Question

balance the redox reaction mno^-4+h^+ +i^- yeld mno2+ h2o+I2

Answer 1

To balance the redox reaction between permanganate ion (\( \text{MnO}_4^- \)) and iodide ion (\( \text{I}^- \)), which yields manganese dioxide (\( \text{MnO}_2 \)), water (\( \text{H}_2\text{O} \)), and iodine (\( \text{I}_2 \)), we can follow these steps.

1. **Identify the oxidation states:**
- In \( \text{MnO}_4^- \), Mn has an oxidation state of +7.
- In \( \text{MnO}_2 \), Mn has an oxidation state of +4.
- In iodide (\( \text{I}^- \)), I has an oxidation state of -1.
- In iodine (\( \text{I}_2 \)), I has an oxidation state of 0.

2. **Determine what is being oxidized and what is being reduced:**
- \( \text{MnO}_4^- \) is reduced to \( \text{MnO}_2 \) (Mn from +7 to +4).
- \( \text{I}^- \) is oxidized to \( \text{I}_2 \) (I from -1 to 0).

3. **Write the half-reactions:**
- Reduction half-reaction:
\[
\text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{MnO}_2 + 4 \text{H}_2\text{O}
\]
- Oxidation half-reaction:
\[
2 \text{I}^- \rightarrow \text{I}_2 + 2 \text{e}^-
\]

4. **Balance the electrons:**
- We need to multiply the oxidation half-reaction by 5 to equalize the number of electrons exchanged:
\[
10 \text{I}^- \rightarrow 5 \text{I}_2 + 10 \text{e}^-
\]

5. **Combine the half-reactions:**
- Now we can add the two half-reactions together:
\[
\text{MnO}_4^- + 8 \text{H}^+ + 10 \text{I}^- \rightarrow \text{MnO}_2 + 4 \text{H}_2\text{O} + 5 \text{I}_2
\]

6. **Final balanced equation:**
\[
\text{MnO}_4^- + 8 \text{H}^+ + 10 \text{I}^- \rightarrow \text{MnO}_2 + 4 \text{H}_2\text{O} + 5 \text{I}_2
\]

This ensures that the number of atoms and charges is balanced on both sides of the equation.