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How do we track and balance electron transfer?

Oxidation and reduction in terms of electron transfer and oxidation number, oxidising and reducing agents, redox half equations, and redox titrations.

A focused answer to WJEC A-Level Chemistry Unit 3, covering oxidation and reduction as electron transfer and changes in oxidation number, oxidising and reducing agents, constructing redox half equations, and redox titration calculations.

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What this dot point is asking

WJEC wants you to define oxidation and reduction in terms of electrons and oxidation number, identify oxidising and reducing agents, build redox half equations, and carry out redox titration calculations.

The answer

Oxidation, reduction and agents

Building half equations

Balance atoms first, then oxygen with water, hydrogen with H+\text{H}^+, and finally charge with electrons. Combine two half equations by matching electron numbers so they cancel.

Redox titrations

Building a balanced redox equation

The reliable method is to write each half equation, balance it fully, then scale and add. Balance the atoms being oxidised or reduced first, then balance oxygen with water, then hydrogen with H+\text{H}^+, then balance the charge by adding electrons. Multiply the two half equations so the electrons match, then add them and cancel the electrons and any species appearing on both sides. This routine works for any acidified redox reaction, including the dichromate and manganate systems that recur in titrations.

Common oxidising and reducing agents

You should recognise the standard reagents. Strong oxidising agents include acidified manganate(VII) (purple to colourless, reduced to Mn2+\text{Mn}^{2+}) and acidified dichromate(VI) (orange to green, reduced to Cr3+\text{Cr}^{3+}). Common reducing agents include iron(II) ions (oxidised to iron(III)), iodide ions (oxidised to iodine) and thiosulfate (oxidised in iodine titrations). Knowing the colour change each undergoes lets you both identify the reaction and judge the end point of a titration without a separate indicator.

Examples in context

Analysing iron supplements. Manganate(VII) titration is a standard quality-control method to check the iron content of tablets, exactly the calculation above. Breathalysers. Early breathalysers used the orange-to-green reduction of dichromate by ethanol, a redox reaction whose colour change measured alcohol concentration.

Try this

Q1. State, in terms of electrons, what happens to a reducing agent. [1 mark]

  • Cue. It loses (donates) electrons and is itself oxidised.

Q2. State the colour change at the end point of an iron(II) versus manganate(VII) titration. [1 mark]

  • Cue. Colourless to the first permanent pink.

Q3. Balance the half equation Fe2+Fe3+\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}. [1 mark]

  • Cue. Fe2+Fe3++e\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-.

Q4. State the colour change when acidified dichromate(VI) acts as an oxidising agent. [1 mark]

  • Cue. Orange to green.

Q5. State the order of steps for balancing an acidified half equation. [1 mark]

  • Cue. Balance the key atoms, then oxygen with water, then hydrogen with H+\text{H}^+, then charge with electrons.

Exam-style practice questions

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

WJEC 20205 marksA 25.0 cm325.0\ \text{cm}^3 sample of an iron(II) solution was titrated with 0.0200 mol dm30.0200\ \text{mol dm}^{-3} potassium manganate(VII) and required 24.0 cm324.0\ \text{cm}^3 to reach the end point. Calculate the concentration of the iron(II) ions. The equation is MnO4+8H++5Fe2+Mn2++4H2O+5Fe3+\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}.
Show worked answer →

Moles of MnO4\text{MnO}_4^- used =0.0200×(24.0/1000)=4.80×104= 0.0200 \times (24.0/1000) = 4.80 \times 10^{-4} mol.

The ratio of MnO4\text{MnO}_4^- to Fe2+\text{Fe}^{2+} is 1:51:5, so moles of Fe2+=5×4.80×104=2.40×103\text{Fe}^{2+} = 5 \times 4.80 \times 10^{-4} = 2.40 \times 10^{-3} mol.

Concentration of Fe2+=n/V=2.40×103/(25.0/1000)=0.0960\text{Fe}^{2+} = n/V = 2.40 \times 10^{-3} / (25.0/1000) = 0.0960 mol dm-3.

Markers reward moles of manganate, the 1:51:5 ratio, and the iron concentration with units (the end point is the first permanent pink, needing no indicator).

WJEC 20183 marksDefine an oxidising agent and explain, in terms of oxidation number, why MnO4\text{MnO}_4^- acts as an oxidising agent when it is reduced to Mn2+\text{Mn}^{2+}.
Show worked answer →

An oxidising agent is a species that accepts electrons (it oxidises another species) and is itself reduced.

In MnO4\text{MnO}_4^- manganese has oxidation number +7+7; in Mn2+\text{Mn}^{2+} it is +2+2.

The oxidation number falls by 55, meaning manganese gains 55 electrons, so MnO4\text{MnO}_4^- is reduced and therefore acts as an oxidising agent.

Markers reward the definition, both oxidation numbers, and the decrease of 55 identifying it as the oxidising agent.

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