Moles of Iodine Released from Reaction of Potassium Permanganate with Potassium Iodide

To determine the number of moles of iodine ($\text{I}_2$) released, we must analyze the redox reaction between potassium permanganate ($\text{KMnO}_4$) and potassium iodide ($\text{KI}$) under the specified reaction conditions.

1. Oxidation-Reduction Reactions of $\text{KMnO}_4$

The outcome of the reaction between $\text{KMnO}_4$ and $\text{KI}$ depends strongly on the $\text{pH}$ of the solution:

• In acidic medium: $\text{MnO}_4^- \text{ oxidizes } \text{I}^- \text{ to elemental iodine } (\text{I}_2):$ $2\text{MnO}_4^- + 10\text{I}^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{I}_2 + 8\text{H}_2\text{O}$

• In neutral or weakly basic medium: $\text{MnO}_4^- \text{ is reduced to manganese dioxide } (\text{MnO}_2), \text{ and oxidizes } \text{I}^- \text{ to iodate ion } (\text{IO}_3^-), \text{ NOT } \text{I}_2.$

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2. Half-Reactions in Weakly Basic Medium

• Reduction Half-Reaction: Permanganate ion ($\text{MnO}_4^-$) is reduced from an oxidation state of $+7$ to $+4$: $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$

• Oxidation Half-Reaction: Iodide ion ($\text{I}^-$) is oxidized from an oxidation state of $-1$ to $+5$: $\text{I}^- + 6\text{OH}^- \rightarrow \text{IO}_3^- + 3\text{H}_2\text{O} + 6e^-$

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3. Overall Balanced Chemical Equation

Multiplying the reduction half-reaction by $2$ to balance the electrons gives: $2\text{MnO}_4^- + \text{H}_2\text{O} + \text{I}^- \rightarrow 2\text{MnO}_2 + 2\text{OH}^- + \text{IO}_3^-$

In complete molecular form: $2\text{KMnO}_4 + \text{H}_2\text{O} + \text{KI} \rightarrow 2\text{MnO}_2 + 2\text{KOH} + \text{KIO}_3$

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4. Stoichiometric Analysis

From the balanced chemical equation:

• $2\text{ moles of KMnO}_4$ produce $1\text{ mole of KIO}_3$ and $0\text{ moles of I}_2$.
• $4\text{ moles of KMnO}_4$ will produce $2\text{ moles of KIO}_3$ and $0\text{ moles of I}_2$.

Thus, no elemental iodine ($\text{I}_2$) is liberated in a weakly basic medium.

Final Answer: The number of moles of $\text{I}_2$ released is $0$.