To write the balanced ionic equation for the oxidation of the iodide ion (I\(^-\)) by the permanganate ion (MnO\(_4^-\)) in basic solution to produce molecular iodine (I\(_2\)) and manganese(IV) oxide (MnO\(_2\)), we can follow these steps:
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Identify the half-reactions:
- Iodide ion (I\(^-\)) is oxidized to iodine (I\(_2\)).
- Permanganate ion (MnO\(_4^-\)) is reduced to manganese(IV) oxide (MnO\(_2\)).
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Write the half-reactions:
- Oxidation half-reaction: \[ 2 \text{I}^- \rightarrow \text{I}_2 + 2 \text{e}^- \]
- Reduction half-reaction: \[ \text{MnO}_4^- + 4 \text{e}^- + 2 \text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4 \text{OH}^- \]
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Balance the electrons: To combine the half-reactions, we need to balance the charge and the number of electrons. The oxidation half-reaction loses 2 electrons, while the reduction half-reaction uses 4 electrons. We can multiply the oxidation half-reaction by 2 to equalize the number of electrons: \[ 4 \text{I}^- \rightarrow 2 \text{I}_2 + 4 \text{e}^- \]
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Combine the half-reactions: \[ 4 \text{I}^- + \text{MnO}_4^- + 2 \text{H}_2\text{O} \rightarrow 2 \text{I}_2 + \text{MnO}_2 + 4 \text{OH}^- \]
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Final balanced ionic equation: Thus, the balanced ionic equation representing the reaction in basic solution is: \[ 4 \text{I}^- + \text{MnO}_4^- + 2 \text{H}_2\text{O} \rightarrow 2 \text{I}_2 + \text{MnO}_2 + 4 \text{OH}^- \]
This equation shows the oxidation of iodide ions to iodine and the reduction of permanganate to manganese(IV) oxide in a basic medium.