Balancing redox half-equations: TCE Chemistry (Tasmania)

Syllabus dot point

“Construct and combine oxidation and reduction half-equations to balance redox reactions.”

9 min readUpdated 2 Jun 2026

Quick answer

Split a redox reaction into an oxidation half-equation and a reduction half-equation. Balance each for atoms (using $\text{H}_2\text{O}$ for oxygen and $\text{H}^+$ for hydrogen in acidic solution) and for charge (using electrons), then scale so the electrons lost equal the electrons gained and add the halves together so the electrons cancel.

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:

Balance the atoms being oxidised or reduced.
Balance oxygen by adding $\text{H}_2\text{O}$ .
Balance hydrogen by adding $\text{H}^+$ .
Balance charge by adding electrons $e^-$ .

Electrons appear on the left of a reduction half-equation (gained) and on the right of an oxidation half-equation (lost).

Half-equation

An equation showing either the oxidation or the reduction part of a redox reaction on its own, including the electrons lost or gained, balanced for both atoms and charge.

Combining the halves

Multiply each half-equation so the number of electrons lost equals the number gained (use the lowest common multiple of the electron counts). When you add the two halves, the electrons cancel exactly and must not appear in the final equation; cancel any species that appears on both sides too.

Permanganate oxidising iron(II)

Balance the reaction between permanganate ions and iron(II) ions in acidic solution by constructing and combining two half-equations.

Step 1: Write the oxidation half-equation

Iron(II) loses one electron to become iron(III). Since there is only one iron atom and no oxygen or hydrogen involved, this half-equation is straightforward. Electrons appear on the right because they are released by the species being oxidised:

\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-.

Check: charge on the left is $+2$ ; charge on the right is $+3 - 1 = +2$ . Balanced.

Step 2: Write the reduction half-equation

Permanganate contains four oxygen atoms that must be removed. Work through the balancing steps in order. First, manganese is already balanced (one on each side). Next, add four water molecules on the right to account for the four oxygens. Then add eight $\text{H}^+$ on the left to supply the hydrogen for those water molecules. Finally, add electrons on the left to balance the charge:

\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l).

Charge check: left $= -1 + 8 - 5 = +2$ ; right $= +2$ . Balanced.

Step 3: Scale so electrons cancel

The reduction half consumes $5e^-$ and the oxidation half releases only $1e^-$ . Multiply the oxidation half-equation by $5$ so that both halves involve the same number of electrons:

5\text{Fe}^{2+}(aq) \rightarrow 5\text{Fe}^{3+}(aq) + 5e^-.

Step 4: Add the halves and cancel the electrons

Combine the two scaled half-equations. The $5e^-$ appear on both sides and cancel exactly. No other species appears on both sides, so the final equation is:

\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5\text{Fe}^{2+}(aq) \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) + 5\text{Fe}^{3+}(aq).

Final answer: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$

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 $\text{H}^+$ and $\text{H}_2\text{O}$ ), then neutralise the $\text{H}^+$ by adding the same number of $\text{OH}^-$ to both sides. The $\text{H}^+$ and $\text{OH}^-$ 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 $\text{OH}^-$ and $\text{H}_2\text{O}$ but no $\text{H}^+$ , 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.

Leaving electrons in the final equation

After scaling and adding the two halves, the electrons must cancel completely. If electrons still appear, you have not scaled correctly. The final ionic equation should never contain free electrons, and the total charge on each side must match. A second slip is balancing O and H before scaling; always finish each half-equation fully before combining.

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 questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

Vanadium(III) ions,

\text{V}^{3+}

, react with nitrate,

\text{NO}_3^-

, in acidic solution to form vanadyl ions,

\text{VO}^{2+}

, and nitric oxide gas,

\text{NO}

. Determine the balanced net ionic equation using the half-equation method.

Show worked answer →

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

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

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

Practice3 marks

When molten, the chlorate ion

\text{ClO}_3^-

disproportionates to perchlorate

\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 →

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

(b) Reduction half ( $\text{Cl}$ from $+5$ to $-1$ , gains $6e^-$ ): $\text{ClO}_3^- + 6\text{H}^+ + 6e^- \rightarrow \text{Cl}^- + 3\text{H}_2\text{O}$ . Multiply the oxidation half by $3$ , add, and cancel $6e^-$ , $6\text{H}^+$ and $3\text{H}_2\text{O}$ : $4\text{ClO}_3^- \rightarrow 3\text{ClO}_4^- + \text{Cl}^-$ . (2 marks) Check: $4\ \text{Cl}$ and $12\ \text{O}$ each side, total charge $-4$ each side.

Practise this

Sources & how we know this

Chemistry Level 4 (CHM415115) course document (2015 onwards) — TASC