The reaction between iodide (\(\text{I}^-\)) and persulphate ions (\(\text{S}_2\text{O}_8^{2-}\)) produces iodine (\(\text{I}_2\)) and sulphate ions (\(\text{SO}_4^{2-}\)):
\[
2 \text{I}^- + \text{S}_2\text{O}_8^{2-} \rightarrow \text{I}_2 + 2 \text{SO}_4^{2-}
\]
This is a redox reaction where iodide ions are oxidized to iodine. The persulphate ion (\(\text{S}_2\text{O}_8^{2-}\)) is reduced to sulphate ions. This reaction typically requires a catalyst to proceed efficiently.
Among the options given, \(\text{Fe}^{3+}\) is the known catalyst for this reaction. The \(\text{Fe}^{3+}\) ion acts as an oxidizing agent, facilitating the oxidation of iodide ions to iodine.
Thus, the correct catalyst is \(\text{Fe}^{3+}\), which is option (B)