5.2.3(d)-(e) - Redox titrations

What redox titrations measure

In a redox titration, one reacting species is oxidised and another is reduced. A measured volume of one solution is placed in a conical flask, and a standard solution is delivered from a burette until the reaction is just complete.

The equivalence point is the chemical point where the reacting amounts are exactly in the mole ratio from the balanced redox equation. The endpoint is the observed colour change used to judge that the equivalence point has just been reached. These are close, but not identical, so practical technique matters.

The two named OCR systems in this lesson are:

• $\ce{Fe^2+}$ with acidified $\ce{MnO4-}$, where $\ce{Fe^2+}$ is oxidised to $\ce{Fe^3+}$ and $\ce{MnO4-}$ is reduced to $\ce{Mn^2+}$.
• $\ce{I2}$ with $\ce{S2O3^2-}$, where iodine is reduced to iodide and thiosulfate is oxidised to tetrathionate.

[DIAGRAM: asset_name: Lesson 5.2.3d-e: Redox titrations - diagram 01; asset_slug: 05_02_03b_redox_titrations__diagram_01; recommended_method: drawn_chem; description: Endpoint signal flow for Fe2+/MnO4- and I2/S2O3^2- redox titrations, including colour sequence, indicator timing and core ionic equations.]

Manganate(VII) with iron(II)

Potassium manganate(VII), $\ce{KMnO4(aq)}$, contains $\ce{MnO4-}$ ions. In acidic solution, $\ce{MnO4-}$ is a strong oxidising agent and is reduced to $\ce{Mn^2+}$.

Iron(II) ions are reducing agents:

Multiplying the iron(II) half-equation by 5 and adding gives:

The practical sequence is usually:

1. Pipette a known volume of $\ce{Fe^2+}$ solution into a conical flask.
2. Add excess dilute sulfuric acid.
3. Fill the burette with standard $\ce{KMnO4(aq)}$.
4. Titrate while swirling until the first permanent pale pink colour remains.

No separate indicator is needed because manganate(VII) is self-indicating. While $\ce{Fe^2+}$ is present, each drop of purple $\ce{MnO4-}$ is reduced to very pale $\ce{Mn^2+}$ and the colour disappears. Once all $\ce{Fe^2+}$ has reacted, the next tiny excess of $\ce{MnO4-}$ leaves a permanent pale pink colour.

The acid must provide enough $\ce{H+}$ for $\ce{MnO4-}$ to reduce to $\ce{Mn^2+}$. If the mixture is not acidic enough, brown $\ce{MnO2}$ can form and the stoichiometry is no longer the clean 1:5 ratio. Dilute sulfuric acid is used because it provides $\ce{H+}$ without adding an ion that is easily oxidised by manganate(VII) under these conditions.

Iodine with thiosulfate

Sodium thiosulfate, $\ce{Na2S2O3(aq)}$, reduces iodine to iodide ions. Thiosulfate ions are oxidised to tetrathionate ions.

This equation gives the key mole ratio:

The endpoint is handled differently from the manganate(VII) titration. Iodine is brown/yellow in solution. As thiosulfate is added, iodine is reduced to colourless iodide, so the brown colour fades. When the mixture becomes pale straw coloured, add starch. The starch forms a blue-black colour with the remaining iodine. Continue adding thiosulfate dropwise until the blue-black colour just disappears.

Starch is added near the endpoint, not at the beginning. If it is added too early, the iodine-starch colour can be slow to disappear and the endpoint is harder to judge accurately.

Worked example: 25.00 $\,\mathrm{cm}^{3}$ of iodine solution is titrated with 0.100 $\,\mathrm{mol} \,\mathrm{dm}^{-3}$ sodium thiosulfate. The mean titre is 22.60 $\,\mathrm{cm}^{3}$. Calculate the concentration of iodine.

First find moles of thiosulfate:

From the equation, 2 $\,\mathrm{mol}$ $\ce{S2O3^2-}$ react with 1 $\,\mathrm{mol}$ $\ce{I2}$:

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

Structured calculations

Keep unrounded calculator values through mole-ratio and dilution steps, then round the final answer appropriately. Premature rounding can change the last reported digit.

Redox titration calculations use the same amount-of-substance tools as acid-base titrations, but the mole ratio often comes from half-equations rather than a simple neutralisation equation.

[DIAGRAM: asset_name: Lesson 5.2.3d-e: Redox titrations - diagram 02; asset_slug: 05_02_03b_redox_titrations__diagram_02; recommended_method: drawn_chem; description: Flow diagram for processing redox titration calculations from concordant mean titre through n = cV, redox mole ratio, aliquot scaling and final units/significant figures.]

A reliable route is:

1. Select the concordant titres and calculate a mean, excluding any rough titre or clear anomaly.
2. Convert the mean titre from $\,\mathrm{cm}^{3}$ to $\,\mathrm{dm}^{3}$.
3. Use $n = cV$ for the solution of known concentration.
4. Use the balanced redox equation to convert to moles of the unknown species.
5. Divide by the aliquot volume, or scale up if the aliquot came from a larger volumetric flask.
6. Give appropriate units and significant figures.

Worked example: 25.00 $\,\mathrm{cm}^{3}$ of $\ce{Fe^2+(aq)}$ is acidified and titrated with 0.0200 $\,\mathrm{mol} \,\mathrm{dm}^{-3}$ $\ce{KMnO4(aq)}$. The mean titre is 18.40 $\,\mathrm{cm}^{3}$. Calculate the concentration of $\ce{Fe^2+(aq)}$.

Equation:

Moles of $\ce{MnO4-}$ used:

Use the 1:5 ratio:

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

OCR-style questions often include a titre table. Record burette readings to two decimal places, normally ending in 0 or 5. A titre is a difference between two readings, so a burette titre carries uncertainty from both the initial and final reading.

A mean titre and its uncertainty

Suppose the rough titre is $20.60\ \mathrm{cm^3}$, followed by $20.20$, $20.25$ and $20.20\ \mathrm{cm^3}$. The last three agree within $0.05\ \mathrm{cm^3}$; exclude the rough and use their unrounded mean $20.2166\ldots\ \mathrm{cm^3}$ in calculations (report $20.22\ \mathrm{cm^3}$). Use the concordance criterion specified by the method; a common school criterion is within $0.10\ \mathrm{cm^3}$.