Question

I am having trouble balancing one of the half reactions:

AsO3^(-3) --> AsO4^(-4)

Which side should OH and H2O go?
Thanks.

Answer 1

(this is in a basic solution, by the way).
Thanks you.

Answer 2

Memorize these steps. They work.
1. Determine the oxidation state
As is +3 on the left; +5 on the right
and add electrons to the appropriate side to balance the change in oxdn state.
AsO3^3- ==> AsO4^3- + 2e

2. Count up the charge on left and right and add OH^- to the appropriate side to balance the charge.
Charge on left is 3-; on right is 5- so add 2OH^- to the left.
AsO3^3- + 2OH^- ==> AsO4^3- + 2e

3.Add H2O to balance the H and O (usually to the side opposite to where you added the OH^- but not always).

4. AsO3^3- + 2OH^- ==>AsO4^3- + 2e + H2O
5. Check it for
a.atoms. 1As both sides, 5 O both sides, 2 H both sides.
b. charge. -5 on the left; -5 on the right.
c. change in oxidation state. Change from +3 to +5 which is change of 2e.

Hope this helps.
By the way, if this were an ACID solution, you need change only one part of the above. In part 2 above, add J^+ instead of OH^- AND I would add a 3.5 part of "Cancel any ions common to both sides."

Answer 3

In the last paragraph one would add H^+ (I'm not sure what a J^+ is :-)] instead of OH^-.