Peroxodisulfate ions, S2O8^2-, react with iodide ions in aqueous solution. S2O8^2-(aq) + 2I-(aq) → 2SO4^2-(aq) + I2(aq) The table shows how the relative rate of this reaction changes when different concentrations of peroxodisulfate ions and iodide ions are used. experiment | concentration of S2O8^2- in mol/dm³ | concentration of I- in mol/dm³ | relative rate of reaction ---|---|---|--- 1 | 0.008 | 0.02 | 1.7 2 | 0.016 | 0.02 | 3.3 3 | 0.032 | 0.02 | 6.8 4 | 0.008 | 0.04 | 3.4 5 | 0.008 | 0.08 | 6.9

1. (a)Use the information in the table to describe how increasing the concentration of each of these ions affects the relative rate of reaction. peroxodisulfate ions iodide ions[2]
2. (b)Iron(III) ions, Fe3+, catalyse this reaction. Explain how catalysts increase the rate of a reaction.[1]
3. (c)(i)Iron(III) ions react with iodide ions. 2I-(aq) + 2Fe3+(aq) → I2(aq) + 2Fe2+(aq) Explain how iron(III) ions are acting as an oxidising agent in this reaction.[1]
4. (c)(ii)What colour change is observed when this reaction happens?[1]
5. (c)(iii)Describe a test for iron(III) ions. test observation[2]
6. (d)Iron(II) ions react with peroxodisulfate ions. The products are iron(III) ions and sulfate ions. Construct the equation for this reaction.[1]