Titrations

Identify which species is the oxidant and which is the reductant. In iodometry, the analyte oxidant (e.g., Cu²⁺) oxidises iodide: I⁻ loses an electron (ox. state −1 → 0). Iodine (I₂) is then the product to be determined. Simultaneously Cu²⁺ (ox. state +2) gains an electron to become Cu⁺ (ox. state +1), precipitated as CuI. Every mole of Cu²⁺ liberates half a mole of I₂ via: 2Cu²⁺ + 4I⁻ → 2CuI(s) + I₂.

A solution of sodium thiosulfate (Na₂S₂O₃) is prepared near the required concentration (~0.1 mol dm⁻³) and standardised against a primary standard such as potassium iodate (KIO₃) or potassium dichromate. In the thiosulfate–iodine reaction, thiosulfate (S₂O₃²⁻, average S oxidation state +2) is oxidised to tetrathionate (S₄O₆²⁻, average S oxidation state +2.5), while I₂ is reduced to 2I⁻. The stoichiometry is 1 mol I₂ : 2 mol Na₂S₂O₃.

Add a large excess of KI (typically 1–2 g) to the acidified analyte solution in a conical flask. The analyte oxidant quantitatively liberates I₂ from I⁻. Seal the flask and allow the reaction to proceed in the dark for ~5 minutes to ensure complete reaction. Acidification (dilute H₂SO₄) is necessary to dissolve precipitates and to push the equilibrium toward I₂ formation.

Transfer the liberated I₂ solution to a burette if large, or titrate directly from a burette of standard Na₂S₂O₃ into the flask. The dark brown colour of I₂/I₃⁻ fades to pale yellow as the endpoint nears. Add 2 cm³ of freshly prepared 1% starch solution at this point. The solution turns deep blue-black. Continue adding Na₂S₂O₃ dropwise; the endpoint is the first permanent disappearance of the blue-black colour (lasting ≥30 s), indicating all I₂ has been reduced to I⁻.

Use the stoichiometric chain: moles of Na₂S₂O₃ used → moles of I₂ liberated → moles of analyte. For copper: n(Na₂S₂O₃) = c × V; n(I₂) = n(Na₂S₂O₃)/2; n(Cu²⁺) = 2 × n(I₂). Multiply by molar mass for mass, or divide by volume for concentration. Always record titres to 4 significant figures and average concordant results (within ±0.1 cm³).

• \ce{I_2} — Iodine (oxidant in this half-reaction) (mol): The liberated iodine whose amount is determined; each I atom is reduced from oxidation state 0 to −1.
• \ce{S_2O_3^{2-}} — Thiosulfate ion (mol): The reducing titrant; sulfur is oxidised from average state +2 to +2.5 in tetrathionate.
• \ce{I^-} — Iodide ion (mol): Product of iodine reduction; its formation causes the blue-black starch complex to decolourise at the endpoint.
• \ce{S_4O_6^{2-}} — Tetrathionate ion (mol): Oxidised sulfur product; two thiosulfate units are coupled by a new S–S bond.

Problem. A 25.00 cm³ sample of a copper sulfate solution was acidified and treated with excess KI. The liberated iodine required 18.40 cm³ of 0.1000 mol dm⁻³ Na₂S₂O₃ for complete reaction at the starch endpoint. Calculate the concentration of Cu²⁺ in the original solution.

Answer. The concentration of Cu²⁺ is 0.07360 mol dm⁻³.

• Oxidation States — Assigning oxidation states to each atom is the essential first step in writing balanced half-equations for any redox titration.
• Half-Equations and Electron Transfer — Iodometric calculations depend on correct electron-transfer stoichiometry derived from balanced oxidation and reduction half-equations.
• Standard Electrode Potentials — The feasibility of the I₂/I⁻ and S₄O₆²⁻/S₂O₃²⁻ couples is predicted by comparing their standard reduction potentials (E° values).
• Acid-Base Titrations — Redox titrations share the same core technique — equivalence points, indicators, and stoichiometric calculations — as acid-base titrations, making this a natural comparison.
• Concentration and Molarity — Converting between moles, volumes, and concentrations (c = n/V) is applied at every calculation step of a redox titration.