Redox equations and titration calculations

What you’ll learn

• Balance redox equations by combining half-equations.
• Use permanganate and iodine–thiosulfate stoichiometry.
• Link titres to concentrations and composition.

Understand the idea

Balance a half-equation by conserving atoms and charge. In acidic aqueous equations, balance oxygen with water, hydrogen with hydrogen ions and charge with electrons. Multiply entire half-equations to cancel equal numbers of electrons, then add and cancel species common to both sides. Check charge as well as atoms in the final equation.

A redox titration uses the mole ratio from that balanced equation. Acidic manganate(VII) ions accept five electrons each when forming manganese(II), so one mole reacts with five moles of iron(II). Iodine consumes two moles of thiosulfate per mole of iodine. Convert volumes to cubic decimetres before using concentration, and account for any dilution or aliquot before reporting the original sample composition. Endpoint observations support the stoichiometry but do not replace it.

Key formulae

$$\mathrm{MnO_4^-+8H^++5e^-\longrightarrow Mn^{2+}+4H_2O}$$

$$\mathrm{Fe^{2+}\longrightarrow Fe^{3+}+e^-}$$

$$\mathrm{I_2+2S_2O_3^{2-}\longrightarrow2I^-+S_4O_6^{2-}}$$

$$n=cV\quad(V\text{ in }\mathrm{dm^3})$$

Worked example

A $25.0\,\mathrm{cm^3}$ iron(II) aliquot reacts with $20.0\,\mathrm{cm^3}$ of $0.0200\,\mathrm{mol\,dm^{-3}}$ manganate(VII) in acidic solution. Find the iron(II) concentration.

1. Moles of manganate(VII) are $0.0200\times0.0200=4.00\times10^{-4}\,\mathrm{mol}$.
2. The 1:5 ratio gives $2.00\times10^{-3}\,\mathrm{mol}$ of iron(II).
3. Divide by the aliquot volume $0.0250\,\mathrm{dm^3}$.

Answer: $[\mathrm{Fe^{2+}}]=0.0800\,\mathrm{mol\,dm^{-3}}$.