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In the course reader and in lecture we discussed how sometimes two half reactions will both be reductions, leading us to flip the sign and order of one of the chemical equations. In lecture we did this with zinc and copper and when we flipped the copper reaction to make it the anode we got E=-0.77 and E for zinc stayed at 0.34. To calculate E of the whole process we did .77-.34 but since E standard is Ecathode-Eanode wouldn't it be 0.34-(-0.77)? I am very confused on this and I believe the same thing was done in the gold reaction a couple of pages later.
Professor Lavelle mentioned the two distinct methods during lecture. One is to use the equation Ecell=Ecathode-Eanode, the other is to add the standard reduction potentials for the two half equations after you flip the anode reaction (and sign for Eanode) to account for oxidation. Both methods give you the same answer, but you can only use one. The subtraction in the equation already accounts for the sign change, so there is no need to flip the sign before plugging it into the equation.
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