Dr. Bob222

Question

I still didn't understand which is the reduction half reaction and its products and oxidation half reaction...and why is it O3 to IO3-...0 to -6 and then
Please help...entrance exam soon
Write the oxidation and reduction half reactions(showing e- lost or gained) for the foll
O3 + I- -> IO3- + O2 (acidic)

Answer 1

Maria--I tried to find your initial post and my response but failed. Here is what I remember about it. This is the most head scratching question I've seen in my career. The I^- is oxidized to IO3^- so you can write that half reaction easily. It is
I^- + 3H2O ==> IO3^- + 6e + 6H^+
That means I^- is oxidized to IO3^-. Everyone knows that ozone (O3) is an oxidizing agent; however, if you write
2O3 ==> 3O2 there is absolutely no evidence at all of an electron change from that equation. O3 is a neutral molecule; O2 is a neutral molecule so both are zero. I went searching on the web. There is an alternative definition of redox that people used 50-60 years ago (and I remember it from my freshman years in college) that says that reduction is the loss of oxygen so we write the O3 ==> O2 this way to show it loses oxygen.
O3 + 2H^+ + 2e ==> H2O + O2 and that is balanced.(I can go through that step by step if necessary). The ozone is reduced (it no only gains electrons but it loses oxygen and that satisfies BOTH definitions) So you multiply the O3 equation by 3 and the I^- equation by 1, add them, cancel those items that appear on both sides of the equation, and it all balances in the end. That will be
I^- + 3H2O + 3O3 + 6H^+ 6e ==> IO3^- + 6e + 6H^+ + 3H2O + 3O2

3H2O on the left cancels with 3H2O on the right. 6H^+ on the left cancels with 6H^+ on the right. 6e on the left cancels with 6e on the right and you are left with
I^- + 3O3 ==> IO3^- + 3O2

Check:
one I on left and right.
9 O on left and right.
-1 charge on left and right.

Everything balances.
So O3 is reduced to O2; I^- is oxidized to IO3^-.

I hope this is clear.
I hope this helps.