Problem 14.23 (b)

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Gavin Kellerman 1D
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Joined: Sat Jul 09, 2016 3:00 am

Problem 14.23 (b)

Postby Gavin Kellerman 1D » Mon Feb 19, 2018 2:55 pm

For this problem, how are we figure out that Br is the anode and Hg is the cathode?

melissa carey 1f
Posts: 53
Joined: Fri Sep 29, 2017 7:06 am

Re: Problem 14.23 (b)

Postby melissa carey 1f » Mon Feb 19, 2018 2:59 pm

Br is the anode since its losing electrons, going from Br- to Br2(l). Br2(l) has an oxidation number of 0, while Br- has an oxidation number of -1. The reverse logic can be used on Hg.

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