Write the chemical reactions that show that 1 mol of I₂ is required for 1 mol of H₂O in a Karl Fischer titration.

The major application of Karl Fischer titration, is in the measurement of traces of water in transformer oil, polymers, solvents, foods, and other substances etc. This procedure is expected to be performed half a million times each day.

The titration is usually performed by delivering titrant from an automated burette or by coulometric generation of titrant. For large amount of water (but can go as low as$~1{\mathrm{mgH}}_2\mathrm{O}$), the volumetric procedure is considered to be suitable whereas for very little amount of water the coulometric procedure is suitable.

The anode solution (base, alcohol, ${\mathrm{SO}}_2,{\mathrm{I}}^{-}$) is poured in main section in above figure and cathodic solution which has reagents to be reduced at cathode is poured in coulometric generator. Current is passed till the end point. An unknown solution is fed via septum. After that the consumption of moisture is observed by coulometer. When the ratio of water and iodine is 1: 1, then 2 moles of ${\mathrm{e}}^{-}$corresponds to one mole of water.

$$\begin{gathered} \mathrm{ROH}+{\mathrm{SO}}_2+\mathrm{B}\to {\mathrm{BH}}^{+}+{\mathrm{ROSO}}_2^{-} \\ {\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}_2+{\mathrm{ROSO}}_2^{-}+2\mathrm{B}\to {\mathrm{ROSO}}_3^{-}+2{\mathrm{BH}}^{+}{\mathrm{I}}^{-} \end{gathered}$$

lodine molecule is generated on oxidizing in anode compartment. Then Iodine molecule oxidizes ${\mathrm{SO}}_2$to form ${{\mathrm{ROSO}}_3}^{-}$. Therefore, one mole of iodine molecule is required to consume one mole of water.