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Spec Mapping — OCR H432 Module 5.2.3 — Redox titrations, covering the principles and practice of quantitative redox analysis: acidified manganate(VII) titrations (MnO4−/Mn2+, E∘=+1.51 V), with the 1:5 stoichiometry against Fe2+, dilute H2SO4 as the acid of choice (HCl and HNO3 rejected on redox-interference grounds), and the self-indicating purple-to-colourless end-point at first permanent pale pink; iodine–thiosulfate titrations (I2/S2O32− at E∘=+0.54/+0.08 V), with the 1:2 I2:S2O32− stoichiometry, starch indicator added near the end-point (blue-black to colourless), and the indirect method for analysing oxidising agents (Cu2+, ClO−, peroxide) via excess KI generating I2; iodometric determination of copper(II) via 2Cu2+ + 4I− → 2CuI + I2, with the 2:1 Cu:I2 stoichiometry collapsing to 1:1 Cu:S2O32− overall; full quantitative calculations (moles of titrant → moles of analyte → scale to flask → percentage); standardisation of KMnO4 against the primary standard ammonium iron(II) sulphate (NH4)2Fe(SO4)2·6H2O; and PAG 2 (acid–base titration) cross-link for technique (burette reading, parallax, swirling, white tile under flask) (refer to the official OCR H432 specification document for exact wording).
Lessons 8 and 9 built the electrochemical framework: standard electrode potentials, IUPAC cell notation, the electrochemical series, and the Ecell∘=Ecathode∘−Eanode∘ calculation. Lesson 10 closes the module by applying this framework to redox titrations — the quantitative analytical technique by which an unknown amount of a reducing agent is determined by reaction with a known volume of a calibrated oxidising agent (or vice versa). The three workhorse systems at OCR A-Level are acidified KMnO4 (self-indicating purple→colourless), I2/Na2S2O3 (starch-indicated blue→colourless near the end-point), and the iodometric route for Cu2+ (Cu2+ liberates I2 from KI; the I2 is then back-titrated with thiosulfate). Five worked examples cover standardisation of KMnO4 against (NH4)2Fe(SO4)2·6H2O, the iron content of an iron tablet, the iron content of an iron-ore sample, and the copper content of a brass alloy. The lesson closes by linking the chemistry to PAG 2 technique standards and to the synoptic Born–Haber/Gibbs/E∘ framework that has run through all ten lessons of this course.
Key Equations: the three core redox-titration stoichiometries are MnO4−+8H++5Fe2+→Mn2++4H2O+5Fe3+(1:5 MnO4−:Fe2+) I2+2S2O32−→2I−+S4O62−(1:2 I2:S2O32−) 2Cu2++4I−→2CuI(s)+I2combined gives 1:1 Cu2+:S2O32−
By the end of this lesson you should be able to:
A redox titration is a quantitative analytical method in which one redox reagent (typically the oxidant) is delivered from a calibrated burette into a flask containing a measured volume of the other (typically the reductant). The end-point — the moment of stoichiometric equivalence — is signalled by a colour change, either of one of the reacting species itself ("self-indicator") or of an external indicator added near the end-point. The known volume of titrant at the end-point, combined with its known concentration and the stoichiometric ratio from the balanced equation, gives the moles of analyte and (via the flask-scale-up) the mass or concentration of analyte in the original sample.
The three workhorse systems at OCR A-Level are:
All three rest on the framework of Lesson 9 — half-equations, Ecell∘>0 for feasibility, and stoichiometry determined by the electron-balance constraint.
graph TD
A[Analyte in flask] --> B{Reducing agent?}
B -->|Yes -- direct titration| C[Add oxidant from burette]
C --> D[Acidified KMnO4 or K2Cr2O7]
D --> E[Colour change at end-point -- record volume]
B -->|No -- analyte is OXIDANT| F[Indirect method via I2/S2O3]
F --> G[Add excess KI -- oxidant liberates I2]
G --> H[Back-titrate I2 with Na2S2O3 -- starch indicator]
H --> E
E --> I[Moles of titrant -- apply stoichiometric ratio -- scale to flask -- compute mass/percentage]
Potassium manganate(VII) (KMnO4, "potassium permanganate") is a deeply coloured purple solution and a powerful oxidising agent in acidic conditions. Under the half-equation:
MnO4−(aq)+8H+(aq)+5e−⇌Mn2+(aq)+4H2O(l),E∘=+1.51V
it accepts 5 electrons per ion, with the colour changing from intense purple (MnO4−, λmax≈525 nm) to essentially colourless (Mn2+ is the very pale pink octahedral hexaaquaion).
