Half equations:

As2O3 + 2H20 --> As2O5 + 4H+ + 4e-

I2 + 2e- --> 2I-

So to balance use twice the second equation and once of the first: (balancing out the electrons)

to get:

As2O3 + 2H20 + 2I2 ---> As2O5 + 4H+ + 4I-
Edited to try to be correct.

Q. Balance using oxidation numbers.
As2O3 + I2 + H2O ---> As2O5 + H+ + I-

A. First step is to get rid of any spectator ions and to wirte an ionic equation. To do this we conside the oxidation states of each of the elements before and after the reaction.

As: +3 --> +5
O: -2 --> -2
I: 0 --> -1
H: +1 --> +1

Eliminate the spectators giving (adding a two infront of the Iodine to balance the ionic equation):

As3+ + I2 --> As5+ + 2I-.

This is the ionic equation, and it can be used to generate the e- half equations.

As3+ obviously loses 2e- so:

As3+ --> As5+ + 2e-

But there are two As3+ ions per molecule so double everything up:

2As3+ --> 2As5+ + 4e-

I2 however clearly gains 2 electrons (1 per atom), so:

I2 + 2e- --> 2I-

So we are faced with Iodine gaining 2 and As gaining 4. The electrons therefore need to be balanced. So times the iodine one by two to give:

2I2 + 4e- --> 4I-

Now we can merge the equations.

So:

2As3+ + 2I2 --> 2As5+ + 4I-

Now we can begin compiling the full equation:

As2O3 + 2I2 --> As2O5 + 4I-

We need 2 more oxygen on the left for balance, we add this is terms of water.

As2O3 + 2I2 + 2H2O --> As2O5 + 4I-

Now to balance the extra water we need to add some H+ to the right.

As2O3 + 2I2 + 2H2O --> As2O5 + 4I- + 4H+

I think that should now be right. Thanks OM.
samd

As2O3 + I2 + 2H2O --> As2O5 + 2I- + 4H+

This seems to conflict with the work of oxymoron above. Oxymoron: where have I gone wrong? How do you figure that there are 4e- in your first electron half equation. Surely As has just gone up in oxidation state by two (i.e. lost two e-)?

This equation can't be right because your charges do not balance. You have a net charge of +2 on the right and 0 on the left.

Sorry I should have explained - you are right that the oxidation state of As goes up by two, but in the half equation I wrote, there are two atoms of As, each going up by 2 - total loss of 4 electrons (2 for each atom of As).
You have to use the actual molecules reacting in the half equation (ie As2O3 rather than just As3+) in order to get the right answer out.

Does that make sense?
As2 O3 + 2H20 + 2I2 ---> 4I- + As2 O5 + 4H+

This is the answer I think.
[QUOTE='[Joe]']As2 O3 + 2H20 + 2I2 ---> 4I- + As2 O5 + 4H+

This is the answer I think.

Yes it is! Exactly as I wrote above
oxymoron
This equation can't be right because your charges do not balance. You have a net charge of +2 on the right and 0 on the left.

Sorry I should have explained - you are right that the oxidation state of As goes up by two, but in the half equation I wrote, there are two atoms of As, each going up by 2 - total loss of 4 electrons (2 for each atom of As).
You have to use the actual molecules reacting in the half equation (ie As2O3 rather than just As3+) in order to get the right answer out.

Does that make sense?

Yes I think so. That is funny that I missed that, because I remembered the fact that there were two Iodine atoms in I2.

I have modified my explanation/answer above and I think it is now correct.
samd
Yes I think so. That is funny that I missed that, because I remembered the fact that there were two Iodine atoms in I2.

I have modified my explanation/answer above and I think it is now correct.

Yes - looks good to me (although I still think it is easier to write out the whole half equations including waters and H+ and then no balancing is needed after combining the two half equations - I think it's easier to avoid mistakes that way!)