Chem II Oxidation Reduction
posted by Ken on .
I2 + KIO3 + HCl > KCl + ICl
Ok, this is my thinking . . . .
I^+5 + 4e^ > I^+1
Cl^1 +1e^ > 2Cl^1
I know that this is wrong, because the e^ are all on the same side. What am I doing wrong?

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.