Redox titrations: iron, manganate(VII), iodine and thiosulfate

What you’ll learn

• Apply redox mole ratios to analytical titration data.
• Explain endpoint evidence, aliquot scaling and systematic dilution errors.

Understand the idea

Core practical 13 includes iron(II) with manganate(VII), and iodine with thiosulfate. In the acidified iron titration, one manganate(VII) ion accepts five electrons and reacts with five iron(II) ions. The purple titrant provides its own endpoint evidence: a slight persistent excess remains coloured after the iron(II) has reacted. The specified acid conditions are essential because changing the medium can change the manganese reduction product and the stoichiometry.

In an iodine–thiosulfate titration, one iodine molecule reacts with two thiosulfate ions. Starch supplies a sensitive indicator near the endpoint, where disappearance of the iodine–starch colour signals iodine consumption. For indirect iodine determinations, first connect the original analyte to the iodine produced, and only then apply the iodine-to-thiosulfate ratio. Keep track of any aliquot and dilution factors; the titration usually analyses only part of the original sample.

Use concordant accurate titres and distinguish endpoint uncertainty from systematic concentration error. For example, residual water that dilutes a burette titrant makes a larger volume necessary; using the nominal, higher concentration then overestimates the analyte amount. In supervised work, acidified oxidants and iodine solutions require the stated splash, ventilation and waste precautions. The analytical argument should use the supplied reaction and observations, without guessing an alternative preparation.

Key formulae

$$\mathrm{MnO_4^-}+8\mathrm{H^+}+5\mathrm{Fe^{2+}}\rightarrow\mathrm{Mn^{2+}}+4\mathrm{H_2O}+5\mathrm{Fe^{3+}}$$

$$\mathrm{I_2}+2\mathrm{S_2O_3^{2-}}\rightarrow2\mathrm{I^-}+\mathrm{S_4O_6^{2-}}$$

$$n(\mathrm{Fe^{2+}})=5n(\mathrm{MnO_4^-})$$

$$n(\mathrm{I_2})=\frac12n(\mathrm{S_2O_3^{2-}})$$

Worked example

A $25.00\ \mathrm{cm^3}$ iron(II) aliquot requires $18.60\ \mathrm{cm^3}$ of $0.0200\ \mathrm{mol\,dm^{-3}}$ manganate(VII) in the stated acidic conditions. Calculate the iron(II) concentration.

1. Manganate(VII) amount is $0.0200\times0.01860=3.72\times10^{-4}\ \mathrm{mol}$.
2. The equation requires five times this amount of iron(II): $n(\mathrm{Fe^{2+}})=1.86\times10^{-3}\ \mathrm{mol}$.
3. Divide by the aliquot volume: $c=1.86\times10^{-3}/0.02500=0.0744\ \mathrm{mol\,dm^{-3}}$.

Answer: $0.0744\ \mathrm{mol\,dm^{-3}}$.