The standard reduction potentials for the reduction of fumarate and FAD are give

Question

The standard reduction potentials for the reduction of fumarate and FAD are given in the formula sheet, which is

FAD + (2H+) + (2e-) --> FADH2 -0.212 V

-OOCCH=CHCOO- + (2H+) + (2e-) --> -OOCCH2CH2COO- +0.045 V

Where delta rG = -49.6 kJ/mol

1) What would the concentrations of fumarate, succinate, FAD and FADH2 be when equilibrium is reached?

The answers are FAD = 2M, succinate = 2M, FADH2 = 9.04*10^-5M and fumarate = 9.04*10^-5M. However, I need help as to how they got those answers. Please and thank you.

Explanation / Answer

:FAD + (2H+) + (2e-) --> FADH2 -0.212 V,Eox=0.212V

-OOCCH=CHCOO- + (2H+) + (2e-) --> -OOCCH2CH2COO- +0.045 V

DELTAG=-nFE

-49.6=-2*96500*E

E=2.56*10^-4V

E=(Eox+Ered)+(0.059/n)log[( -OOCCH2CH2COO-)/(FAD )]

2.56*10^-4V=(0.212+0.045)+0.059/2log[( -OOCCH2CH2COO-)/(FAD )]

-8.7=log[( -OOCCH2CH2COO-)/(FAD )]

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