Chemical Forums
Specialty Chemistry Forums => Biochemistry and Chemical Biology Forum => Topic started by: billsauce on April 30, 2013, 12:05:58 AM
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construct an electron transport chain using 3 carriers listed, so that electrons transfer spontaneously?
CoQ/CoQH2 = +0.10V; Fe3+/Fe2+ = +0.771 V; FAD/FADH2 = -0.18; NO2-/NH4+ = +0.44 V;
CO2/glucose = -0.43 V; CO2/CH4 = -0.24V; NAD+/NADH = -0.32 V; H+/H2 = -0.42 V;
O2/H2O = +0.815 V; pyruvate/lactate = -0.19 V; S0/H2S = -0.27 V.
Construct an electron transport chain using 3 carriers listed above, so that electrons are transferred spontaneously from one carrier to the next:
Carrier #1 _____________________________, Carrier #2 ___________________________, Carrier #3 ________________________________
I am not exactly understanding what the question asks. This is what I came up with.
Fe3+ +e- -------> Fe2+ +.771V
FADH2 ---------> FAD+ +2e- +.18V
NAD+ + e- -----> NADH -.32V
Total: all electrons cancel, total V = +.631
Is this correct? so my carriers are Fe3+/Fe2+, FAD+/FADH2, NAD+/NADH?
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Just glancing at this, I don't think you are adding a subtracting your standard half-cell potentials correctly. I am also not certain that you have a good grasp of what the half-cell potentials tell one about whether or not the reaction will proceed as written.
Suppose I have two half reactions
A1ox + e- :rarrow: A1red E°' = 0.200
A2ox + e- :rarrow: A2red E°' = 0.300
Would the following reaction be thermodynamically spontaneous?
A1ox + A2red :rarrow: A1red + A2red
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how am I not adding them together correctly?
and no that is not spontaneous
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billsauce, I believe the question is obliquely asking you to identify two reactions, each of which is spontaneous, via the form:
A :rarrow: B, ΔG < 0
B :rarrow: C, ΔG < 0
Your three carriers would be A, B, C, which would effectively (spontaneously) shuttle electrons from one place to another because the overall reaction is "thermodynamically downhill".
All you've done is identify half-reactions, but electron transport requires redox activity between two species: an oxidized form that becomes reduced and a reduced form that becomes oxidized. You need to identify combinations that lead to a spontaneous chain (ΔG < 0), and you need to be clever about it because the product of one of them has to react spontaneously with the reactant of the other.
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billsauce, I believe the question is obliquely asking you to identify two reactions, each of which is spontaneous, via the form:
A :rarrow: B, ΔG < 0
B :rarrow: C, ΔG < 0
Your three carriers would be A, B, C, which would effectively (spontaneously) shuttle electrons from one place to another because the overall reaction is "thermodynamically downhill".
All you've done is identify half-reactions, but electron transport requires redox activity between two species: an oxidized form that becomes reduced and a reduced form that becomes oxidized. You need to identify combinations that lead to a spontaneous chain (ΔG < 0), and you need to be clever about it because the product of one of them has to react spontaneously with the reactant of the other.
ok, that makes sense, but what would be "B" in your A, B, C equation because none of them are the same?
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so would this work:
CoQ + 2e- ------> CoQH2 +.10V
FADH2 ---------> FAD+ +2e- +.18V
-----------------------------------------
CoQ + FADH2 -------> CoQH2 + FAD+ +.28V
acceptor donor
Then:
FAD+ + 2e- -------> FADH2 -.18V
H2 ------------> 2e- + H+ +.42V
-------------------------------------
FAD+ + H2 ---------> H+ + FADH2 +.24V
acceptor donor
which would make my carriers CoQ/CoQH2, FAD+/FADH2, H+/H2 and a total of +.52V so its spontaneous
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I didn't check your numbers but you've got the right approach now.
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so i can basically take any 3 carriers I want and just use the A, B, C approach and make A to C spontaneous?
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and i can write my carriers like this and it would make sense that this is spontaneous?
carrier 1 CoQ/CoQH2
carrier 2 FAD+/FADH2
carrier 3 H+/H2
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so i can basically take any 3 carriers I want and just use the A, B, C approach and make A to C spontaneous?
Yes it's possible there is more than one correct answer. As long as each reaction has a negative Gibbs energy change, you have satisfied the question.
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do i have to reverse the way carrier 2 and carrier 3 are written so I show that i reversed the half reaction?
for example:
CoQ + 2e- ------> CoQH2 +.10V
FADH2 ---------> FAD+ +2e- +.18V <----------------------- this one i had to reverse
-----------------------------------------
CoQ + FADH2 -------> CoQH2 + FAD+ +.28V
acceptor donor
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billsauce, I may not following you here. Are the electrons in your system flowing from hydrogen gas to CoQ, or are they flowing from CoQH2 to protons?
EDT
The electrons seemed to be flowing the wrong way in your first attempt at this problem.
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Personally I don't think it really matters, but it can't hurt to include the full reactions in your written answers. This way you remove all ambiguity.