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Posted: Mon Feb 19, 2018 6:00 pm
by Kayla Danesh 1F
14.31 Identify the reactions with K>1 in the following list and, for each such reaction, identify the oxidizing agent and calculate the standard cell potential.

(c) 2 Pb^2(aq)--->Pb(s)+ Pb^4+(aq)

Can someone please explain how to determine which of the two half reactions is for the anode and which is for the cathode? The half reactions I got are Pb(s)--->Pb^2+ + 2e- (+0.13V, anode) and Pb^4+ + 2e- ---> Pb^2+ (+1.67 V, cathode), but the textbook says that the anode and cathode should be the opposite, and that I should have -0.13V, not +0.13V. Thank you!

Re: 14.31c

Posted: Wed Feb 21, 2018 3:11 pm
by Curtis Wong 2D
Pb^4+ + 2e- ---> Pb^2+ should be the anode where Pb^4+ will be on the right side of the half-reaction. And this is mainly because of the fact that in the final equation, we can see hat Pb^4+ should be on the right side. Thus it is the anode.

2 Pb^2(aq)--->Pb(s)+ Pb^4+(aq)

Now for Pb(s)--->Pb^2+ + 2e- (+0.13V), I am assuming that you flipped the sign of the voltage because it's no longer being reduced, but being oxidized. Thus, when its the reverse half reaction it will be -.13V.