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I think that, in this case, we cannot add the E*'s together because the number of moles of electrons are not the same in the half-reactions (1e- vs. 3e-). According to the solutions manual, we have to use the deltaG*=-nFE* equation, to find the deltaG* of each individual half-reaction. Then we add these together to find the overall reaction's delta G*, which can be replaced by -4 (4 as the number of moles of e- for the reaction given in the problem) times Faraday's constant (F) times E*total (what we are trying to find). We divide the added half-reaction deltaG*'s by -4 and F to get the final answer of E*total.
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