Redox potential (Using the Nernst Equation)

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The redox midpoint potential (Eo’) for NAD+/NADH at pH 7.0 is -320 mV, with a total concentration of 1.5 mM for both species. The cellular redox potential is approximately -200 mV, and under oxidative stress, it increases to -100 mV. The confusion arises regarding the terms "oxidized" and "reduced" in the Nernst equation, which refers to NAD+ and NADH, respectively. Issues with calculations may stem from incorrect units for R and F, as they should yield about 60 mV/n, which is crucial for accurate concentration estimates.
Gannon
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The redox midpoint potential (Eo’) of NAD+/NADH at pH 7.0 is -320 mV. If the total concentration of {[NAD+] + [NADH]} is 1.5 mM in cells, and the cellular redox potential is about -200 mV, what are the concentrations of [NAD+] and [NADH], respectively? Under certain oxidative stress conditions, the cellular redox potential is increased to -100 mV. What will be the concentrations of [NAD+] and [NADH] in this cell under the oxidative stress conditions? (Assume pH 7.0 is not changed under oxidative stress conditions).

Eh = Eo' - (RT/nF)ln(reduced/oxidized) but what does he mean by "oxidized," the molecule being oxidized (NADH) or what has been oxidized (NAD+)?

When I do this problem I keep getting an extremely small numbers that do me no good in the end; what am I doing wrong here? I'll just show the first part of the question (Eh = -200mV).

Eh = Eo' - (RT/nF)ln(reduced/oxidized)
Eh = -200E-3V, Eo' = -320E-3V
-200E-3V = -320E-3V - [(1.987E-3)(298)/(2)(23.1)]ln([red]/[ox])
120E-3V = (-0.01174)ln([red]/[ox])
-10.2 = ln([red]/[ox])
e^(-10.2) = [red]/[ox]
6.30957344E-11[ox] = [red]

[ox] + [red] = 1.5E-3M (total)
6.30957344E-11[ox] +[ox] = [red]
From here, because the coefficient for [ox] is so small I get [ox] = [red].

Any help is greatly appreciated!
 
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"reduced/oxidized" means the concentration of the reduced form of the ion or atom or specie divided by the oxidized form of the ion or atom or specie.
 
Gannon said:
The redox midpoint potential (Eo’) of NAD+/NADH at pH 7.0 is -320 mV. If the total concentration of {[NAD+] + [NADH]} is 1.5 mM in cells, and the cellular redox potential is about -200 mV, what are the concentrations of [NAD+] and [NADH], respectively? Under certain oxidative stress conditions, the cellular redox potential is increased to -100 mV. What will be the concentrations of [NAD+] and [NADH] in this cell under the oxidative stress conditions? (Assume pH 7.0 is not changed under oxidative stress conditions).

Eh = Eo' - (RT/nF)ln(reduced/oxidized) but what does he mean by "oxidized," the molecule being oxidized (NADH) or what has been oxidized (NAD+)?

When I do this problem I keep getting an extremely small numbers that do me no good in the end; what am I doing wrong here? I'll just show the first part of the question (Eh = -200mV).

Eh = Eo' - (RT/nF)ln(reduced/oxidized)
Eh = -200E-3V, Eo' = -320E-3V
-200E-3V = -320E-3V - [(1.987E-3)(298)/(2)(23.1)]ln([red]/[ox])
120E-3V = (-0.01174)ln([red]/[ox])
-10.2 = ln([red]/[ox])
e^(-10.2) = [red]/[ox]
6.30957344E-11[ox] = [red]

[ox] + [red] = 1.5E-3M (total)
6.30957344E-11[ox] +[ox] = [red]
From here, because the coefficient for [ox] is so small I get [ox] = [red].

Any help is greatly appreciated!

What units are you using for R and F?
 
R is 1.987E -3 kcal/mol/degree; F is 23.1 kcal/V/mol.
 
RT/nF should give about 60 mV/n, it doesn't.
 
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