PLEASE ANSWER ALL PARTS SO I CAN UNDERSTAND THESE PROBLEMS AS A WHOLE. THANK YOU
ID: 1007723 • Letter: P
Question
PLEASE ANSWER ALL PARTS SO I CAN UNDERSTAND THESE PROBLEMS AS A WHOLE. THANK YOU!!!
NAD+ + H+ + 2e- --> NADH (E^o = -0.3 V)
FAD+ + 2H+ + 2e- --> FADH2 (E^o = -0.1 V)
1/2O2 + 2H+ + 2e- --> H2O (E^o = 0.8 V)
(1) Which would be the favorable spontaneous reaction that would occur with the two half-reactions involved would be:
a. FAD + 1/2O2 2H+ --> FADH2 + H2O
b. FADH2 + 1/2O2 --> FAD + H2O
c. FADH2 + H2O --> FAD + 1/2O2 + 2H+
d. FAD + H2O --> FADH2 + 1/2O2
e. More than one is correct, OR there is no way to answer this with the info given
(2) According to the information, which of these is the best reducing agent?
a. NADH
b. FADH2
c. H2O2
d. O2
e. H2O
(3) The reduction of FAD and O2 have E^o values of -0.1 and 0.8 V respectively. If deltaG^o = -nF(deltaE^o), how much energy is released when electrons move from FAD to oxygen via the ETC at 27 degrees Celcius? (n= # of electrons and F= 23 kcal/mole)
a. 1.5
b. 2.5
c. around 30
d. around 50
e. around 700
Explanation / Answer
(1)
As we know,
E = Ereduction - Eoxidation
as well we know that,
delta G = -nFE where G= Gibb's free energy F= faraday constant n= no of electrons
If G value is negative, we can say the reaction is spontaneous.
Therefore we have to find the E values of each reaction.
* E for reaction a)
E = Ereduction - Eoxidation
E = (-0.1)-0.8 = -0.9 V
* E for reaction b)
E= 0.8- (-0.1) = 0.9 V
* E for reaction c)
E = (-1)- 0.8 = -0.9 V
* E for reaction d)
E =(-1)- 0.8 = -0.9 V
For a spontaneous reaction, E should be positive. so the answer is (b).
(2)
NADH has a more positive oxidation potential and hence is a better reducing agent than others.The answer is (a).
(3)
E= (-0.1) - 0.8 =-0.9 V
Delta G = - 2 x 23 x (-0.9)
= 41.4 kcal/mol
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