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Problem 14.23 (b)

Posted: Mon Feb 19, 2018 2:55 pm
by Gavin Kellerman 1D
For this problem, how are we figure out that Br is the anode and Hg is the cathode?

Re: Problem 14.23 (b)

Posted: Mon Feb 19, 2018 2:59 pm
by melissa carey 1f
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.