234 marksVanadium(III) ions, V3+\text{V}^{3+}, react with nitrate, NO3βˆ’\text{NO}_3^-, in acidic solution to form vanadyl ions, VO2+\text{VO}^{2+}, and nitric oxide gas, NO\text{NO}. Determine the balanced net ionic equation using the half-equation method.
Show worked answer β†’

Oxidation half: V\text{V} goes from +3+3 (V3+\text{V}^{3+}) to +4+4 (VO2+\text{VO}^{2+}). Balance O with water, H with H+\text{H}^+, charge with electrons: V3++H2Oβ†’VO2++2H++eβˆ’\text{V}^{3+} + \text{H}_2\text{O} \rightarrow \text{VO}^{2+} + 2\text{H}^+ + e^-. (Charge: left +3+3, right +2+2βˆ’1=+3+2 + 2 - 1 = +3.) (1 mark)

Reduction half: N\text{N} goes from +5+5 (NO3βˆ’\text{NO}_3^-) to +2+2 (NO\text{NO}): NO3βˆ’+4H++3eβˆ’β†’NO+2H2O\text{NO}_3^- + 4\text{H}^+ + 3e^- \rightarrow \text{NO} + 2\text{H}_2\text{O}. (Charge: left βˆ’1+4βˆ’3=0-1 + 4 - 3 = 0, right 00.) (1 mark)

Multiply the oxidation half by 33 to supply 33 electrons, add to the reduction half, and cancel 3eβˆ’3e^- and common H+\text{H}^+ and H2O\text{H}_2\text{O}: 3V3+(aq)+NO3βˆ’(aq)+2H+(aq)β†’3VO2+(aq)+NO(g)+H2O(l)3\text{V}^{3+}(aq) + \text{NO}_3^-(aq) + 2\text{H}^+(aq) \rightarrow 3\text{VO}^{2+}(aq) + \text{NO}(g) + \text{H}_2\text{O}(l). (2 marks)

TCE 20223 marksWhen molten, the chlorate ion ClO3βˆ’\text{ClO}_3^- disproportionates to perchlorate ClO4βˆ’\text{ClO}_4^- and chloride. (a) Write the half-equation for chlorate forming perchlorate. (b) Write the balanced net ionic equation for the disproportionation.
Show worked answer β†’

(a) Oxidation half (Cl\text{Cl} from +5+5 to +7+7, loses 2eβˆ’2e^-): ClO3βˆ’+H2Oβ†’ClO4βˆ’+2H++2eβˆ’\text{ClO}_3^- + \text{H}_2\text{O} \rightarrow \text{ClO}_4^- + 2\text{H}^+ + 2e^-. (1 mark)

(b) Reduction half (Cl\text{Cl} from +5+5 to βˆ’1-1, gains 6eβˆ’6e^-): ClO3βˆ’+6H++6eβˆ’β†’Clβˆ’+3H2O\text{ClO}_3^- + 6\text{H}^+ + 6e^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}. Multiply the oxidation half by 33, add, and cancel 6eβˆ’6e^-, 6H+6\text{H}^+ and 3H2O3\text{H}_2\text{O}: 4ClO3βˆ’β†’3ClO4βˆ’+Clβˆ’4\text{ClO}_3^- \rightarrow 3\text{ClO}_4^- + \text{Cl}^-. (2 marks) Check: 4Β Cl4\ \text{Cl} and 12Β O12\ \text{O} each side, total charge βˆ’4-4 each side.

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