# Redox titrations

The quantity of oxidising or reducing agent present in a redox reaction can be calculated from the results of a titration by using the balanced redox equation - this tells us the molar ratio of reactants, and the formula triangle linking number of moles with concentration and volume.

Acidified permanganate is a common chemical used in redox titrations. An indicator is not required, as purple permanganate solution turns colourless when reduced.

The mass of vitamin C in a tablet can be determined by redox titration. An iodine solution of known concentration and starch indicator are used.

A vitamin C tablet is dissolved in deionised water in a beaker and transferred to a 250 cm3 standard flask. This is made up to the graduation mark using deionised water (and the washings of the beaker).

25 cm3 of this solution is transferred by pipette, to a conical flask and starch indicator is added. The iodine can be added by burette until the vitamin C solution turns blue/black.

This can be repeated and an average taken.

## Example

The results of the titration described above show that 17.6 cm3 of 0.031 mol l-1 iodine solution was required to reach the end-point with 25 cm3 of vitamin C solution.

Calculate the mass of vitamin C (C6H8O6) in the original tablet.

The balanced redox equation for the process is:

$C_{6}H_{8}O_{6} + I_{2} \rightarrow C_{6}H_{6}O_{6} + 2H^{+}+ 2I^{-}$

$1\,mole\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,mole$

This shows that one mole of vitamin C will react with one mole of iodine. You can used $$n = c \times v$$ to calculate how many moles of iodine were required from the average volume and concentration in the question.

$n = c \times v\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, c = 0.031\, mol\, l^{-1}\,\,\,\,\,\,\,\, v = 0.0176 l$

$\,\,\,\,\,\, = 0.031 \times 0.0176$

$\,\,\,\,\,\, = 5.456 \times 10^{-4}\, moles$

$1\,\, mole\,\, of\,\, I_{2} \rightarrow 1\,\, mole\,\, C_{6}H_{8}O_{6}$

$5.456 \times 10^{-4}\,\, moles\,\, of\,\, I_{2} \rightarrow 5.5456 \times 10^{-4}\,\, moles\,\, of\,\, C_{6}H_{8}O_{6}$

This means that a 25 cm3 portion of the vitamin C solution contains 5.5456 x 10-4 moles of vitamin C. The original vitamin C tablet was dissolved in 250 cm3 of water, so:

$25cm^{3} \,of\, Vit.\, C\, solution\,\,\,\,\,\,\,\,\,= 5.456 \times 10^{-4}\, moles\, Vit.\, C$

$250cm^{3} \,of\, Vit.\, C\, solution\,\,\,\,\,\,\,\,\,= 5.456 \times 10^{-4}\times 10$

$= 5.456 \times 10^{-3} \,moles\, Vit.\, C\, in\, one\, tablet$

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