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Determined Member
Nov 28, 2012
Okay so in the official MCAT guide Gen Chem section q 27 asks

Given the half reactions
Ag+ + e ---> Ag E0= .80V
Cu----> Cu 2+ + 2e E0=-.34 V

Overall Rxn
Cu + 2Ag+---> Cu2+ + 2e- What is the E Cell

I said Cu is the anode because its getting oxidized and Ag is the cathode becuase it is getting reduced. So Cathode-Anode should give the E0 Cell to be .8-(-.34)= 1.14 V.

However, the correct answer is .46 V. Which would result from just adding the two together.

So is there a difference between when you should subtract Cathode-Anode or use Oxidation +Reduction? This is driving me nuts!!
Dec 23, 2013
yes most of this is right. Except, when using Ecell= Ecathode- Eanode both E are the reduction potential.. so therefore Cu--->Cu2+ + 2e- is the oxidation potential, and you need to reverse it Cu2+ +2e- --> Cu to get the reduction potential which changes -.34 V to positive .34V when you reverse it. Thus, using the equation:
Ecell= Ecathode- Eanode
Ecell= 0.80-0.34= 0.46 V
You can only use this equation with reduction potentials, so make sure everything is written as a reduction.
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