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Topic: Calculate reduction potential help  (Read 1160 times)

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Offline MJF

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Calculate reduction potential help
« on: August 02, 2013, 11:55:38 AM »
In lecture, we were talking about electrochemistry and calculating the reduction potential. Its been a bit confusing since there are two situations here that our professor illustrated on the board:

Lets say in a voltaic cell consisting of Zn as the anode and Cu as the cathode:

Cu2+ + 2e- --> Cu(s)  E°cathode = +0.34v
Zn2+ + 2e- --> Zn(s)  E°anode =     -0.76v

E°cell = 0.34 - (-.76) = 1.10v

Another situation he presented is slightly different but obviously yielding the same result:

Cu2+ + 2e- --> Cu(s) E°1 = +0.34
Zn(s) --> Zn2+ + 2e-  E°2 = +0.76 (flipping the equation means changing the sign of the voltage
E°cell = 0.34 + 0.76 = 1.10v

I asked this question on another site and someone recommended that I don't change the sign even if the reaction is flipped, as it will complicate calculations. However there were times in class when we subtracted potentials and when we added the potentials (pretty much when we were dealing with redox equilibria problems)

Its sort of a silly question but I'm not sure when to do one and to do the other.

Here's another example:

Fe2+ + 2e- --> Fe(s) E° = -0.44
Al3+ + 3e- --> Al(s) E° = -1.18

According to our professor, the Fe will be the cathode because it is the most positive of the two. So when we calculate: E°cell = -0.44 - (-1.18) = 0.74

How will I know when to add or subtract?

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