To balance the redox reaction \( \text{MnO}_4^- + \text{SO}_3^{2-} \to \text{Mn}^{2+} + \text{SO}_4^{2-} \) in basic medium using the ion-electron method, follow these steps:
1. **Separate into half-reactions**:
Oxidation half-reaction: \( \text{SO}_3^{2-} \to \text{SO}_4^{2-} \)
Reduction half-reaction: \( \text{MnO}_4^- \to \text{Mn}^{2+} \)
2. **Balance atoms other than H and O**:
For the oxidation half-reaction:
\( \text{SO}_3^{2-} \to \text{SO}_4^{2-} \)
For the reduction half-reaction:
\( \text{MnO}_4^- \to \text{Mn}^{2+} \)
3. **Balance oxygen atoms by adding \( \text{H}_2\text{O} \)**:
For the oxidation half-reaction:
\( \text{SO}_3^{2-} \to \text{SO}_4^{2-} \) already has the same number of O atoms.
For the reduction half-reaction:
\( \text{MnO}_4^- \to \text{Mn}^{2+} \)
Add water to balance the oxygens:
\( \text{MnO}_4^- \to \text{Mn}^{2+} + 4 \text{H}_2\text{O} \)
4. **Balance hydrogen atoms by adding \( \text{OH}^- \)** (since it's in basic medium):
For the reduction half-reaction:
\( \text{MnO}_4^- + 4 \text{H}_2\text{O} \to \text{Mn}^{2+} \)
Add OH to both sides to balance H atoms:
\( \text{MnO}_4^- + 4 \text{H}_2\text{O} \to \text{Mn}^{2+} + 8 \text{OH}^- \)
5. **Balance charge by adding electrons (\( \text{e}^- \))**:
For the oxidation half-reaction (where \( \text{SO}_3^{2-} \to \text{SO}_4^{2-} \)):
\( \text{SO}_3^{2-} \to \text{SO}_4^{2-} \)
The \( \text{SO}_3^{2-} \) oxidation state is conserved in \( \text{SO}_4^{2-} \), but an electron needs to be added to match the charges:
\( \text{SO}_3^{2-} \to \text{SO}_4^{2-} + 2 \text{e}^- \)
For the reduction half-reaction:
\( \text{MnO}_4^- + 4 \text{H}_2\text{O} + 3 \text{e}^- \to \text{Mn}^{2+} + 8 \text{OH}^- \)
6. **Equalize the number of electrons transferred in both half-reactions**:
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to equalize the electron transfer:
Oxidation: \( 3 \text{SO}_3^{2-} \to 3 \text{SO}_4^{2-} + 6 \text{e}^- \)
Reduction: \( 2 \text{MnO}_4^- + 8 \text{H}_2\text{O} + 6 \text{e}^- \to 2 \text{Mn}^{2+} + 16 \text{OH}^- \)
7. **Add the half-reactions together**:
Adding the half-reactions:
\( 2 \text{MnO}_4^- + 8 \text{H}_2\text{O} + 3 \text{SO}_3^{2-} \to 2 \text{Mn}^{2+} + 16 \text{OH}^- + 3 \text{SO}_4^{2-} \)
8. **Verify that everything is balanced**:
Check atoms and charges:
- Mn: 2 on both sides
- S: 3 on both sides
- O: \( 2 \times 4 + 8 + 3 \times 3 = 31 \) oxygens on the left, \( 2 + 16 + 3 \times 4 = 31 \) on the right
- H: \( 8 \times 2 = 16 \) hydrogens on the left, \( 16 \) on the right
- Charge: \( 2 \times (-1) + 3 \times (-2) = -8 \) on the left, \( 2 \times 2 + 16 \times (-1) + 3 \times (-2) = -8 \) on the right
Thus, the balanced equation in basic medium is:
\[ 2 \text{MnO}_4^- + 3 \text{SO}_3^{2-} + 8 \text{H}_2\text{O} \to 2 \text{Mn}^{2+} + 3 \text{SO}_4^{2-} + 16 \text{OH}^- \]
Balance the following equation by ion electron method in basic medium
MnO4- + so3-² gives Mn+2 + so4-²
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