Redox titrationsEdexcel A-Level Chemistry: Revision notes

A redox titration finds the concentration of an oxidising or reducing agent by measuring the volume of a standard solution that reacts completely with it. The key steps are:

1. Write the balanced equation to find the mole ratio.
2. Calculate moles of the known solution: $n = c \times V$, with $V$ in dm³ (cm³ ÷ 1000).
3. Use the ratio to find moles of the unknown.
4. Scale up for any dilution (for example a 25.0 cm³ portion of a 250.0 cm³ solution is ÷10).
5. Calculate the required quantity: concentration, mass or percentage purity.

The end-point is when the reactants have just reacted exactly. Use the mean of concordant titres (within 0.10 cm³).

Acidified manganate(VII) is a strong oxidising agent that oxidises Fe²⁺ to Fe³⁺:

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

The mole ratio is MnO₄⁻ : Fe²⁺ = 1 : 5.

• The potassium manganate(VII) is put in the burette and the iron(II) solution in the conical flask.
• The flask is acidified with an excess of dilute sulfuric acid. Hydrochloric acid is unsuitable because MnO₄⁻ would oxidise Cl⁻ to Cl₂ (giving a titre that is too large); nitric acid is itself an oxidising agent.
• No indicator is needed: MnO₄⁻ is purple and is decolourised as it is reduced to colourless Mn²⁺. The end-point is the first permanent pale pink.
• Read the top of the meniscus because the purple solution is too dark to see through.

Worked example. 25.0 cm³ Fe²⁺ needs 22.40 cm³ of 0.0200 mol dm⁻³ MnO₄⁻. $n(\mathrm{MnO_4^-}) = 0.0200 \times 0.02240 = 4.48 \times 10^{-4}$ mol; $n(\mathrm{Fe^{2+}}) = 5 \times 4.48 \times 10^{-4} = 2.24 \times 10^{-3}$ mol; $c = 2.24 \times 10^{-3} / 0.0250 = 0.0896$ mol dm⁻³.

Many oxidising agents are analysed indirectly. An excess of iodide ions is added, the oxidising agent oxidises iodide to iodine, and the iodine is titrated with sodium thiosulfate:

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

The mole ratio is I₂ : S₂O₃²⁻ = 1 : 2. Examples of the first step:

• Copper(II): $\mathrm{2Cu^{2+} + 4I^- \rightarrow 2CuI + I_2}$ (so Cu²⁺ : S₂O₃²⁻ = 1 : 1)
• Chlorate(I): $\mathrm{OCl^- + 2I^- + 2H^+ \rightarrow I_2 + Cl^- + H_2O}$

End-point. Titrate until the brown iodine fades to pale yellow, then add starch, which gives a blue-black colour. Continue adding thiosulfate dropwise until the blue-black just disappears. Starch is added late because at high iodine concentrations the starch–iodine complex breaks down slowly, which blurs the end-point.

For an indirect titration, trace the amount through each equation.

Worked example (bleach). 10.0 cm³ bleach is diluted to 250.0 cm³; 25.0 cm³ of this is treated with excess KI and acid and titrated with 0.100 mol dm⁻³ thiosulfate (mean titre 19.40 cm³).

• $n(\mathrm{S_2O_3^{2-}}) = 0.100 \times 0.01940 = 1.94 \times 10^{-3}$ mol
• $n(\mathrm{I_2}) = 1.94 \times 10^{-3}/2 = 9.70 \times 10^{-4}$ mol
• $n(\mathrm{OCl^-})$ in 25.0 cm³ = $9.70 \times 10^{-4}$ mol
• in 250.0 cm³: $\times 10 = 9.70 \times 10^{-3}$ mol, from 10.0 cm³ of bleach
• $c = 9.70 \times 10^{-3}/0.0100 = 0.970$ mol dm⁻³

For a percentage purity, find the mass of pure substance from the moles, then divide by the mass of sample used and multiply by 100.

To obtain accurate results:

• Rinse the burette with the solution it will hold, fill above zero and run through the jet to remove air bubbles; remove the funnel.
• Use a pipette for the fixed volume (25.0 cm³) of the solution in the flask.
• Add the titrant while swirling, slowly near the end-point, and wash down the flask walls with distilled water if needed.
• Do a rough titration first, then repeat until two titres are concordant; take the mean of concordant titres only.

Sources of error include air oxidation of Fe²⁺ in standing solutions (so make up and titrate promptly), overshooting the end-point, and misreading the meniscus.