CHEMISTRY COMMUNITY

Ok so I know that there's a typo for 6K3.D where the one of the products is meant to be 2Cl^- rather than Cl₂.

For the oxidation reaction I have:
Cl₂ + H₂O -> HclO + H⁺ + e^-

For the reduction reaction I have:
Cl₂ + 2e^- -> 2Cl^-

so when balancing the electrons you multiply the oxidation reaction by two (to cancel the electrons), and then when you add them together I get:
3Cl₂ + 2H₂O -> 2HclO + 2Cl^- + 2H⁺

This is definitely not the right answer but I'm not sure where I went wrong. Is anyone able to help me out here?

Check your oxidation half-reaction; it's not balanced. You have a 2 chlorines on the reactant side and 1 chlorine on the product side.

Can someone explain why we can write Cl2 -> 2 Cl- please?

I believe you write Cl2 --> 2Cl- in the products because this is the only way to go about doing the reduction. If you leave it as Cl2 --> Cl2 nothing is really happening in the reaction and rewriting it is equivalent but now you have a charge to work with.

so fixing the balancing error I did get the right answer.

also, we can do Cl2 -> 2 Cl- because there was a typo in the book and the equation was supposed to be
Cl2 -> HClO + Cl- where we had to balance the extra H+ and waters.