6K.3 Part D

[Chloe Alviz 1E]

"Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in acidic solution. Identify the oxidizing agent and reducing agent in each reaction."

d) Cl₂(g) ––> HClO(aq) + Cl₂(g)

For this reaction, why would Cl₂ be both the reducing and oxidizing agent if it has a zero charge?

[Savannah Mance 4G]

The half rxn of Cl2--->Cl goes from a charge of 0 to a charge of -1 meaning the Cl is reduced. And in the other half rxn of Cl2--->HClO the chlorine is going from a 0 charge to a +1 charge and is oxidized. This is why it is both oxidized and reduced.

[Sue Bin Park 2I]

to add on, the charge of the reactant on its own doesn't indicate anything about whether or not it is an oxidising or reducing agent; it is rather its relation to the charge of the counterpart in the products that indicates it.
here, the Cl becomes both positively and negatively charged after the arrow, so we assume that one of the Cls has effectively donated an electron to the other Cl (a redox process).

[Brittney Hun 2C]

So how does the charge of cl go from 0 to -1, don't both the cl2 have an oxidation number of 0?

[Sue Bin Park 2I]

So how does the charge of cl go from 0 to -1, don't both the cl2 have an oxidation number of 0?

the way i figured was the e- had to come from somewhere. this takes place in an acidic solution so theres no free OH-s to take from, it must go to Cl-
but thats a valid question, honestly i made the leap and the answer key confirmed it. i'm 75% sure its just a typo from 2Cl- into Cl2. do you have the student solutions manual by any chance? i know the answer key in the back has a lot of typos

[Savannah Mance 4G]

Yeah, I'm pretty sure it's a typo and it's supposed to be Cl2 to Cl- not Cl2 to Cl2.