balance the following redox reaction in acidic solution

Question

Complete and balance the following redox reaction in acidic solution

MnO₄⁻(aq) + S₂O₃²⁻(aq) → S₄O₆²⁻(aq) + Mn²⁺(aq)

Answer 1

MnO₄⁻(aq) + S₂O₃²⁻(aq) → S₄O₆²⁻(aq) + Mn²⁺(aq) ... unbalanced equation

MnO₄^- ==> Mn²⁺ ... unbalanced reduction half reaction (Mn goes from +7 to +2)

MnO₄^- ==> Mn²⁺ + 4H2O .. balanced for Mn and O

MnO₄^- + 8H⁺ ==> Mn²⁺ + 4H2O .. balanced for Mn, O and H (using acid, H+)

MnO₄^- + 8H⁺ + 5e- ==> Mn²⁺ + 4H2O .. balanced for Mn, O, H and charge = BALANCED HALF REACTION

S₂O₃^2- ==> S₄O₆^2- .. unbalanced oxidation half reaction (S goes from +2 to +2.5)

2S₂O₃^2- ==> S₄O₆^2- .. balanced for S and O

2S₂O₃^2- ==> S₄O₆^2- + 2e- .. balanced for S, O and charge = BALANCED HALF REACTION

Multiply reduction half reaction by 2 and oxidation half reaction by 5 to equalize electrons:

2MnO₄^- + 16H⁺ + 10e- ==> 2Mn²⁺ + 8H2O

10S₂O₃^2- ==> 5S₄O₆^2- + 10e-

Add together and combine/cancel like terms to get final balanced redox equation:

2MnO₄^- + 16H⁺ + 10S₂O₃^2- ==> 2Mn²⁺ + 8H2O + 5S₄O₆^2- ... BALANCED REDOX EQUATION