The value for n depends upon how you solve the problem. If you use the total balanced equation as you have it written, then n is the total number of electrons (6 in this case) exchanged. If you decide to solve it by calculating the half cell potential for EACH, then adding the two half cells to obtain the total reaction (in which case you DON'T multiply Cu E by 3 and Cr E by 2) then n is 2e for the copper half and 3e for the Cr half. As for the anode/cathode part,
if you work using the entire equation, if E is +, then Cr will be the anode (anode is where oxidation occurs and Cr is losing electrons which makes it oxidized) but if E is -, the Cu will be the anode and Cr the cathode.
Calculate the Gibbs free energy change for the reaction below when initial concentrations of CR3+ and Cu2+ are 1 M (F= 96500 J/V mole e-)
2Cr3+ (aq) + 3Cu (s) --> 2Cr (s) + 3Cu2+ (aq)
I assume we use the deltaG= -nFE equation, but how do you know n when Cr is using 3 e- and Cu is using 2 e-. and how do you find E? which is the cathode/anode?
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