6K3(d) 7th edition

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6K3(d) 7th edition

Postby paytonm1H » Sun Feb 24, 2019 11:37 pm

I'm working on part d of 6K3. It asks to balance the reaction of chlorine in water
Cl2(g) --> HClO(aq) + Cl2(g)

I got the oxidation half rxn as Cl2 + 2H20 --> HClO + 2H+

the final answer from the key is H20(l) + Cl2(g) --> HOCl(aq) + H+(aq) + Cl1(aq)

I don't understand why the Cl2 in the reactants changed to Cl- in the final product. Can someone explain this please? Thanks!

Matthew Choi 2H
Posts: 59
Joined: Fri Sep 28, 2018 12:18 am

Re: 6K3(d) 7th edition

Postby Matthew Choi 2H » Mon Feb 25, 2019 1:48 am

When Cl2 is broken up into the constituent Cl ions, then one will obviously react with water to create HClO. The remaining Cl will react with the H+ ion leftover from the creation of HClO to form HCl. But since HCl is a strong acid, it immediately dissociates into H+ and Cl-. This, I believe, is why Cl- exists on the products side.

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