# Chem II Oxidation Reduction

posted by Ken

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?

1. DrBob222

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.

2. Ken

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!

3. DrBob222

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.

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