Balance the following redox reaction in basic solution

Question

Complete and balance the following redox reaction in basic solution. Be sure to include the proper phases for all species within the reaction.

TeO₃²⁻(aq) + As(s) → Te(s) + H₂AsO₄⁻(aq)

Answer 1

I realize this is an old question, but left unanswered, so if anyone else sees this, hope it still helps!

To balance a redox reaction in a basic solution, you use the same exact method as balancing the redox reaction, in an acidic solution, but with a couple of extra steps!

It's systematic, and if you can do it once, for one reaction, the same steps apply for others as well and you should be able to get the balanced equation. The steps to this question are as follows:

1. We can split the reaction we have into the following two half-reactions, keeping together the two species that share the same element (not including H or O)
2. TeO₃^2-(aq) -> Te(s)
3. As(s) -> H₂AsO₄^-(aq)
4. We now balance all atoms that are not H or O. However, in this particular case, the two atoms, Te and As, are both balanced, in their respective half-reactions. So no need to do anything else.
5. Now we balance the O atoms in both half-reactions, by adding H₂O to the side that has lesser O's, ensuring only to add the amount of H₂O's that would make the number of O's equal on both sides.
6. Initially, the Te half-reaction has 3 O's on the left, and no O's on the right, so we add 3 H₂O's on the right side.
7. The As half-reaction, had no O's on the left, and 4 O's on the right, so we add 4 H₂O on the left side.
8. TeO₃^2-(aq) -> Te(s) + 3 H₂O(l)
9. 4 H₂O(l) + As(s) -> H₂AsO₄^-(aq)
10. Now, we do similar, in balancing H atoms, in both half-reactions, by adding H⁺ ions to the side that has the lesser amount of H's, again only adding enough to make both sides have equal numbers of H's in each half-reaction.
11. In the Te equation, there were 6 H's on the right, and 0 on the left, so we add 6 H⁺ on the left side
12. In the As equation, there were 8 H's on the left, and 2 on the right, so we add 6 H⁺ to the right side, to make the total H count, 8 on both sides
13. 6 H⁺(aq) + TeO₃^2-(aq) -> Te(s) + 3 H₂O(l)
14. 4 H₂O(l) + As(s) -> H₂AsO₄^-(aq) + 6 H⁺(aq)
15. Now we check the total charges (the sum of charges), on both sides of each half-reaction, adding electrons to the side that is more positive, to reduce the charge, and ensure they're the same on both sides.
16. In the Te reaction, the current total charge is 4+ on the left side, and 0 on the right side, so we add 4 e^- to the left side, to bring the charge down to 0
17. In the As reaction, the total charge is 0 on the left, and 5+ on the right, so we add 5 e^- to make both sides equal to 0
18. 4 e^- + 6 H⁺(aq) + TeO₃^2-(aq) -> Te(s) + 3 H₂O(l)
19. 4 H₂O(l) + As(s) -> H₂AsO₄^-(aq) + 6 H⁺(aq) + 5e^-
20. Now we need to make sure, that the number of electrons, gained by one of the half-reactions, is equal to the electrons lost, in the other half-reaction.
21. The Te reaction, gains 4 electrons, while the As reaction, loses 5 electrons in the products
22. To make sure they're equal, we need to multiply the coefficients in both equations by some number to make them equal. The easiest way to do that is to take the coefficient in front of the e^- in one of the equations and multiply the coefficients of the other equation by that number. Specifically here, the As equation, has 5 electrons, so I multiply the entire Te coefficients by 5. Similarly, since the Te equation has 4 electrons, I multiply all the coefficients in the As equation by 4, giving:
23. 5 ( 4e^- + 6 H⁺(aq) + TeO₃^2-(aq) -> Te(s) + 3 H₂O(l) )
24. 4( 4 H₂O(l) + As(s) -> H₂AsO₄^-(aq) + 6 H⁺(aq) + 5 e^- )
25. Executing those multiplies, we get:
26. 20 e^- + 30H⁺(aq) + 5TeO₃^2-(aq) -> 5Te(s) + 15H₂O(l)
27. 16 H₂O(l) + 4 As(s) -> 4 H₂AsO₄^-(aq) + 24 H⁺(aq) + 20 e^-
28. Now both sides are balanced, and we can now combine the half-reactions into one large reaction. By this I mean just throw all the reactants together on one side, and all the products together on the right side:
29. 20 e^- + 30H⁺(aq) + 5 TeO₃^2-(aq) + 16 H₂O(l) + 4 As(s)
30. ->
31. 5Te(s) + 15 H₂O(l) + 4 H₂AsO₄^-(aq) + 24 H⁺(aq) + 20 e^-
32. (I split them into three lines, to avoid having this run over, but the top most is the reactants, and the bottom is the products)
33. Now we go ahead and cancel out anything that appears on both sides
34. By this, the 20 e^- goes away on both sides
35. We also have 15 H₂O on the products, and 16 H₂O on the reactants, so canceling them out, leaves 1 H₂O on the reactants
36. Finally, there are 30 H⁺ on the reactants and 24 H⁺ on the products, so canceling those out, we are left with 6 H⁺ ion the reactants, yielding now:
37. 6 H⁺(aq) + 5 TeO₃^2-(aq) + H₂O(l) + 4 As(s) -> 5 Te(s) + 4 H₂AsO₄^-(aq)
38. If you check the number of atoms, and the charges, it should be the same on both sides. However, this only is complete, if this is a reaction in an acidic solution, we now must add the same amount of OH^- to both sides of the equation, as there are H⁺
39. So, in our reaction, since there are 6 H⁺ in the equation, we go ahead and add 6 OH^- to both sides of the equation
40. 6 OH^-(aq) + 6 H⁺(aq) + 5 TeO₃^2-(aq) + H₂O(l) + 4 As(s)
41. ->
42. 5 Te(s) + 4 H₂AsO₄^-(aq) + 6 OH^-(aq)
43. In the side that has the OH^- and H⁺, we combine them to form H₂O, giving:
44. 7 H₂O(l) + 5 TeO₃^2-(aq) + 4 As(s) -> 5 Te(s) + 4 H₂AsO₄^-(aq) + 6 OH^-(aq)
45. The final step would be to cancel again, the number of H₂O's that appear on both sides, but since H₂O only appears on one side, there is no need, so this should be the complete answer ^^

It is a long response, but hopefully the steps make sense, and of course feel free to comment, if there are any questions!