lodide ions in aqueous solution are oxidised to iodine by a variety of oxidising agents. One of these is the peroxodisulfate ion, $$\(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}\)$$, which reacts as shown. $$\[ 2 \mathrm{I}^{-}(\mathrm{aq})+\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq}) \rightarrow \mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{SO}_{4}{ }^{2-}(\mathrm{aq}) \]$$ Sodium thiosulfate is added to the reaction mixture to react with iodine as it is produced. When all of the thiosulfate has reacted, further iodine produced reacts with starch indicator to give a dark colour. You will carry out two experiments to investigate how the rate of this reaction is affected by changing the concentration of the peroxodisulfate ion. FA 1 is $$\(0.0200 \mathrm{moldm}^{-3}\)$$ potassium peroxodisulfate, $$\(\mathrm{K}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}\)$$. FA 2 is $$\(0.00500 \mathrm{~mol} \mathrm{dm}^{-3}\)$$ sodium thiosulfate, $$\(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\)$$. FA 3 is $$\(1.00 \mathrm{moldm}^{-3}\)$$ potassium iodide, KI . FA 4 is starch indicator. The potassium iodide is in a large excess in Experiments 1 and 2. Suggest why a large excess of iodide ions is needed in these experiments. ............................................................................................................................................... . .........................................................................................................................................

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