To write the balanced ionic equation representing the oxidation of iodide ion (I<sup>-</sup>) by permanganate ion (MnO<sub>4</sub><sup>-</sup>) in basic solution to yield molecular iodine (I<sub>2</sub>) and manganese(IV) oxide (MnO<sub>2</sub>), we can follow these steps:
Step 1: Write the half-reactions.
Oxidation half-reaction (Iodide ion to molecular iodine):
\[ 2 I^- \rightarrow I_2 + 2 e^- \]
Reduction half-reaction (Permanganate ion to manganese(IV) oxide):
In basic solution, permanganate (MnO<sub>4</sub><sup>-</sup>) is reduced to manganese(IV) oxide (MnO<sub>2</sub>). This reaction can be simplified as follows:
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Start with the permanganate ion reduction: \[ MnO_4^- + e^- \rightarrow MnO_2 \]
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Since we are in a basic environment, we will need to add hydroxide ions (OH<sup>-</sup>) and water molecules to balance the equation. The reduction of permanganate to MnO<sub>2</sub> in basic solution is represented as: \[ MnO_4^- + 2 e^- + 2 H_2O \rightarrow MnO_2 + 4 OH^- \]
Step 2: Balance the half-reactions.
Now we have the two half-reactions:
Oxidation: \[ 2 I^- \rightarrow I_2 + 2 e^- \]
Reduction: \[ MnO_4^- + 2 e^- + 2 H_2O \rightarrow MnO_2 + 4 OH^- \]
Step 3: Equalize the number of electrons.
Both half-reactions involve 2 electrons, so they are already balanced for charge.
Step 4: Combine the half-reactions.
Now we can combine the balanced half-reactions, ensuring that all the terms are represented on one side or the other:
\[ 2 I^- + MnO_4^- + 2 H_2O \rightarrow I_2 + MnO_2 + 4 OH^- \]
Step 5: Final balanced ionic equation.
The final balanced ionic equation for the oxidation of iodide ion by permanganate ion in basic solution is:
\[ 2 I^- + MnO_4^- + 2 H_2O \rightarrow I_2 + MnO_2 + 4 OH^- \]
This equation is balanced in terms of atoms and charge, representing the overall redox reaction taking place in a basic medium.