Part 1: Electron transfer and redox titrations

Balance unfamiliar redox changes, connect the electron ratio to titration stoichiometry and explain each important procedural choice.

Track both electron changes

Oxidation is electron loss and an increase in oxidation number; reduction is electron gain and a decrease. An oxidising agent accepts electrons and is reduced. A reducing agent donates electrons and is oxidised. Naming the agent without stating what happens to it often conceals a reversed argument.

To balance a half-equation in acidic solution, balance the element changing oxidation state, add water to balance oxygen, H⁺ to balance hydrogen, then electrons to balance charge. Multiply half-equations until electron loss equals electron gain, add them and cancel electrons. Verify atoms and total charge in the final equation.

For permanganate in acid, manganese falls from +7 to +2 and gains five electrons. Iron(II) loses one electron per ion. Therefore one permanganate reacts with five Fe²⁺ ions, not one. The ionic equation supplies the mole ratio for the calculation.

Why acid and the endpoint matter

Pipette a measured Fe²⁺ aliquot into a conical flask and add sufficient dilute sulfuric acid. Titrate with standard permanganate, swirling and adding dropwise near the endpoint. In acid, added purple MnO₄⁻ is reduced until Fe²⁺ is exhausted; the first slight persistent pink from excess permanganate signals the endpoint. No separate indicator is needed.

Sulfuric acid supplies the H⁺ required by the half-equation. Inadequate acidity can give a different manganese product, such as brown MnO₂, and invalidate the intended ratio. Hydrochloric acid can introduce chloride that is oxidised under these conditions, while nitric acid can oxidise the analyte. Choose reagents for their chemistry, not simply because they are all acids.

Prepare and analyse Fe²⁺ solutions promptly where air oxidation is relevant. Repeat concordant titres and calculate using the mean of suitable results. An unwashed burette containing water dilutes titrant and can bias the inferred amount if the calculation still uses the stated stock concentration.

Worked example: ratio, dilution and molar mass

An original illustrative 25.0 cm³ aliquot requires 18.60 cm³ of 0.0200 mol dm⁻³ MnO₄⁻. The titrant amount is 0.01860 × 0.0200 = 3.72 × 10⁻⁴ mol. Fe²⁺ in that aliquot is five times this, 1.86 × 10⁻³ mol.

If the original sample was made up to 250.0 cm³, multiply by 250.0/25.0 = 10 to obtain 0.0186 mol Fe²⁺ in the whole flask. With Fe molar mass 55.8 g mol⁻¹, that corresponds to 1.04 g of iron. If asked for a hydrated salt instead, multiply by that salt’s molar mass rather than 55.8. Label every amount so the tenfold dilution factor and the fivefold reaction factor remain distinct.

Iodometry uses two linked reactions

Iodine is reduced to iodide by thiosulfate, while thiosulfate forms tetrathionate. One mole I₂ consumes two moles S₂O₃²⁻. Add starch when the iodine solution is pale straw-coloured near the endpoint, then continue until the blue-black colour disappears. Adding starch at a high iodine concentration can retain iodine strongly and make the endpoint sluggish.

An oxidising analyte can first liberate iodine from excess iodide. For copper(II), 2Cu²⁺ + 4I⁻ → 2CuI(s) + I₂. Combining this with the thiosulfate reaction shows n(Cu²⁺) = n(S₂O₃²⁻). The equality emerges from two ratios; do not assume a universal 1:1 redox relationship.

For an illustrative titre of 12.40 cm³ of 0.0500 mol dm⁻³ thiosulfate, n(S₂O₃²⁻) = 6.20 × 10⁻⁴ mol, n(I₂) = 3.10 × 10⁻⁴ mol and n(Cu²⁺) = 6.20 × 10⁻⁴ mol. Iodine loss before titration would reduce the titre and underestimate copper in this model.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources