#14.13 Part C (Sixth Edition)

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gabbym
Posts: 76
Joined: Wed Apr 11, 2018 3:00 am

#14.13 Part C (Sixth Edition)

Postby gabbym » Tue Feb 26, 2019 1:48 pm

Can someone help me with #14.13 Part C? It asks you to write the half-reactions, the balanced equation for the cell reaction, and the cell diagram for each of the following
skeletal equations:
(c) Cl2(g) / H2(g) --> HCl(aq)
I don't understand how they are getting the half reactions. Can someone explain it to me?

Lynsea_Southwick_2K
Posts: 55
Joined: Fri Sep 28, 2018 12:25 am

Re: #14.13 Part C (Sixth Edition)

Postby Lynsea_Southwick_2K » Tue Feb 26, 2019 4:13 pm

I did one half rxn in regards to the Cl2(g) and then the other in regards to H2(g), then when you combine the half rxns you get 2H+(aq) and 2Cl-(aq) on the product side which is also 2HCl

gabbym
Posts: 76
Joined: Wed Apr 11, 2018 3:00 am

Re: #14.13 Part C (Sixth Edition)

Postby gabbym » Wed Feb 27, 2019 2:14 pm

But I'm not understanding how you arrive at that conclusion for the half reactions. How do you know to separate the HCl into Cl- and H+? I would have just put that Cl2 goes to HCl and then H2 goes to HCl

annabel 2A
Posts: 67
Joined: Fri Sep 28, 2018 12:18 am

Re: #14.13 Part C (Sixth Edition)

Postby annabel 2A » Sat Mar 09, 2019 10:02 pm

I think that you're supposed to assume that HCl dissociates into its ions in an aqueous solution because it is a strong acid, so you're supposed to see the equation as Cl2(g) + H2(g) --> 2H+(aq) + 2Cl-(aq)


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