Answer:
The coefficients should be: 2, 5, 2, 5
Explanation:
Given redox reaction: MnO₄⁻ + SO₃²⁻ → Mn²⁺+ SO₄²⁻
To balance the given redox reaction in acidic medium, the oxidation and the reduction half-reactions should be balanced first.
<u>Reduction half-reaction:</u> MnO₄⁻ → Mn²⁺
Oxidation state of Mn in MnO₄⁻ is +7 and the oxidation state of Mn in Mn²⁺ is +2. Therefore, Mn accepts 5e⁻ to get reduced from +7 to +2 oxidation state.
⇒ MnO₄⁻ + 5e⁻ → Mn²⁺
Now the total charge on reactant side is (-6) and the total charge on product side is +2. Therefore, to balance the total charge, 8H⁺ must be added to the reactant side.
⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺
To balance the number of hydrogen and oxygen atoms, 4H₂O must be added to the product side.
⇒ MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O .....equation 1
<u>Oxidation half-reaction:</u> SO₃²⁻ → SO₄²⁻
Oxidation state of S in SO₃²⁻ is +4 and the oxidation state of S in SO₄²⁻ is +6. Therefore, S loses 2e⁻ to get oxidized from +4 to +6 oxidation state.
⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻
Now the total charge on reactant side is (-2) and the total charge on product side is (-4). Therefore, to balance the total charge, 2H⁺ must be added to the product side.
⇒ SO₃²⁻ → SO₄²⁻ + 2e⁻ + 2H⁺
To balance the number of hydrogen and oxygen atoms, 1 H₂O must be added to the reactant side.
⇒ SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺ .....equation 2
<u>Now, to cancel the electrons transferred, equation (1) is multiplied by 2 and equation (2) is multiplied by 5.</u>
<u>Balanced Reduction half-reaction:</u>
MnO₄⁻ + 5e⁻ + 8H⁺ → Mn²⁺ + 4H₂O ] × 2
⇒ 2MnO₄⁻ + 10e⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O .....equation 3
<u>Balanced Oxidation half-reaction:</u>
SO₃²⁻ + H₂O → SO₄²⁻ + 2e⁻ + 2H⁺ ] × 5
⇒ 5SO₃²⁻ + 5H₂O → 5SO₄²⁻ + 10e⁻ + 10H⁺ .....equation 4
Now adding equation 3 and 4, to obtain the <u>overall balanced redox reaction:</u>
2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 5SO₄²⁻ + 3H₂O
<u><em>Therefore, the coefficients should be: 2, 5, 2, 5 </em></u>
