Chem II Oxidation Reduction

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?

asked by Ken
  1. 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.

    posted by DrBob222
  2. 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!

    posted by Ken
  3. 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.

    posted by DrBob222

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