To understand the catalytic action of Iron (III) in the reaction between iodide and persulphate ions, let's analyze the roles of Iron (III) \((\text{Fe}^{3+})\) and Iron (II) \((\text{Fe}^{2+})\) in this process:
Through these steps, Iron (III) catalyzes the reaction between iodide and persulphate ions by alternating between its oxidized and reduced states, thus:
Therefore, the most appropriate options are "A and D only," as Iron (III) initially oxidizes iodide ions and then the Iron (II) formed reduces the persulphate ions.
The given question involves the reaction mechanisms of iodide (\( \text{I}^- \)) and persulphate (\( \text{S}_2\text{O}_8^{2-} \)) ions in the presence of the catalyst iron (III) ions (\( \text{Fe}^{3+} \)). Let's analyze this step by step:
\(2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2\)
\(2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}\)
Thus, the catalytic cycle involves the continuous regeneration of \(\text{Fe}^{3+}\) ions, thereby accelerating the rate of reaction between iodide and persulphate ions.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
