Deduce the oxidation number of sulfur in H_2SO_3. [1 mark]

In Zn + CuSO_4 → ZnSO_4 + Cu, identify the reducing agent and justify your answer. [2 marks]

OCR OCR 2019-style practice: (a) Deduce the oxidation number of manganese in MnO_4^- and of nitrogen in NO_3^-. (b) In the reaction Cl_2 + 2KBr → 2KCl + Br_2, identify the species oxidised and the species reduced, in terms of oxidation number change.

(a) In MnO_4^-, oxygen is -2, so x + 4(-2) = -1, giving Mn = +7 (1). In NO_3^-, x + 3(-2) = -1, giving N = +5 (1). (b) Bromine is oxidised: Br goes from -1 in KBr to 0 in Br_2 (1). Chlorine is reduced: Cl goes from 0 in Cl_2 to -1 in KCl (1). Markers reward both oxidation numbers and identifying oxidation or reduction by the direction of change.

OCR OCR 2021-style practice: Acidified manganate(VII) ions react with iron(II) ions. The half-equations are MnO_4^- + 8H^+ + 5e^- → Mn^2+ + 4H_2O and Fe^2+ → Fe^3+ + e^-. (a) Construct the overall ionic equation. (b) State which species is the oxidising agent.

(a) Multiply the iron half-equation by 5 to balance electrons, then add: MnO_4^- + 8H^+ + 5Fe^2+ → Mn^2+ + 4H_2O + 5Fe^3+ (2: balancing electrons, then combining). (b) The manganate(VII) ion MnO_4^- is the oxidising agent because it gains electrons and is reduced (it oxidises the iron(II)) (2). Markers reward balancing the 5e^- correctly and identifying the oxidising agent with reasoning.

EnglandChemistrySyllabus dot point

How do oxidation numbers let us track electron transfer in chemical reactions?

Oxidation numbers and the rules for assigning them, oxidation and reduction as loss and gain of electrons, oxidising and reducing agents, and the construction of half-equations and overall redox equations.

An OCR H432 module 2 answer covering oxidation number rules, oxidation and reduction as electron transfer, oxidising and reducing agents, and building half-equations and balanced redox equations.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Oxidation numbers
Oxidation and reduction
Oxidising and reducing agents
Half-equations and overall equations
Examples in context
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What this topic is asking

OCR specification point 2.1.5 wants you to assign oxidation numbers using the standard rules, define oxidation and reduction as loss and gain of electrons, identify oxidising and reducing agents, and combine half-equations into balanced overall redox equations.

Oxidation numbers

For example, in the sulfate ion $\text{SO}_4^{2-}$ , oxygen is $-2$ , so $x + 4(-2) = -2$ gives sulfur as $+6$ . Roman numerals in names, such as iron(III), give the oxidation number directly.

Oxidation and reduction

A useful check is that oxidation and reduction always happen together: electrons lost by one species are gained by another.

Oxidising and reducing agents

Half-equations and overall equations

A half-equation shows either the oxidation or the reduction part, including the electrons. To combine two half-equations, balance the number of electrons in each, then add and cancel.

Combining half-equations for a redox reaction

Acidified dichromate(VI) oxidises iron(II). The half-equations are $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ and $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ . Build the overall equation.

step 1: Balance the electrons

The dichromate half gains $6e^-$ ; the iron half loses $1e^-$ . Multiply the iron half-equation by $6$ .

step 2: Write the scaled iron half-equation

$6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-$ .

step 3: Add the two half-equations and cancel electrons

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}$ .

step 4: Check the balance

Atoms balance and the total charge is $+24$ on each side, confirming the overall redox equation.

Examples in context

Example 1. Iron(II) tablets analysis. The iron(II) content of dietary supplements is found by titrating a dissolved tablet against acidified potassium manganate(VII). The purple colour vanishes as $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ , and the first persistent pink tinge marks the end point, a self-indicating redox titration built on the half-equations above.

Example 2. Disproportionation of chlorine in water. When chlorine is added to water, the same element is both oxidised and reduced: $\text{Cl}_2 + \text{H}_2\text{O} \rightarrow \text{HClO} + \text{HCl}$ , with chlorine going from $0$ to $+1$ and to $-1$ . Tracking oxidation numbers identifies this as disproportionation, the chemistry behind chlorinating drinking water.

Try this

Q1. Deduce the oxidation number of sulfur in $\text{H}_2\text{SO}_3$ . [1 mark]

Cue. Hydrogen $+1$ , oxygen $-2$ : $2(+1) + x + 3(-2) = 0$ , so $x = +4$ .

Q2. In $\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$ , identify the reducing agent and justify your answer. [2 marks]

Cue. Zinc is the reducing agent; it loses electrons (oxidation number $0$ to $+2$ ) and reduces the copper(II) ion.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marks(a) Deduce the oxidation number of manganese in

\text{MnO}_4^-

and of nitrogen in

\text{NO}_3^-

. (b) In the reaction

\text{Cl}_2 + 2\text{KBr} \rightarrow 2\text{KCl} + \text{Br}_2

, identify the species oxidised and the species reduced, in terms of oxidation number change.

Show worked answer →

(a) In $\text{MnO}_4^-$ , oxygen is $-2$ , so $x + 4(-2) = -1$ , giving Mn $= +7$ (1). In $\text{NO}_3^-$ , $x + 3(-2) = -1$ , giving N $= +5$ (1).

(b) Bromine is oxidised: $\text{Br}$ goes from $-1$ in $\text{KBr}$ to $0$ in $\text{Br}_2$ (1). Chlorine is reduced: $\text{Cl}$ goes from $0$ in $\text{Cl}_2$ to $-1$ in $\text{KCl}$ (1).

Markers reward both oxidation numbers and identifying oxidation or reduction by the direction of change.

OCR 20214 marksAcidified manganate(VII) ions react with iron(II) ions. The half-equations are

\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}

and

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-

. (a) Construct the overall ionic equation. (b) State which species is the oxidising agent.

Show worked answer →

(a) Multiply the iron half-equation by $5$ to balance electrons, then add: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$ (2: balancing electrons, then combining).

(b) The manganate(VII) ion $\text{MnO}_4^-$ is the oxidising agent because it gains electrons and is reduced (it oxidises the iron(II)) (2).

Markers reward balancing the $5e^-$ correctly and identifying the oxidising agent with reasoning.

Related dot points

Sources & how we know this

OCR A-Level Chemistry A (H432) specification — OCR (2015)