C5H12(g)+O2(g)→5CO2(g)+6H2O(l) ΔG°=[(−394.4×5)+(−237.2×6)]−[(−8.2)+(8×0)] =−3387 kJ Note that the standard free energy change of elementary substances is taken as zero. For the fuel cell, the complete cell reaction is : C5H12(g)+8O2(g)→5CO2(g)+6H2O(l) Which is the combination of the following two half reactions: C5H12(g)+10H2O(l)→5CO2(g)+32H++32eand 8O2(g)+32H++32e→16H2O(l) Therefore, the number of electrons exchanged is 32 here, means n = 32. This is the trickiest part of the problem. ΔG°=nFE° −3387×103J=−32×96500