Write the reduction half reaction for the redox reaction below.

ClO-(aq) + I-(aq) => Cl-(aq) + I2(s)

Is I- being reduced since Cl is losing an O?

No, its being oxidized. It has a lower reduction potential.

These are the half reactions from a standard half reaction chart:

I2 + 2e- -> 2I-

ClO- + 2e- -> Cl-

**I2 is being oxidized, so you need to reverse the reaction:

2I^- <- - -> I2 + 2e

ClO- + 2e- -> Cl-

Electrons are balanced.

Ok I had the wrong idea, thanks so much!

I posted an answer/solution to another problem that you had. I think that it explains the concept of what you are suppose to be doing a little bit better, or it can be used as an additional resource to help you with similar problems.

I would like to add here that you had the right idea if you had followed through. Your statement that ClO is losing O is ok but since it is LOSING O is must be reduced. Gaining O is oxidation. Of course if O is nowhere in the oxidation it's tough to use that as a definition which is why you should stick to the definition that oxidation is the loss of electrons and reduction is the gain of electrons. Note that Cl is +1 oxidation state and Cl^- is -1 so the Cl has gained two electrons which makes it reduced.

To determine which species is being reduced in the given redox reaction, we need to compare the oxidation states of iodine (I) in the reactants and products.

In the reactant, iodine has an oxidation state of -1, as it is present as the iodide ion (I-). In the product, iodine has an oxidation state of 0, as it is present as elemental iodine (I2).

Since the oxidation state of iodine has decreased from -1 to 0, we can conclude that iodine is being reduced in this reaction.

Now, let's write the reduction half reaction by focusing on the species being reduced:

I-(aq) => I2(s)

In the reduction half reaction, the iodide ion (I-) is gaining two electrons (2e-) and being converted to elemental iodine (I2).

Therefore, the reduction half reaction for the given redox reaction is:

I-(aq) => I2(s)