Balance each net ionic equation by the half-reactions method in acidic solutions. S2O3^2- +OCL^- --> Cl^- + S4O6^2-?

Reduction half-equation: 2S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻ …… [1]
Oxidation half-equation: OCl⁻ + 2H⁺ + 2e⁻ → Cl⁻ + H₂O …… [2]

[1] + [2], and cancel 2e⁻ on the both sides. The overall net ionic equation is:
2S₂O₃²⁻ + OCl⁻ + 2H⁺ →S₄O₆²⁻ + Cl⁻ + H₂O