Self-indicator: while reducing agent remains in the flask, every drop of MnO4− from the burette is reduced essentially instantaneously to Mn2+ and the purple colour is discharged. At the end-point — when the last molecule of reducing agent has reacted — the next drop of MnO4− remains unreacted, and the flask takes on the first permanent pale pink colour. There is no need for an external indicator: KMnO4 is its own indicator.
Acid choice — dilute H2SO4:
Stoichiometric ratio: combining MnO4− reduction (5 e−) with Fe2+ oxidation (1 e− each) gives the 1 : 5 ratio:
MnO4−(aq)+8H+(aq)+5Fe2+(aq)→Mn2+(aq)+4H2O(l)+5Fe3+(aq)
Ecell∘ = +1.51−(+0.77)=+0.74 V (Lesson 9 Worked Example 4), ΔG∘=−357 kJ mol−1 — strongly negative, so the reaction is essentially quantitative.
KMnO4 is not a primary standard — solid KMnO4 contains traces of MnO2 from slow disproportionation, so its mass cannot be relied on for direct preparation of standard solutions. Instead, the prepared KMnO4 solution is standardised by titration against a known mass of a true primary standard: ammonium iron(II) sulphate hexahydrate, (NH4)2Fe(SO4)2·6H2O, Mr=392.1. This salt is a true primary standard because it (i) can be obtained in high purity, (ii) is not hygroscopic, and (iii) is stable on storage.
Procedure: dissolve a known mass (say 3.92 g, exactly 1.00 × 10−2 mol) of (NH4)2Fe(SO4)2·6H2O in dilute H2SO4 and make up to 250 cm3 in a volumetric flask. Pipette 25.0 cm3 into a conical flask, add ~10 cm3 of dilute H2SO4, and titrate against the unknown KMnO4 solution until the first permanent pale pink colour persists for 30 s. Suppose the titre is 19.50 cm3.
Step 1: n(Fe2+) in the aliquot = 1.00×10−2×(25.0/250)=1.00×10−3 mol.
Step 2: Use 1 MnO4− : 5 Fe2+. n(MnO4−) = (1/5)×1.00×10−3=2.00×10−4 mol.
Step 3: [KMnO4] = n/V=2.00×10−4/(19.50×10−3)=1.026×10−2 mol dm−3, or 0.0103 mol dm−3 to 3 s.f.
The titrated concentration of KMnO4 is now known to about 0.5 % accuracy — much better than weighing crystalline KMnO4 would allow. Replicate titrations within 0.10 cm3 are required for the result to be reported.
A typical commercial iron tablet contains about 200 mg of "elemental iron" present as FeSO4 (the iron(II) salt is well-absorbed in the duodenum). An anaemia patient wants to verify the label claim.
Procedure: one tablet is crushed and dissolved in 100 cm3 of dilute H2SO4 in a conical flask, then made up to 250 cm3 in a volumetric flask. A 25.0 cm3 aliquot is titrated against 0.0100 mol dm−3 KMnO4, requiring 14.30 cm3 to reach the first permanent pale pink.
Step 1 — moles of MnO4− used:
n(MnO4−)=C×V=0.0100×(14.30/1000)=1.43×10−4mol
Step 2 — moles of Fe2+ in the aliquot (1 : 5 ratio):
n(Fe2+)=5×1.43×10−4=7.15×10−4mol
Step 3 — moles of Fe2+ in the whole flask (scale factor 250/25=10):
n(Fe2+,total)=7.15×10−4×10=7.15×10−3mol
Step 4 — mass of Fe in the tablet (Ar(Fe) = 55.8):
m(Fe)=7.15×10−3×55.8=0.399g=399mg
Step 5 — comparison with label. The label claim is 200 mg of "elemental iron" per tablet. The measured value is 399 mg — roughly double, suggesting either a manufacturing batch error or, more likely, that the tablet is a 400-mg dose taken in error. (Sanity check: a 400-mg dose is at the upper end of clinical iron-supplement formulations; daily limits in the UK are ~200 mg/day in divided doses to avoid GI upset.)
This is a real-world quality-control application of the KMnO4 titration.
A 0.420 g sample of hematite ore (containing Fe2O3 with silicate impurities) is reduced to Fe2+ in dilute H2SO4 using a zinc reductor column, then made up to 100 cm3 in a volumetric flask. A 25.0 cm3 aliquot is titrated against 0.0100 mol dm−3 KMnO4, requiring 18.60 cm3. Find the %Fe2O3 in the ore. Mr(Fe2O3) = 159.7.
