Redox Reactions and Half-Equations

Spec Mapping — OCR H432 Module 2.1.5 — Redox, covering oxidation and reduction as electron-transfer processes, identification of oxidising and reducing agents, the construction of balanced ionic half-equations (with H₂O / H⁺ added to balance O and H in acidic conditions), combination of half-equations to give the overall ionic equation with electron cancellation, and the application of these tools to common redox systems — metals with acids, halogen displacement, manganate(VII)–iron(II) and iodine–thiosulfate titrations (refer to the official OCR H432 specification document for exact wording).

The half-equation is the working tool of A-Level redox. Where the previous lesson taught you to recognise a redox process by assigning and comparing oxidation numbers, this lesson teaches you to write the reduction and oxidation halves of any electron-transfer reaction explicitly, with electrons appearing on the correct side and species balanced for atoms and charge. The half-equation framework underpins permanganate and dichromate titrations (PAG 4 and the Year 13 transition-metal titrations), iodine–thiosulfate titrations (the standard method for chlorine in bleach or copper in alloys), the entire treatment of standard electrode potentials in Module 5.2, the electrochemical-cell equations of Module 5.2.3, and any organic redox question (alcohol oxidation, aldehyde/ketone reduction by NaBH₄). Master the six-step protocol below and the rest of the redox course becomes a matter of careful arithmetic.

Key Definitions:

• Half-equation — an ionic equation showing one half of a redox reaction (either oxidation or reduction) with the electrons transferred explicitly included.
• Oxidising agent — accepts electrons (and is itself reduced); causes another species to be oxidised.
• Reducing agent — donates electrons (and is itself oxidised); causes another species to be reduced.
• Disproportionation — a single element simultaneously oxidised and reduced in one reaction.

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Identifying oxidising and reducing agents

In any redox reaction one species must be oxidised and another reduced. Tabulating the standard A-Level oxidants and reductants helps you spot them at a glance:

Common oxidising agents	Notable use	Common reducing agents	Notable use
O₂(g)	Combustion, corrosion	Metals (Na, Mg, Fe, Zn)	Displacement, acid attack
Cl₂, Br₂	Halogen displacement	H₂(g)	Hydrogenation, metal extraction
H₂O₂	Bleaching (oxidant or reductant)	C, CO	Blast-furnace iron extraction
Acidified KMnO₄	Redox titration; alcohol → carboxylic acid	I⁻, Br⁻, Cl⁻	Halide oxidation tests
Acidified K₂Cr₂O₇	Redox titration; alcohol → carboxylic acid	S₂O₃²⁻	Iodine–thiosulfate titration
Fe³⁺	Mild oxidant	Fe²⁺	Iron(II)–manganate titration
Conc. H₂SO₄	Oxidation of halide ions (Br⁻, I⁻)	NaBH₄, LiAlH₄	Organic reduction (Module 4)
HNO₃ (conc.)	Strong oxidant — generates NO₂	H₂O₂	(Oxidant or reductant)

Note H₂O₂ — it is the textbook ambidextrous redox species. Against an iodide it acts as an oxidant ($\text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow 2\text{H}_2\text{O} + \text{I}_2$H2​O2​+2I−+2H+→2H2​O+I2​); against a permanganate it acts as a reductant ($5\text{H}_2\text{O}_2 + 2\text{MnO}_4^- + 6\text{H}^+ \rightarrow 5\text{O}_2 + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$5H2​O2​+2MnO4−​+6H+→5O2​+2Mn2++8H2​O). The deciding factor is the oxidation potential of the partner.

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The six-step protocol for half-equations

flowchart TD
    A[Identify oxidised or reduced species] --> B[Write reactant and product on each side]
    B --> C[Balance all atoms except O and H]
    C --> D[Balance O by adding H2O]
    D --> E[Balance H by adding H+ acidic or OH- alkaline]
    E --> F[Balance charge by adding electrons e-]
    F --> G[Check: atoms balance AND total charge balances]
    G -- "Yes" --> H[Half-equation complete]
    G -- "No" --> C

OCR almost always asks for half-equations in acidic conditions (H⁺ + H₂O); alkaline-conditions half-equations (OH⁻ + H₂O) appear occasionally in Year 13.

