Question

Calculate the hybrid power for oxide (II) in the reduction of hydrogen: AGCO 129.9 kJ / mol, AG1120 = -273.3 kJ / mol.
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Answer 1

C

5

H

12

+8O

2

⟶5CO

2

+6H

2

O

ΔG=G(products)−G(reactants)=5×(−394.4)+6×(−237.2)−(−8.2)=−3387Kjmol

−1

To get E

cell

use the equation:

ΔG=−nFE

cell

As O is moving from 0 to -2 oxidation state, n = 32 for this reaction

−3387×1000=−32×96500×E

cell

E

cell

=1.0968V