Question
Write a balanced ionic equation to represent the oxidation of iodine ion (1-) by permanganate ion (MnO4-) in basic solutions to yield molecular iodine (l2) and manganese (iv) oxide (MnO2).
Answers
Answered by
GPT-4o mini
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:
1. **Identify the half-reactions:**
- Iodide ion (I\(^-\)) is oxidized to iodine (I\(_2\)).
- Permanganate ion (MnO\(_4^-\)) is reduced to manganese(IV) oxide (MnO\(_2\)).
2. **Write the half-reactions:**
- Oxidation half-reaction:
\[
2 \text{I}^- \rightarrow \text{I}_2 + 2 \text{e}^-
\]
- Reduction half-reaction:
\[
\text{MnO}_4^- + 4 \text{e}^- + 2 \text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4 \text{OH}^-
\]
3. **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:
\[
4 \text{I}^- \rightarrow 2 \text{I}_2 + 4 \text{e}^-
\]
4. **Combine the half-reactions:**
\[
4 \text{I}^- + \text{MnO}_4^- + 2 \text{H}_2\text{O} \rightarrow 2 \text{I}_2 + \text{MnO}_2 + 4 \text{OH}^-
\]
5. **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.
1. **Identify the half-reactions:**
- Iodide ion (I\(^-\)) is oxidized to iodine (I\(_2\)).
- Permanganate ion (MnO\(_4^-\)) is reduced to manganese(IV) oxide (MnO\(_2\)).
2. **Write the half-reactions:**
- Oxidation half-reaction:
\[
2 \text{I}^- \rightarrow \text{I}_2 + 2 \text{e}^-
\]
- Reduction half-reaction:
\[
\text{MnO}_4^- + 4 \text{e}^- + 2 \text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4 \text{OH}^-
\]
3. **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:
\[
4 \text{I}^- \rightarrow 2 \text{I}_2 + 4 \text{e}^-
\]
4. **Combine the half-reactions:**
\[
4 \text{I}^- + \text{MnO}_4^- + 2 \text{H}_2\text{O} \rightarrow 2 \text{I}_2 + \text{MnO}_2 + 4 \text{OH}^-
\]
5. **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.