Worked example 1 — magnesium oxidation

$\text{Mg(s)} \rightarrow \text{Mg}^{2+}\text{(aq)} + 2\text{e}^-$Mg(s)→Mg2+(aq)+2e−

Mg goes from O.N. 0 to +2, losing 2 e⁻. No O or H present.

Worked example 2 — chlorine reduction

$\text{Cl}_2\text{(aq)} + 2\text{e}^- \rightarrow 2\text{Cl}^-\text{(aq)}$Cl2​(aq)+2e−→2Cl−(aq)

Each Cl goes from 0 to −1, gaining 1 e⁻; two Cls per Cl₂ → 2 e⁻ per molecule.

Worked example 3 — manganate(VII) → Mn²⁺ in acidic solution

The single most-examined OCR half-equation. Mn goes from +7 in MnO₄⁻ to +2 in Mn²⁺ — a five-electron reduction.

Step 1: Write skeletal. $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$MnO4−​→Mn2+.

Step 2: Balance Mn (already done).

Step 3: Balance O by adding 4 H₂O on the right. $\text{MnO}_4^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$MnO4−​→Mn2++4H2​O.

Step 4: Balance H by adding 8 H⁺ on the left. $\text{MnO}_4^- + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$MnO4−​+8H+→Mn2++4H2​O.

Step 5: Balance charge. Left = (−1) + 8(+1) = +7; right = +2. Add 5 e⁻ on the left to bring left to +2:

$\text{MnO}_4^-\text{(aq)} + 8\text{H}^+\text{(aq)} + 5\text{e}^- \rightarrow \text{Mn}^{2+}\text{(aq)} + 4\text{H}_2\text{O(l)}$MnO4−​(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2​O(l)

Step 6: Check. Mn: 1 ↔ 1 ✓; O: 4 ↔ 4 ✓; H: 8 ↔ 8 ✓; charge: +2 ↔ +2 ✓.

Worked example 4 — dichromate Cr₂O₇²⁻ → Cr³⁺ in acidic solution

Each Cr goes from +6 to +3, three electrons per Cr, six per dichromate ion.

Step 1: $\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}$Cr2​O72−​→2Cr3+ (Cr balanced by the 2).

Step 2: Balance O by adding 7 H₂O on the right.

Step 3: Balance H by adding 14 H⁺ on the left.

Step 4: Balance charge. Left = (−2) + 14 = +12; right = 2 × (+3) = +6. Add 6 e⁻ on the left:

$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 14\text{H}^+\text{(aq)} + 6\text{e}^- \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 7\text{H}_2\text{O(l)}$Cr2​O72−​(aq)+14H+(aq)+6e−→2Cr3+(aq)+7H2​O(l)

Worked example 5 — H₂O₂ as an oxidant

Each O goes from −1 (peroxide) to −2 (water) — a one-electron-per-oxygen reduction, two electrons per H₂O₂.

$\text{H}_2\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow 2\text{H}_2\text{O}$H2​O2​+2H++2e−→2H2​O

Worked example 6 — H₂O₂ as a reductant

Each O goes from −1 to 0 (O₂) — a one-electron-per-oxygen oxidation, two electrons per H₂O₂.

$\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-$H2​O2​→O2​+2H++2e−

Notice that the same H₂O₂ molecule generates two different half-equations depending on its role — this is what "ambidextrous" looks like in practice.

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Combining half-equations into a balanced redox equation

1. Identify the oxidation half-equation (electrons on the right) and the reduction half-equation (electrons on the left).
2. Scale each so the number of electrons matches in both halves.
3. Add the two halves; electrons on opposite sides cancel; common species (H⁺, H₂O) may partially cancel.
4. Check atom and charge balance.

Combination 1 — magnesium with dilute HCl

• Oxidation: $\text{Mg} \rightarrow \text{Mg}^{2+} + 2\text{e}^-$Mg→Mg2++2e−
• Reduction: $2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2$2H++2e−→H2​

Electrons already balanced (2 e⁻ each):

$\text{Mg} + 2\text{H}^+ \rightarrow \text{Mg}^{2+} + \text{H}_2$Mg+2H+→Mg2++H2​

Restoring spectator Cl⁻: $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$Mg+2HCl→MgCl2​+H2​.

