Question

Using the values listed in the table provided, determine the deltaG associated with teh reaction listed below:

NADPH + 1/2O2 + H+ --> NADP+ +H2O

(V) Half-Reaction 0.815 O, 2 H 2 e H3O SO 2 H+ 2 e SO H2O 0.48 NO3 2 H 2 e NO3 0.42 Cytochrome a3 (Fe3+) e cytochrome a (Fe2+) 0.385 O2 (g) 2 H+ 2 e 2O2. 0.295 Cytochrome a (Fes e cytochrome a (Fe2+) 0.29 Cytochrome c (Fe3+) e cytochrome c (Fe 0.235 tochrome c (Fe3+) e cytochrome cr (Fe2+) 0.22 Cytochrome b (Fest) e cytochrome b (Fe2+) mitochondrial 0.077 Ubiquinone 2 H 2 e ubiquinol 0.045 Fumarate 2 H+ 2 e succinate 0.031 FAD 2 H+ 2 e FADH2 (in flavoproteins) Oxaloacetate 2 H 2 e malate 0.166 Pyruvate 2 H 2 e lactate 0.185 Acetaldehyde 2 H 2 e ethanol 0.197 FAD 2 H 2 e FADH ree coenzyme) 0.219 S 2 H+ 2 e H S 0.23 Lipoic acid 2 H 2 e dihydrolipoic acid 0.29 NAD+ H 2 e NADH 0.315 NADP H+ 2 e NADPH 0.320 Cystine 2 H+ 2 e 2 cysteine 0.340 Acetoacetate 2 H+ 2 e B-hydroxybutyrate 0.346 0.421 e Acetate 3 H 2 e acetaldehyde H20 0.581

Answer 1

1/2 O₂ + H⁺ +2e^- = H₂O E⁰ = 0.815

this reaction is electron acceptor reaction

NADP⁺ + H⁺ + 2e^- --> NADPH E⁰ = -0.320

this reaction is electron donor reaction

E°cell = E°cathode - E°anode.

E^o = E (acceptor) - E (donor)

=0.815 - (-0.320)

= 1.135

∆G=-nFE°

n = number of elcetron

F= faraday constant = 96458 C/mol e^-

E^o = 1.135

∆G=-nFE°

- 2* 96458​ * 1.135

= - 218959 kJ

So the ∆G = -218959kJ