Redox titration

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Redox titration (also called oxidation-reduction titration) is a type of titration based on a redox reaction between the analyte and titrant.

Redox titration may involve the use of a redox indicator and/or a potentiometer.

[edit] Example

An example of a redox titration is treating a solution of iodine with a reducing agent and using starch as indicator. Iodine forms an intensely blue complex with starch. Iodine (I₂) can be reduced to iodide (I⁻) by e.g. thiosulfate (S₂O₃²⁻), and when all iodine is spent the blue colour disappears. This is called iodometric titration.

Most often the reduction of iodine to iodide is the last step in a series of reactions where the initial reactions are used to convert an unknown amount of the analyte (the substance you want to analyse) to an equivalent amount of iodine, which may then be titrated. Sometimes other halogens than iodine are used in the intermediate reactions because they are available in better measurable standard solutions and/or react more readily with the analyte. The extra steps in iodometric titration may be worth while because the equivalence point, where the blue turns colourless, is more distinct than some other analytical methods.

Such a series of reactions in aqueous solution with resorcinol as analyte, bromate as standard, thiosulfate as titrant and starch as indicator, can be as follows. Oxidation numbers are given in parentheses.

$BrO_3^-(5) + 5Br^-(-1) + 6H^+ \ \to \ 3Br_2(0) + 3H_2O$
Bromate + surplus of bromide in acidic solution → Bromine + water. Bromine continues to ii.
$3Br_2(0) + resorcinol \ \to \ 3HBr(-1) + tribromo(+1)resorcinol$
Bromine + resorcinol → Hydrogen bromide + tribromoresorcinol. Unreacted bromine continues to iii.
$Br_2(0) + 2I^-(-1) \ \to \ 2Br^-(-1) + I_2(0)$
Bromine + surplus of iodide → Bromide + iodine. Iodine continues to iv.
$I_2(0) + 2SSO_3^{2-}(-1,5) \ \to \ 2I^-(-1) + (SSO_3)_2^{2-}(0,5)$
Iodine + thiosulfate → Iodide + tetrathionate.

Let us call the initial amounts of resorcinol and bromate R and B, and the amount of thiosulfate T, where B og T are known and R is unknown – but we have to make sure that R is less than B. (All amounts are in mole.) Then R = B - T/6