Q.Explain how the variable oxidation states of iron allow $\text{Fe}^{3+}$ to catalyse the reaction between iodide ions and peroxodisulfate ions, $2\text{I}^- + \text{S}_2\text{O}_8^{2-} \rightarrow \text{I}_2 + 2\text{SO}_4^{2-}$, giving the two intermediate steps.

Direct reaction between iodide ions and peroxodisulfate ions is slow, because both are negatively charged and repel each other electrostatically as they approach. $\text{Fe}^{3+}$ (or $\text{Fe}^{2+}$) catalyses the reaction by providing an alternative pathway that avoids a direct anion-anion collision, using its ability to cycle between the $+3$ and $+2$ oxidation states.

Step 1. $\text{Fe}^{3+}$, a positively charged ion, is attracted to and readily oxidizes iodide ion:

Iron is reduced from $+3$ to $+2$ in this step, while iodide is oxidized to iodine.

Step 2. The $\text{Fe}^{2+}$ produced in Step 1 is then, in turn, oxidized by peroxodisulfate ion, regenerating $\text{Fe}^{3+}$:

Iron is oxidized back from $+2$ to $+3$ in this step, exactly restoring its original oxidation state and completing the catalytic cycle -- the $\text{Fe}^{3+}$ that emerges from Step 2 is now available to catalyse Step 1 again with a fresh pair of iodide ions.

Overall reaction. Adding Step 1 and Step 2 together, the iron species cancel out completely (2 $\text{Fe}^{3+}$ consumed in Step 1, 2 $\text{Fe}^{3+}$ regenerated in Step 2):

$2\text{I}^- + \text{S}_2\text{O}_8^{2-} \longrightarrow \text{I}_2 + 2\text{SO}_4^{2-}$ …

Write the two half-cycle equations (Fe3+ oxidizing I-, then S2O8^2- re-oxidizing the resulting Fe2+ back to Fe3+), add them to confirm iron cancels …

Writing only one of the two steps (missing that the catalyst must be BOTH consumed AND regenerated for it to be a true catalytic cycle); forgetting to verify that adding the two …