Dr. Bob222

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)

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.

Sure, I can help you understand the oxidation and reduction half-reactions in the given chemical equation.

To identify the oxidation and reduction half-reactions, you need to determine which species is losing electrons (oxidation) and which is gaining electrons (reduction).

Let's break down the given equation:

O3 + I- -> IO3- + O2

In this equation, oxygen (O) has different oxidation states on both sides of the equation: O in O3 has an oxidation state of 0, and O in O2 has an oxidation state of -2. This means that oxygen is undergoing a change in oxidation state and is therefore being oxidized.

On the other hand, iodine (I) has an oxidation state of -1 on both sides of the equation. So, iodine is not undergoing any change in oxidation state and is neither oxidized nor reduced.

Now, I will explain the reduction half-reaction and the oxidation half-reaction step by step:

Reduction half-reaction:
The reduction half-reaction involves the species that gains electrons. In this case, IO3- gains electrons to form O2.

To write the reduction half-reaction, we need to balance the number of atoms and the overall charge. In the reduction half-reaction, the IO3- ion is reduced to form O2. Since the IO3- ion gains 6 electrons (e-) in this process, we need to balance the equation accordingly:

IO3- + 6e- -> 3O2

So, the reduction half-reaction is IO3- + 6e- -> 3O2, indicating that IO3- gains 6 electrons.

Oxidation half-reaction:
The oxidation half-reaction involves the species that loses electrons. In this case, O3 loses electrons to form O2.

To write the oxidation half-reaction, we need to balance the number of atoms and the overall charge. In the oxidation half-reaction, O3 is oxidized to form O2. Since O3 loses 2 electrons (e-) in this process, we need to balance the equation accordingly:

O3 -> 2O2 + 2e-

So, the oxidation half-reaction is O3 -> 2O2 + 2e-, indicating that O3 loses 2 electrons.

Therefore, the overall balanced equation is obtained by combining the reduction and oxidation half-reactions:

O3 + I- -> IO3- + O2 (acidic)

Oxidation half-reaction: O3 -> 2O2 + 2e-
Reduction half-reaction: IO3- + 6e- -> 3O2

I hope this explanation helps you understand how to identify the oxidation and reduction half-reactions for a given chemical equation. Good luck with your entrance exam!