14.18-14.19 - Redox Titrations

What A Redox Titration Measures

A redox titration uses a measured volume of an oxidising agent or reducing agent to find the amount of another reacting species. The colour change or indicator tells you when the redox reaction has just finished, but the calculation comes from the balanced redox equation.

The route is the same disciplined route as any titration, with one extra redox decision:

1. Identify the redox equation or construct it from half-equations.

2. Identify the known concentration and measured titre.

3. Convert $\mathrm{cm^{3}}$ to $\mathrm{dm^{3}}$.

4. Calculate moles of the known solution.

5. Use the redox mole ratio from the balanced equation.

6. Link to the aliquot volume, dilution or excess reagent in the question.

7. Give concentration, mass or percentage with units and sensible significant figures.

The key step is the mole ratio. In an acid-base titration the ratio might be found from $\ce{H+}$ and $\ce{OH-}$. In a redox titration it comes from electron transfer. If one ion accepts five electrons, it must react with enough reducing agent to supply five electrons.

Common-error contrast. Do not assume that endpoint means equal moles. Endpoint means the reacting species have met in the stoichiometric ratio from the balanced equation. For $\ce{MnO4-}$ with $\ce{Fe^{2+}}$, the ratio is $1:5$, not $1:1$.

Manganate(VII) And Iron(II)

Potassium manganate(VII), $\ce{KMnO4}$, is a common oxidising agent in redox titrations. In acidic solution, purple $\ce{MnO4-}$ ions are reduced to very pale pink, almost colourless, $\ce{Mn^{2+}}$ ions:

Iron(II) ions are oxidised to iron(III) ions:

The manganate(VII) half-equation needs five electrons, and each $\ce{Fe^{2+}}$ ion supplies one electron. Multiply the iron half-equation by 5 and combine:

So the calculation ratio is:

1 mol $\ce{MnO4-}$ reacts with 5 mol $\ce{Fe^{2+}}$

The solution must be acidified, usually with dilute sulfuric acid. Without enough $\ce{H+}$, $\ce{MnO4-}$ is not reliably reduced to $\ce{Mn^{2+}}$, so the equation and ratio above no longer describe the titration. Hydrochloric acid is avoided because chloride ions can be oxidised by manganate(VII); nitric acid is avoided because it is itself an oxidising acid.

Endpoint logic. Manganate(VII) is self-indicating. While $\ce{Fe^{2+}}$ remains in the flask, each drop of purple $\ce{MnO4-}$ is decolourised as it is reduced. The endpoint is the first permanent faint pink colour, showing that a tiny excess of $\ce{MnO4-}$ remains.

Worked example: direct $\ce{Fe^{2+}}$ determination.

$25.00 \mathrm{cm^{3}}$ of an acidified $\ce{Fe^{2+}}$ solution is titrated with $0.02000 \mathrm{mol\,dm^{-3}} \ce{KMnO4(aq)}$. The mean titre is $23.60 \mathrm{cm^{3}}$. Calculate the concentration of $\ce{Fe^{2+}}$ in the original solution.

First calculate moles of $\ce{MnO4-}$ added:

Use the redox ratio, 1 mol $\ce{MnO4-}$ : 5 mol $\ce{Fe^2+}$:

This amount was in $25.00\,\mathrm{cm^{3}} = 0.02500\,\mathrm{dm^{3}}$:

$c(\ce{Fe^2+}) = 2.360\times 10^{-3} / 0.02500 = 0.0944\,\mathrm{mol\,dm^{-3}}$ (3 s.f.)

The chemical sense check is that the iron(II) concentration is larger than the manganate(VII) concentration because five moles of $\ce{Fe^{2+}}$ are needed for each mole of $\ce{MnO4-}$.

Iodine And Thiosulfate

Iodine-thiosulfate titrations use a different redox pair. Iodine is reduced to iodide ions:

$\ce{I2 + 2e- -> 2I-}$

Thiosulfate ions, $\ce{S2O3^{2-}}$, are oxidised to tetrathionate ions, $\ce{S4O6^{2-}}$:

$\ce{2S2O3^{2-} -> S4O6^{2-} + 2e-}$

Combining the half-equations gives:

$\ce{I2 + 2S2O3^{2-} -> 2I- + S4O6^{2-}}$

So the calculation ratio is:

1 mol $\ce{I2}$ reacts with 2 mol $\ce{S2O3^{2-}}$

Iodine is brown or yellow-brown in solution. Starch is used as the indicator because it forms a blue-black colour with iodine. Add starch near the endpoint, when the iodine colour has faded to pale straw, then continue adding thiosulfate dropwise until the blue-black colour just disappears.

Worked example: iodine solution.

$25.00 \mathrm{cm^{3}}$ of iodine solution is titrated with $0.1000 \mathrm{mol\,dm^{-3}} \ce{Na2S2O3(aq)}$. The mean titre is $20.40 \mathrm{cm^{3}}$. Calculate the concentration of iodine.

Calculate moles of thiosulfate:

Use the ratio 1 mol $\ce{I2}$ : 2 mol $\ce{S2O3^2-}$:

This iodine was in $25.00\,\mathrm{cm^{3}}$:

Core Practical 11 Method Decisions

Reliable redox titration requires an accurately known aliquot, a sharp endpoint and repeatable delivery of titrant.