Step 1: Identify the redox nature of the reaction
This is a redox reaction in an acidic medium involving:
- \( MnO_4^- \) being reduced to \( Mn^{2+} \)
- \( SO_2 \) being oxidized to \( HSO_4^- \)
Step 2: Assign oxidation states
- In \( MnO_4^- \), Mn is in +7 oxidation state
- In \( Mn^{2+} \), Mn is in +2 oxidation state
⇒ Mn gains 5 electrons (reduction)
- In \( SO_2 \), S is in +4 oxidation state
- In \( HSO_4^- \), S is in +6 oxidation state
⇒ S loses 2 electrons (oxidation)
Step 3: Balance the electron transfer
To balance the electrons, we use the LCM of 5 and 2 = 10
- 2 \( MnO_4^- \) will gain 10 electrons
- 5 \( SO_2 \) will lose 10 electrons
Step 4: Balanced redox reaction in acidic medium
The balanced reaction becomes:
\[
2MnO_4^- + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5HSO_4^- + 2H^+
\]
Step 5: Identify the stoichiometric coefficient of SO₂
From the balanced equation, the coefficient of SO₂ is 5
Final Answer: 5