Pearson Edexcel Edexcel 2021-style practice: Acidified manganate(VII) oxidises 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 balanced ionic equation. (b) Identify the oxidising agent and the reducing agent.

Balance the electrons, add the half-equations, then assign roles. (a) Multiply the iron half-equation by 5 so the electrons cancel: 5Fe^2+ → 5Fe^3+ + 5e^- (1). Add to the manganate half-equation: MnO_4^- + 8H^+ + 5Fe^2+ → Mn^2+ + 4H_2O + 5Fe^3+ (2). (b) The oxidising agent is MnO_4^- (it is reduced, gaining electrons) (1); the reducing agent is Fe^2+ (it is oxidised, losing electrons) (1).

EnglandChemistrySyllabus dot point

How do we track and balance the transfer of electrons in chemical reactions?

Oxidation numbers, oxidation and reduction as electron transfer, oxidising and reducing agents, ionic half-equations and the construction of balanced redox equations including disproportionation.

An Edexcel 9CH0 Topic 3 answer covering oxidation numbers, oxidation and reduction as electron transfer, oxidising and reducing agents, half-equations, and disproportionation.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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

Have a quick question? Jump to the Q&A page

Jump to a section

What this topic is asking
The answer
Examples in context
Try this

What this topic is asking

Edexcel Topic 3 wants you to assign oxidation numbers using the standard rules, define oxidation and reduction in terms of electron transfer, identify oxidising and reducing agents, write ionic half-equations, and combine them into balanced redox equations, including recognising disproportionation.

The answer

Oxidation numbers

To find an unknown oxidation number, set up an equation: the known values plus the unknown must equal the overall charge. For example, in $\text{SO}_4^{2-}$ , oxygen is $-2$ , so $x + 4(-2) = -2$ , giving sulfur $= +6$ .

Oxidation and reduction

Half-equations and balancing

Write a half-equation for the oxidation and one for the reduction. Balance the atoms (using $\text{H}^+$ and $\text{H}_2\text{O}$ for oxygen and hydrogen in acidic conditions), then balance the charge by adding electrons to the more positive side. Finally, scale the two half-equations so the electrons are equal, and add them so the electrons cancel. For example:

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

Combining this (which needs

5

electrons) with the iron(II) oxidation

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

requires multiplying the iron half-equation by five so the electrons cancel.

Disproportionation

Building a balanced redox equation from half-equations

Construct the overall equation for acidified dichromate(VI) oxidising iron(II), given $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$ and $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$ .

step 1: Identify the electrons exchanged

The dichromate half-equation gains $6$ electrons; the iron half-equation loses $1$ electron.

step 2: Scale to match electrons

Multiply the iron half-equation by $6$ : $6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6\text{e}^-$ .

step 3: Add 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 (2 Cr, 7 O, 14 H, 6 Fe) and charge balances: left $= -2 + 14 + 12 = +24$ ; right $= +6 + 0 + 18 = +24$ . The equation is correct.

Examples in context

Example 1. Bleach and swimming pools. Household bleach and pool chlorination chemistry rely on the disproportionation of chlorine in alkali to form the chlorate(I) ion ( $\text{ClO}^-$ ), the active bleaching and disinfecting species. Recognising that chlorine is both oxidised to $+1$ and reduced to $-1$ explains why adding chlorine to cold dilute sodium hydroxide makes bleach, a direct everyday use of the disproportionation concept from Topic 3.

Example 2. Rust and the redox of iron. When iron rusts, iron atoms (oxidation number $0$ ) are oxidised to $\text{Fe}^{3+}$ ( $+3$ ) while oxygen is reduced from $0$ to $-2$ . Assigning oxidation numbers identifies iron as the reducing agent and oxygen as the oxidising agent, and balancing the half-equations gives the overall corrosion equation. This is the same electron-transfer bookkeeping used throughout redox, applied to a costly real-world process.

Try this

Q1. Determine the oxidation number of chromium in the dichromate ion $Cr_2O_7^{2-}$ . [2 marks]

Cue. Oxygen is $-2$ , so $2x + 7(-2) = -2$ , giving $x = +6$ .

Q2. Explain why the reaction of chlorine with cold dilute sodium hydroxide is a disproportionation. [2 marks]

Cue. Chlorine starts at oxidation number $0$ and is both oxidised to $+1$ (in $ClO^-$ ) and reduced to $-1$ (in $Cl^-$ ).

Exam-style practice questions

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

Edexcel 20194 marks(a) Deduce the oxidation number of nitrogen in

\text{NO}_3^-

and in

\text{NH}_4^+

. (b) Hence state, with a reason, whether nitrogen is oxidised or reduced in the change

\text{NO}_3^- \rightarrow \text{NH}_4^+

Show worked answer →

Apply the oxidation-number rules, then compare.

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

(b) Nitrogen goes from $+5$ to $-3$ , a decrease in oxidation number (1), which is a gain of electrons, so nitrogen is reduced (1).

Edexcel 20215 marksAcidified manganate(VII) oxidises iron(II) ions. The half-equations are

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

and

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

. (a) Construct the overall balanced ionic equation. (b) Identify the oxidising agent and the reducing agent.

Show worked answer →

Balance the electrons, add the half-equations, then assign roles.

(a) Multiply the iron half-equation by $5$ so the electrons cancel: $5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5\text{e}^-$ (1). Add to the manganate half-equation: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$ (2).

(b) The oxidising agent is $\text{MnO}_4^-$ (it is reduced, gaining electrons) (1); the reducing agent is $\text{Fe}^{2+}$ (it is oxidised, losing electrons) (1).

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Chemistry (9CH0) specification — Pearson Edexcel (2015)