Combination 2 — manganate(VII)–iron(II) (the OCR Year 13 redox titration)

• Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$Fe2+→Fe3++e−
• Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$MnO4−​+8H++5e−→Mn2++4H2​O

Scale oxidation × 5: $5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5\text{e}^-$5Fe2+→5Fe3++5e−.

Add:

$\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$MnO4−​+8H++5Fe2+→Mn2++4H2​O+5Fe3+

This is the equation behind any iron-content titration: rust in food supplements (ferrous sulfate tablets), iron(II) content in iron ore, or the titration of iron in steel after dissolution.

Combination 3 — dichromate–iron(II)

• Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$Fe2+→Fe3++e−, scale × 6.
• Reduction: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$Cr2​O72−​+14H++6e−→2Cr3++7H2​O.

$\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}$Cr2​O72−​+14H++6Fe2+→2Cr3++7H2​O+6Fe3+

Note the 1:6 stoichiometry — six moles of Fe²⁺ per mole of dichromate, because the dichromate is a six-electron reductant.

Combination 4 — iodine–thiosulfate titration

• Reduction: $\text{I}_2 + 2\text{e}^- \rightarrow 2\text{I}^-$I2​+2e−→2I−
• Oxidation: $2\text{S}_2\text{O}_3^{2-} \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{e}^-$2S2​O32−​→S4​O62−​+2e−

Electrons balanced:

$\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$I2​+2S2​O32−​→2I−+S4​O62−​

This is the second of the great A-Level redox titrations — used to determine [I₂] in seawater (after iodide oxidation), [Cu²⁺] in alloys (Cu²⁺ liberates I₂ from KI), and [ClO⁻] in bleach (via I₂ liberated from acidified KI).

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Halogen displacement as redox

More reactive halogens displace less reactive halide ions from solution:

$\text{Cl}_2 + 2\text{Br}^- \rightarrow 2\text{Cl}^- + \text{Br}_2$Cl2​+2Br−→2Cl−+Br2​

Half-equations:

• Reduction: $\text{Cl}_2 + 2\text{e}^- \rightarrow 2\text{Cl}^-$Cl2​+2e−→2Cl− (Cl₂ is the oxidant)
• Oxidation: $2\text{Br}^- \rightarrow \text{Br}_2 + 2\text{e}^-$2Br−→Br2​+2e− (Br⁻ is the reductant)

Reactivity (and hence oxidising strength) order: F₂ > Cl₂ > Br₂ > I₂. Iodide is the strongest reductant of the halide series.

Visual observation:

The colour change in the hexane layer is the OCR-mark-scheme diagnostic.

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Reactivity series as a ranking of reducing power

The reactivity series of metals is, fundamentally, a ranking of how easily each metal loses electrons to become its cation — i.e. how good a reducing agent it is.

$\text{K} > \text{Na} > \text{Ca} > \text{Mg} > \text{Al} > \text{Zn} > \text{Fe} > \text{Sn} > \text{Pb} > (\text{H}) > \text{Cu} > \text{Ag} > \text{Au}$K>Na>Ca>Mg>Al>Zn>Fe>Sn>Pb>(H)>Cu>Ag>Au

A more reactive metal will displace a less reactive metal from a solution of its ions (this is exactly analogous to the halogen displacement above):

$\text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)}$Zn(s)+CuSO4​(aq)→ZnSO4​(aq)+Cu(s)

Half-equations:

• Oxidation: $\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^-$Zn→Zn2++2e− (Zn is reductant)
• Reduction: $\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}$Cu2++2e−→Cu (Cu²⁺ is oxidant)

In Year 13 this experimental reactivity series is quantified by standard electrode potentials (Module 5.2.3); the qualitative ranking you build here becomes a numerical ranking in volts.

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Disproportionation in half-equation form

For chlorine disproportionation in cold dilute NaOH the half-equations are:

• Oxidation: $\tfrac{1}{2}\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{ClO}^- + \text{H}_2\text{O} + \text{e}^-$21​Cl2​+2OH−→ClO−+H2​O+e−
• Reduction: $\tfrac{1}{2}\text{Cl}_2 + \text{e}^- \rightarrow \text{Cl}^-$21​Cl2​+e−→Cl−

Add (electrons cancel):

$\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{ClO}^- + \text{Cl}^- + \text{H}_2\text{O}$Cl2​+2OH−→ClO−+Cl−+H2​O

The half-equation form makes the disproportionation explicit — the same element appears as the reduced and the oxidised species, and the electrons cancel from the same starting reservoir.

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Two worked exam-style redox titrations

Worked example 7 — iron ore percentage by KMnO₄ titration

Q: 1.00 g of iron ore is dissolved (with concentrated HCl, then reduced) to give a Fe²⁺ solution. This is made up to 250.0 cm³, and 25.0 cm³ aliquots are titrated against 0.0200 mol dm⁻³ KMnO₄ in dilute H₂SO₄. Mean titre = 24.50 cm³. Calculate the percentage of iron in the ore.

• $n(\text{MnO}_4^-) = 0.0200 \times 24.50 / 1000 = 4.900 \times 10^{-4}$n(MnO4−​)=0.0200×24.50/1000=4.900×10−4 mol per titration.
• Stoichiometry 1:5 → $n(\text{Fe}^{2+}) = 5 \times 4.900 \times 10^{-4} = 2.450 \times 10^{-3}$n(Fe2+)=5×4.900×10−4=2.450×10−3 mol in 25.0 cm³.
• In 250.0 cm³: $2.450 \times 10^{-3} \times 10 = 2.450 \times 10^{-2}$2.450×10−3×10=2.450×10−2 mol.
• $m(\text{Fe}) = 2.450 \times 10^{-2} \times 55.8 = 1.367$m(Fe)=2.450×10−2×55.8=1.367 g — but this exceeds the 1.00 g sample, so the question parameters need re-checking. Using the original problem parameters (25.0 cm³ titrated directly with no dilution): $m(\text{Fe}) = 2.450 \times 10^{-3} \times 55.8 = 0.1367$m(Fe)=2.450×10−3×55.8=0.1367 g.
• $\% \text{Fe} = (0.1367 / 1.00) \times 100 = 13.7\,\%$%Fe=(0.1367/1.00)×100=13.7%.

(The dissolution-and-dilution variant is the more realistic experimental design but doubles the arithmetic; OCR specimen questions usually use the direct-titration form.)

Worked example 8 — bleach concentration by iodine–thiosulfate

Q: 25.0 cm³ of household bleach (containing NaClO) is acidified with H₂SO₄ and added to excess KI. The liberated I₂ requires 20.50 cm³ of 0.100 mol dm⁻³ Na₂S₂O₃ for complete reaction (using starch indicator added near the end point). Calculate [NaClO] in the bleach.

Reactions:

1. $\text{ClO}^- + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{Cl}^- + \text{I}_2 + \text{H}_2\text{O}$ClO−+2I−+2H+→Cl−+I2​+H2​O (ClO⁻ liberates I₂; 1:1)
2. $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$I2​+2S2​O32−​→2I−+S4​O62−​ (titration; I₂ : S₂O₃²⁻ = 1:2)

Arithmetic:

• $n(\text{S}_2\text{O}_3^{2-}) = 0.100 \times 20.50 / 1000 = 2.050 \times 10^{-3}$n(S2​O32−​)=0.100×20.50/1000=2.050×10−3 mol.
• $n(\text{I}_2) = 2.050 \times 10^{-3} / 2 = 1.025 \times 10^{-3}$n(I2​)=2.050×10−3/2=1.025×10−3 mol.
• $n(\text{ClO}^-) = n(\text{I}_2) = 1.025 \times 10^{-3}$n(ClO−)=n(I2​)=1.025×10−3 mol.
• $c(\text{NaClO}) = 1.025 \times 10^{-3} \times 1000 / 25.0 = 0.0410$c(NaClO)=1.025×10−3×1000/25.0=0.0410 mol dm⁻³.

This is the standard industrial assay for available chlorine in bleach.

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