6K.3 part d

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804994652
Posts: 68
Joined: Fri Sep 28, 2018 12:26 am

6K.3 part d

Postby 804994652 » Sun Mar 10, 2019 7:46 pm

In the 7th edition, 6K #3 asks that you balance the following redox rxn in an acidic solution and find the oxidizing/reducing agent.
Part d had the equation) Cl2 (g) --> HClO (aq) + Cl2 (g)

The answer is H2O (l) + Cl2 (g) --> HOCl (aq) + H+ (aq) + Cl- (aq)
where Cl2 is both the oxidizing and reducing agent...

How do you do this problem?

Lexie Baughman 2C
Posts: 30
Joined: Sat Oct 06, 2018 12:16 am

Re: 6K.3 part d

Postby Lexie Baughman 2C » Sun Mar 10, 2019 8:44 pm

You do the problem as you would any other redox reaction problem. Obtain the balanced half reactions, multiply the oxidation and reduction half-reactions by appropriate factors that will result in the same number of electrons present, and then add the half reactions. In this case, Cl2 just happens to be oxidized and reduced.

804994652
Posts: 68
Joined: Fri Sep 28, 2018 12:26 am

Re: 6K.3 part d

Postby 804994652 » Sun Mar 10, 2019 11:03 pm

So you would set it up the half reactions like: Cl2 (g) --> HClO (aq) + Cl2 (g)
Cl2 + 2 H2O --> 2 HCLO(g) + 2H+
Cl2 --> Cl2 (g)??


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