Redox titrations

An iron tablet claims 65 mg of iron. How do you check? Dissolve it in dilute sulfuric acid so that all the iron is Fe²⁺, and titrate with purple potassium permanganate: the purple vanishes drop after drop until one extra drop leaves the solution faintly pink — the endpoint. A redox titration has no pH meter to watch; its “curve” is the solution’s electrode potential, fixed point by point by the Nernst equation.

Each drop of titrant reacts completely, and the solution has one potential that satisfies the Nernst equations of both couples:

• Before equivalence both Fe²⁺ and Fe³⁺ are plentiful; use the iron couple, $E = E°'_\ce{Fe} - 0.05916\lg([\ce{Fe^2+}]/[\ce{Fe^3+}])$. Half-way, $E = E°'_\ce{Fe}$.
• After equivalence the excess titrant and its reduced form are both present; use the titrant couple.
• At equivalence add the two equations. With equal electron numbers and no H⁺ involved, $E_\text{eq}$ is the mean of the two formal potentials.

With different electron numbers, or H⁺ in the half-reaction, weight by electrons. For KMnO₄ titrating Fe²⁺ (MnO₄⁻ takes 5 electrons and consumes 8 H⁺):

Indicators. A redox indicator is itself a couple that changes colour over $E°_\text{In} \pm 0.05916/n$ (from 1 part in 11 oxidised to 10 parts in 11), so choose one whose range sits inside the jump. The jump is the change in potential between 99.9 % and 100.1 % titrated — the ±0.1 % a titration is expected to meet. Ferroin (1.147 V, n = 1, red → pale blue) changes between 1.088 and 1.206 V and serves for Ce⁴⁺; diphenylamine sulfonate (0.85 V, colourless → red-violet) serves for dichromate, and because its colour step transfers two electrons and two protons at once (n = 2), its range is half as wide: 0.85 ± 0.030 V. KMnO₄ is its own indicator: one excess drop of 0.02 M solution turns 100 mL faintly pink.

Why phosphoric acid. In 1 M H₂SO₄, E°′(Fe) = 0.68 V, so at 99.9 % titrated $E = 0.68 + 0.05916\lg 999 = 0.857$ V — the foot of the jump. Yet diphenylamine sulfonate starts to change at 0.82 V, when about 0.4 % of the Fe²⁺ is still unoxidised. Phosphate complexes Fe³⁺ and lowers E°′(Fe) to about 0.61 V, which moves the foot of the jump down to 0.79 V and puts the whole colour change (0.82–0.88 V) inside it: when the indicator starts to change, 99.97 % of the iron is oxidised — and the colourless iron(III) phosphate complex removes the yellow of Fe³⁺ as well.

The I₃⁻/I⁻ couple has E° = 0.536 V, a mild go-between. In iodometry the oxidising analyte reacts with excess KI to make an exactly equivalent amount of I₃⁻, which is then titrated with standard Na₂S₂O₃: $\ce{I3-} + \ce{2S2O3^2-} \mathrel{\ce{->}} \ce{3I-} + \ce{S4O6^2-}$. Hypochlorite in bleach, copper(II) ($\ce{2Cu^2+} + \ce{4I-} \mathrel{\ce{->}} \ce{2CuI(s)} + \ce{I2}$) and dissolved oxygen in water (the Winkler method) are all determined this way. Sodium thiosulfate itself is not stable and is standardised with potassium iodate (1 mol IO₃⁻ gives 3 mol I₂, equivalent to 6 mol S₂O₃²⁻).

Starch forms a deep-blue complex with I₃⁻ that absorbs far more strongly than the yellow-brown of I₃⁻ alone, making tiny amounts of iodine visible. But add the starch only when the solution has faded to pale yellow: at high iodine levels much of it binds tightly to the starch and is released slowly, dragging the endpoint.