To determine the correct answer to this question, let's analyze the catalysis mechanism involving Iron (III) in the reaction between iodide ($$\mathrm{I^-}$$) and persulfate ($$\mathrm{S_2O_8^{2-}}$$) ions.

Iron (III), or $$\mathrm{Fe}^{3+}$$, can act as a catalyst by undergoing redox reactions, cycling between $$\mathrm{Fe}^{2+}$$ and $$\mathrm{Fe}^{3+}$$. Here’s how it typically works:

1. $$\mathrm{Fe}^{3+}$$ can oxidize iodide ions to iodine. The reaction will proceed as follows:

$$\mathrm{2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2}$$

2. $$\mathrm{Fe}^{2+}$$, now formed, can be oxidized back to $$\mathrm{Fe}^{3+}$$ by reducing persulfate ions. The reaction will be:

$$\mathrm{2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}}$$

Based on this mechanism, $$\mathrm{Fe}^{3+}$$ is responsible for oxidizing the iodide ions, and $$\mathrm{Fe}^{2+}$$ is responsible for reducing the persulfate ions. Thus, we can conclude:

A. $$\mathrm{Fe}^{3+}$$ oxidises the iodide ion: This statement is correct.

B. $$\mathrm{Fe}^{3+}$$ oxidises the persulphate ion: This statement is incorrect because $$\mathrm{Fe}^{3+}$$ does not oxidize persulfate ions.

C. $$\mathrm{Fe}^{2+}$$ reduces the iodide ion: This statement is incorrect because $$\mathrm{Fe}^{2+}$$ does not reduce iodide ions.

D. $$\mathrm{Fe}^{2+}$$ reduces the persulphate ion: This statement is correct.

Therefore, the most appropriate answer is:

Option C: A and D only