Question

Write the ionic equations for the oxidising action of potassium permanganate for its reaction with \(I^-\) in both acidic and alkaline solutions.

Hint:

Remember the reduction products of \(KMnO_4\):
Acidic: \(Mn^{2+}\) (Pink \(\rightarrow\) Colourless)
Alkaline/Neutral: \(MnO_2\) (Purple \(\rightarrow\) Brown precipitate)

Answer 1

Step 1: Understanding the Concept:
Potassium permanganate (\(KMnO_4\)) is a powerful oxidizing agent. In acidic medium, it is reduced to \(Mn^{2+}\) (+2), while in neutral or alkaline medium, it is reduced to \(MnO_2\) (+4).
Step 2: Detailed Explanation:
1. In Acidic Solution:
Permanganate ion (\(MnO_4^-\)) oxidizes iodide (\(I^-\)) to free iodine (\(I_2\)).
Reduction half-reaction: \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)
Oxidation half-reaction: \(2I^- \rightarrow I_2 + 2e^-\)
Overall balanced ionic equation:
\[ 2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 8H_2O + 5I_2 \]
2. In Alkaline Solution:
Permanganate ion oxidizes iodide (\(I^-\)) specifically to iodate ion (\(IO_3^-\)).
Reduction half-reaction: \(MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-\)
Oxidation half-reaction: \(I^- + 6OH^- \rightarrow IO_3^- + 3H_2O + 6e^-\)
Overall balanced ionic equation:
\[ 2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^- \]
Step 3: Final Answer:
The reaction products of iodide oxidation depend heavily on the pH: \(I_2\) in acid and \(IO_3^-\) in alkali.