Acidified MnO4^2- is added to Na2SO3. When writing the half equations do you use Na oxidising to Na+ or SO3^2- oxidising to SO4^2-?

In the solution, Na⁺ ion exists as spectator ion, and there is no Na metal. In the reaction, MnO₄⁻ ion (acting as oxidizing agent) is reduced to Mn²⁺ ion, while SO₃²⁻ ion (acting as reducing agent) is oxidized to SO₄²⁻ ion.

Reduction half-equation: MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l) …… [1]
Oxidation half-equation: SO₃²⁻(aq) + H₂O(l) → SO₄²⁻(aq) + 2H⁺(aq) + 2e⁻ …… [2]

[1]×2 + [2]×5, and cancel off 10e⁻, 10H⁺(aq) and 5H₂O(l) on the both sides. The balanced overall equation is:
2MnO₄⁻(aq) + 5SO₃²⁻(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 5SO₄²⁻(aq) + 3H₂O(l)

2MnO4^- + 16H^+ + 10S2O3^2- -----> 2Mn^2+ + 5S4O6^2- + 8H2O

are u doing atar

Shhhh