Calculating K and _Go from Eocell
Calculating K: the adjusted half reactions show that 2 mol e- are
transferred per mole of reactant as written, so n = 2:
Eocell = log K = log K = 0.93 V
0.0592 V
n
0.0592 V
2
so, log K = = 31.42 and K = 2.6 x 1031
(0.93 V)(2)
0.0592 V
Calculating _ Go :
_ Go = - nFEocell = x x 0.93 V
2 mol e-
mol rxn
_ Go = - 1.8 x 102 kJ/mol rxn
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