Hi Maia,

I'm not sure exactly what is meant by the "acid reaction" or the base formula; I may have learned this slightly differently, but I'll go through how I was able to solve this (can't be that different!)...

I got the same oxidation reaction: 2I- ==> I2 + 2e-

My reduction reaction was different: 2e- + 2H+ + H2O2 ==> H2O + H2O.  Since there was initially 2 O's on the reactant side and 1 O on the product side (H2O2 ==> H2O), you must add a water molecule to the product side.  This will then give you 4 H's on the product side while the reactant side has 2 H's (H2O2 ==> H2O + H2O), so you must add 2H+ ions to the reactant side.  Then, in order to balance the charges, you must add 2e- to the reactant side.

Since both reactions have the same number of electrons, we do not need to introduce any multipliers.

With both reactions written, we can combine them: 2I- + 2e- + 2H+ + H2O2 ==> I2 + 2e- + 2H2O

The 2e- cancel out, and we're left with: 2I- + 2H+ + H2O2 ==> I2 + 2H2O.  Unless the problem asked for this reaction to be solved under neutral or basic conditions, we are done.  This would be the answer under acidic conditions.

If the problem asked for the reaction to be solved under basic conditions, we would have to get rid of the 2H+ on the reactant side.  Basic conditions require that no hydrogen ions be present at all in the reaction.  To do this, we would add an equal amount of hydroxide ions to both sides:

2OH- + 2I- + 2H+ + H2O2 ==> I2 + 2H2O + 2OH-

This will allow us to combine the hydrogen and hydroxide ions on the reactant side into water:

2H2O + 2I- + H2O2 ==> I2 + 2H2O + 2OH-

The water molecules on both sides then cancel, and we are left with: 2I- + H2O2 ==> I2 + 2OH-.  This would be the answer under basic conditions.  This would also be the answer under neutral conditions, since neutral conditions require that hydrogen and hydroxide ions be absent only on the reactant side.

Hope this helps!