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Writing a full redox answer for the assessment

What a complete answer contains

The achievement criteria describe links between three things: reactions, observations and equations. A complete answer for one reaction contains all of these, in this order:

1. Identify the species that changed, and name each as oxidised or reduced.
2. Give the reason — the electrons lost or gained, or the change in oxidation number.
3. Name the oxidant and the reductant.
4. Write the two balanced half equations.
5. Combine them into the balanced overall equation.
6. Link each observation to the specific species in those equations.
Steps 1–2 alone reach Achieved. Adding 4 and linking to 6 reaches Merit. Making all six read as one connected account reaches Excellence.

A template you can reuse

Fill in the blanks for any reaction and you have the skeleton of a full answer:

The [species] was oxidised, because its oxidation number rose from [a] to [b], meaning each [species] lost [n] electrons. It therefore acted as the reductant: $\text{[oxidation half equation]}$ The [species] was reduced, because its oxidation number fell from [c] to [d], gaining [m] electrons per [species]. It therefore acted as the oxidant: $\text{[reduction half equation]}$ Multiplying by [factors] so that both involve [LCM] electrons and adding gives: $\text{[overall equation]}$ The observation that [what was seen] is explained by [species created or destroyed], which appears in the [which] half equation.

What separates the grades, in practice

Grade	The move that gets you there
Achieved	Never state a change without its reason
Merit	Every equation is followed by a sentence beginning "This is shown by the observation that…"
Excellence	The electron counts, the equations and the observations are shown to be the same fact seen three ways

Excellence is usually reached by doing one of three things: comparing two reactions, evaluating the reliability of an observation, or justifying why one species behaved as it did rather than the other way.

Common ways answers lose marks

Naming the oxidant as oxidised. The oxidant is reduced. Check every time.
Colours without species. "It went from purple to colourless" is an observation; "purple MnO₄⁻ was reduced to colourless Mn²⁺" is a link.
Molecular equations where ionic ones belong. Write MnO₄⁻, not KMnO₄, and leave spectator ions out.
Unbalanced charge. The commonest slip, and the easiest to check.
Leftover electrons in an overall equation.
Missing state symbols. Conventions are named in the criteria at every grade.

Worked ExampleA complete Excellence-level answer

Acidified potassium permanganate solution is added drop by drop to a solution of sodium sulfite. The purple colour disappears on each drop until suddenly a permanent pink colour remains. Give a full account of the chemistry, including half equations, the overall equation, and an explanation of the observations.

Step 1 — Identify the species that change

The two species that change are MnO₄⁻ and SO₃²⁻. The K⁺ and Na⁺ ions are spectators.

Step 2 — Assign oxidation numbers

In MnO₄⁻: oxygen is −2 × 4 = −8, and the ion's charge is −1, so manganese is +7. The product is Mn²⁺, where manganese is +2.

The number falls by 5, so each MnO₄⁻ gains 5 electrons. MnO₄⁻ is reduced and is the oxidant.

In SO₃²⁻: oxygen is −2 × 3 = −6, and the ion's charge is −2, so sulfur is +4. The product is SO₄²⁻, where sulfur is +6.

The number rises by 2, so each SO₃²⁻ loses 2 electrons. SO₃²⁻ is oxidised and is the reductant.

Step 3 — Write the half equations

Reduction, balanced in acid by the five-step method:

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

Oxidation:

$\text{SO}_3^{2-}(aq) + \text{H}_2\text{O}(l) \rightarrow \text{SO}_4^{2-}(aq) + 2\text{H}^+(aq) + 2e^-$

Step 4 — Combine

The lowest common multiple of 5 and 2 is 10. Multiply the first by 2 and the second by 5:

$2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$

$5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-$

Add, cancel the 10e⁻, then cancel 10H⁺ from the 16H⁺ and 5H₂O from the 8H₂O:

$2\text{MnO}_4^-(aq) + 6\text{H}^+(aq) + 5\text{SO}_3^{2-}(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 3\text{H}_2\text{O}(l) + 5\text{SO}_4^{2-}(aq)$

Charge check: left $2(-1) + 6(+1) + 5(-2) = -6$ ; right $2(+2) + 5(-2) = -6$ . Correct.

Step 5 — Explain every observation

The purple colour disappearing on each drop: the purple is MnO₄⁻. While sulfite remains, every permanganate ion added is immediately reduced to Mn²⁺, which is effectively colourless at this concentration — so the colour vanishes as fast as it is added.
The sudden permanent pink: once all the sulfite has been oxidised to sulfate, there is no reductant left. The next drop of MnO₄⁻ is not reduced, so its purple colour persists. The change is sharp because it takes only a tiny excess of the intensely coloured permanganate to be visible.
No visible change in the sulfur species: both SO₃²⁻ and SO₄²⁻ are colourless, so the oxidation half of the reaction is invisible. Everything that is seen comes from the reduction half.

Step 6 — Integrate

The three strands agree. The oxidation numbers say 5 electrons per manganese and 2 per sulfur; the half equations carry those same numbers; the overall equation's 2 : 5 ratio is the direct consequence of matching them at 10; and the colour behaviour follows because only the manganese species are coloured.

Answer: 2MnO₄⁻(aq) + 6H⁺(aq) + 5SO₃²⁻(aq) → 2Mn²⁺(aq) + 3H₂O(l) + 5SO₄²⁻(aq). MnO₄⁻ is the oxidant, reduced from Mn +7 to +2; SO₃²⁻ is the reductant, oxidised from S +4 to +6.