14.37 (c)


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Andy Liao 1B
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Joined: Thu Jul 13, 2017 3:00 am

14.37 (c)

Postby Andy Liao 1B » Sun Feb 18, 2018 12:00 am

Problem:
14.37 Determine the potential of each of the following cells:
(c) Pt(s) | Cl2(g, 250 Torr) | HCl(aq, 1.0M) || HCl(aq, .85M) | H2(g, 125 Torr) | Pt(s)

The cathode half-reaction for this cell reaction is 2H+(aq) + 2e- -> H2(g). I was wondering where the Cl from the HCl went. Can someone please explain why the Cl was omitted?

sahiltelang-Discussion 1J
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

Re: 14.37 (c)

Postby sahiltelang-Discussion 1J » Sun Feb 18, 2018 9:49 am

The Cl is the oxidation step such that Cl- becomes Cl and loses an electron as it forms a covalent bond with another Cl that has also lost an electron.


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