Hi, I am really confused about this problem. I found the charges, and I got that 2 e are lost, and 0 are gained. Is it possible to have 0 gained? Also, if it is possible, then how do you multiply to get it equal to 2? Do you divide 2 by 2? I am a bit confused.

Here is the question:

29.    Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method.
----->( u said don't use the half rxn method, so I didn't)

a. I-(aq) + ClO-(aq) ---> I3-(aq) +Cl-(aq)
   

I would say that you should try to use the other method and see if that helps. I still tried to use the ON method and I think what happened is that you forgot to do one part of a step. Step 2 explains that you must calculate the ON of elements and B atoms undergoing change in ON. If you remove everything undergoing reduction, for now, you will get:

I- --> I3-

Iodine atoms are not balanced and you would have to put a 3 in front of I- (to balance out all of the Iodine atoms). If you add everything back together you would get:

3I-(aq) + ClO-(aq) ---> I3-(aq) + Cl-(aq)

Now you can finish the rest of the steps knowing that there are 2 electrons lost in oxidation and 2 electrons gained in reduction.

yeahh, that's the step I forgot, thank you.

Your welcome .

 

wow !!!!