Posted by Ken on Sunday, July 27, 2008 at 6:43pm.
The I in I2 is zero and it changes to +1 on the right.
The I in KIO3 is +5 and it goes to +1 on the right.
The easiest way to do this is to set up two half equations.
I2 ==> 2ICl
IO3^- ==> ICl
=================
Balance those two, multiply by the appropriate numbers, add them, then cancel anything common to both sides.
Ok, here is my equation on this one.
I2 -----> I^+1 + 1e-
4e- + I^+5 ----> I^+1
4(I2 ----> I^+1 +1e-)
4e- +I^5 ----> I^+1
on this problem, the e- balance, but when I start trying to balance the equation, I am not getting the same amount of I or Cl on each side. I can add H^+ or OH- or H2O, but I don't think I can add more elements. Please help me!
I would do it this way.
I2 ==> 2ICl
IO3^- ==> ICl
===================
You CAN add ions if they are spectator ions (they don't change oxidation state) and you want a molecular equation instead of an ionic equation.
For I2 ==> 2ICl
I2 is zero on the left: +1 on the right(for each) or +2 for both. (That's the first mistake you made). Now we add electrons.
I2 ==> 2ICl + 2e
Now I would add the Cl^- like so.
I2 + 2Cl^- ==> 2ICl + 2e
Adding 2Cl^- is the only way to get them into the equation.
Charge on left is -2 (from 2Cl^-) and on the right is -2 (from 2e) and this half is balanced.
The other one is the following and you need to do it. Here is what you should get.
6H^+ + 4e + Cl^- + IO3^- ==> ICl + 3H2O
Following those steps I wrote in the previous problem will get these things done almost every time. I have found, through the years, one or two that I've had trouble with when I use this method.