Complete and balance the following redox reaction in basic solution. Be sure to include the proper phases for all species within the reaction. IO₃⁻(aq) + S²⁻(aq) → IO⁻(aq) + SO₂(g)

Question

Complete and balance the following redox reaction in basic solution. Be sure to include the proper phases for all species within the reaction. IO₃⁻(aq) + S²⁻(aq) → IO⁻(aq) + SO₂(g)

Answer 1

Tried to submit entire process, but Wyzant won't allow as it says it exceeds 10,000 character, which it didn't. So here's the answer. I'll try to post the full procedure, but it is lengthly.

3IO₃^- + 2H₂O + 2S^2- => 3IO^- + 4OH^- + 2SO₂ .. BALANCED REDOX EQUATION

Edit: Best I can do, is add the essential parts

IO₃^- + 4H₂O + 4e- ==> IO^- + 2H₂O + 4OH^- .. balance for I, O, H and charge = balanced half reaction

S^2- + 2H₂O + 4OH^- ==> SO₂ + 4H₂O + 6e- .. balanced for S, H, O and charge = balanced half reaction

To equalize electrons, multiply reduction reaction by 3 and oxidation reaction by 2 and add them:

3IO₃^- + 12H₂O + 12e- ==> 3IO^- + 6H₂O + 12OH^-

2S^2- + 4H₂O + 8OH^- ==> 2SO₂ + 8H₂O + 12e-

Add them together, combine/cancel like terms:

3IO₃^- + 12H₂O + 12e- + 2S^2- + 4H₂O + 8OH^- ==> 3IO^- + 6H₂O + 12OH^- + 2SO₂ + 8H₂O + 12e-

3IO₃^- + 2H₂O + 2S^2- ==> 3IO^- + 4OH^- + 2SO₂ .. BALANCED REDOX EQUATION

Answer 2

This problem becomes much easier when you count the atoms before and after the reaction. Notice that there is 1 sulfur, 3 oxygens and 1 iodine in S^-2 + IO^-3 moving to the other side of the reaction there are 3 oxygens, 1 sulfur and 1 Iodine in the products. Therefore the coefficients are 1,1,1,1 because the equation is already balanced with sulfur just stealing two oxygens.