quant chemistry

Question

   Balance the following Redox equation in acidic solution

                         ClO₃^- + As₂S₃ →Cl^- + H₂AsO₄^-+ SO₄^2-

Answer 1

ClO₃^- + As₂S₃ →Cl^- + H₂AsO₄^-+ SO₄^2-

ClO₃^- ==> Cl^- ... reduction half reaction

ClO₃^- ==> Cl^- + 3H₂O to balance oxygens

ClO₃^- + 6H⁺ ==> Cl^- + 3H₂O to balance hydrogens (in acid)

ClO₃^- + 6H⁺ + 6e- ==> Cl^- + 3H₂O to balance electrons and the reduction half reaction

As₂S₃ ==> H₂AsO₄^- + SO₄^2- ... oxidation half reaction

As₂S₃ ==> 2H₂AsO₄^- + SO₄^2- to balance As

As₂S₃ ==> 2H₂AsO₄^- + 3SO₄^2- to balance S

As₂S₃ + 20 H₂O ==> 2H₂AsO₄^- + 3SO₄^2- to balance O

As₂S₃ + 20 H₂O ==> 2H₂AsO₄^- + 3SO₄^2- + 36H⁺ to balance H in acid

As₂S₃ + 20 H₂O ==> 2H₂AsO₄^- + 3SO₄^2- + 36H⁺ +28e- to balance electrons and the oxidation reaction

To equalize electrons, multiply reduction reaction by 14 and oxidation reaction by 3. You get...

14ClO₃^- + 84H⁺ + 84e- ==> 14Cl^- + 42H₂O reduction reaction

3As₂S₃ + 60 H₂O ==> 6H₂AsO₄^- + 9SO₄^2- + 108H⁺ +84e- oxidation reaction

Combine the two to get the balanced redox equation:

14ClO₃^- + 3As₂S₃ + 18H₂O ==> 14Cl^- + 6H₂AsO₄^- + 9SO₄^2- + 24H⁺ BALANCED REDOX EQ.