The %Fe in the ore = (2×55.8/159.7)×70.7=49.4 % — realistic for a moderately rich hematite deposit; commercial iron ores typically run 50–65 % Fe. Always sanity-check: if the calculated mass of analyte exceeds the original sample mass, an arithmetic step has been missed.
When the analyte is itself an oxidising agent (Cu2+, MnO2, ClO−, peroxide), the direct titration approach fails because no convenient coloured reducing titrant exists in the burette. The standard workaround is the iodometric method: convert the analyte's oxidising power into a stoichiometrically known amount of I2, then back-titrate that I2 against a calibrated thiosulfate solution.
The thiosulfate half-equation (E∘≈+0.08 V):
2S2O32−(aq)⇌S4O62−(aq)+2e−
(2 thiosulfate ions are oxidised to 1 tetrathionate ion, losing 2 electrons total.)
The iodine half-equation (E∘≈+0.54 V):
I2(aq)+2e−⇌2I−(aq)
Combining (subtract the half-equations, Ecell∘=+0.54−0.08=+0.46 V — feasible):
I2(aq)+2S2O32−(aq)→2I−(aq)+S4O62−(aq)
Stoichiometric ratio: 1 I2 : 2 S2O32−.
Colour change at the end-point: the I2/I− solution is brown (the equilibrium I2 + I− ⇌ I3− shifts colour intensity). As thiosulfate is added, I2 is consumed and the brown colour fades to pale yellow. Near the end-point (when only ~5–10 % of the I2 remains), a few drops of starch solution are added. Starch wraps around the helical I3− chain (amylose helix complex with polyiodide), giving an extremely intense blue-black colour visible at low (~10−5 mol dm−3) iodine concentration. The end-point is the sudden disappearance of the blue colour — one of the sharpest colour changes in volumetric analysis.
Why starch is added late, not at the start: if starch is present from the beginning, the large excess of I2 partially decomposes the starch (slow oxidation of carbohydrate by I2) and irreversibly binds to the amylose helix. Both effects mean the blue colour does not fully discharge at the equivalence point, giving a systematically high (over-titrated) titre. Adding starch only when the solution is already pale yellow keeps the starch–iodine complex within its reversible range and gives a knife-sharp end-point to within one drop (0.05 cm3) of titrant.
The iodometric route to Cu2+ rests on the surprising disproportionation-like reaction in which Cu2+ oxidises some I− to I2 while simultaneously forming a precipitate of insoluble Cu(I) iodide:
2Cu2+(aq)+4I−(aq)→2CuI(s)+I2(aq)
Why does this work despite E∘(Cu2+/Cu+)=+0.15 V and E∘(I2/I−)=+0.54 V predicting that I2 should oxidise Cu+, not the reverse? Because the insolubility of CuI (Ksp≈10−12) shifts the effective potential of the Cu2+/CuI couple to ∼+0.86 V — a Nernst-equation shift caused by the very low [Cu+] in equilibrium with CuI(s). Under standard conditions Cu2+ + e− ⇌ Cu+ is at +0.15 V; with [Cu+] reduced to ~10−7 by precipitation of CuI, the actual potential rises by 0.0592⋅log10(107)=+0.41 V to about +0.56 V — now thermodynamically capable of oxidising I− to I2. The precipitation drives the reaction.
Stoichiometric chain: 2 Cu2+ produce 1 I2; that 1 I2 then reacts with 2 S2O32−:
2Cu2+:1I2:2S2O32−⇒1 Cu2+:1 S2O32−
The chain collapses to a clean 1:1 Cu2+:S2O32− overall, which is what makes this method so tidy in practice.
A 0.500 g sample of brass (a copper–zinc alloy used in plumbing fittings and musical instruments) is dissolved in dilute HNO3. The NOx is boiled off (it would interfere with later iodine-liberation), and excess NaHCO3 added to neutralise residual HNO3. Excess KI is added; the liberated I2 is titrated with 0.100 mol dm−3 Na2S2O3 (starch indicator near the end-point), requiring 38.50 cm3. Find the percentage of copper in the brass. Ar(Cu) = 63.5.
Step 1: n(S2O32−) = 0.100×(38.50/1000)=3.85×10−3 mol.
Step 2: Apply 1:1 Cu2+:S2O32−. n(Cu2+) = 3.85×10−3 mol.
Step 3: Mass of Cu = 3.85×10−3×63.5=0.2445 g.
Step 4: %Cu = (0.2445/0.500)×100= 48.9 %.
(Sanity check: typical "alpha" brass for plumbing is 60–70 % Cu, 30–40 % Zn; a 49 % Cu reading is closer to "Muntz metal" (60/40) brass or to a deliberately zinc-rich variant.)
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