Choose the correct products formed in acidic medium.

${5SO_3}^{-}2+ 2{MnO_4}^- + 6H^+→?$

$2Mn^{2+} + 3H_2O + 5{SO_4}^{2-}$

The correct answer is Option (2) → $2Mn^{2+} + 3H_2O + 5{SO_4}^{2-}$

This is a redox reaction in acidic medium.

Permanganate ion (MnO₄⁻) in acidic medium is reduced to Mn²⁺.

Reduction half reaction:

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Sulphite ion (SO₃²⁻) is oxidised to sulphate (SO₄²⁻).

Oxidation half reaction:

SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻

Balancing electrons and combining both half reactions gives:

5SO₃²⁻ + 2MnO₄⁻ + 6H⁺ → 5SO₄²⁻ + 2Mn²⁺ + 3H₂O

Thus the products are: 2Mn²⁺ + 3H₂O + 5SO₄²⁻ 

Option 1: 2MnO₂ + 6H⁺ + 5SO₄²⁻ MnO₂ forms in neutral or weakly acidic medium, not strongly acidic medium. Incorrect.

Option 2: 2Mn²⁺ + 3H₂O + 5SO₄²⁻ --- Correct products in acidic medium. Correct.

Option 3: 2Mn²⁺ + 5SO₄²⁻ + 6OH⁻ OH⁻ cannot appear in acidic medium. Incorrect.

Option 4: 2MnO₂ + 6H₂O + 5/2 S₂O₃²⁻ Thiosulphate is not formed in this reaction. Incorrect.