Question

MnO4^-1+S¯²--->MnO2+SO3^-²vin bais mediam balance this reaction

Answer 1

To balance the redox reaction \( \text{MnO}_4^- + \text{S}^{2-} \rightarrow \text{MnO}_2 + \text{SO}_3^{2-} \) in basic medium, follow these steps:

1. **Assign Oxidation States:**
- In \( \text{MnO}_4^- \), Mn is in the +7 oxidation state.
- In \( \text{MnO}_2 \), Mn is in the +4 oxidation state.
- In \( \text{S}^{2-} \), sulfur is in the -2 oxidation state.
- In \( \text{SO}_3^{2-} \), sulfur is in the +4 oxidation state.

2. **Identify the Changes:**
- Mn: +7 to +4 (reduction).
- S: -2 to +4 (oxidation).

3. **Write Half-Reactions:**
- **Reduction Half-Reaction:**
\[
\text{MnO}_4^- + 3 \text{e}^- \rightarrow \text{MnO}_2
\]
- **Oxidation Half-Reaction:**
\[
\text{S}^{2-} \rightarrow \text{SO}_3^{2-} + 6 \text{e}^-
\]

4. **Equalize the Number of Electrons:**
- We need 6 electrons in total for the reaction, so we’ll multiply the reduction half-reaction by 2:
\[
2 \text{MnO}_4^- + 6 \text{e}^- \rightarrow 2 \text{MnO}_2
\]

5. **Combine Half-Reactions:**
\[
2 \text{MnO}_4^- + 6 \text{e}^- + \text{S}^{2-} \rightarrow 2 \text{MnO}_2 + \text{SO}_3^{2-} + 6 \text{e}^-
\]
The electrons cancel out, leading to:
\[
2 \text{MnO}_4^- + \text{S}^{2-} \rightarrow 2 \text{MnO}_2 + \text{SO}_3^{2-}
\]

6. **Balance Oxygen with Water (in basic solution):**
- There are 8 oxygen atoms on the left (from 2 \( \text{MnO}_4^- \)) and 5 on the right (2 from \( \text{MnO}_2 \) and 3 from \( \text{SO}_3^{2-} \)).
- Add 3 water molecules to the right side to balance oxygen:
\[
2 \text{MnO}_4^- + \text{S}^{2-} \rightarrow 2 \text{MnO}_2 + \text{SO}_3^{2-} + 3 \text{H}_2\text{O}
\]

7. **Balance Hydrogen with Hydroxide:**
- There are 6 hydrogen atoms on the right (from 3 water molecules).
- Add 6 hydroxide ions to the left:
\[
2 \text{MnO}_4^- + \text{S}^{2-} + 6 \text{OH}^- \rightarrow 2 \text{MnO}_2 + \text{SO}_3^{2-} + 3 \text{H}_2\text{O}
\]

8. **Final Balanced Equation:**
\[
2 \text{MnO}_4^- + \text{S}^{2-} + 6 \text{OH}^- \rightarrow 2 \text{MnO}_2 + \text{SO}_3^{2-} + 3 \text{H}_2\text{O}
\]

This equation is now balanced in basic medium, with the same number of atoms of each element and equal charges on both sides.