Redox potential (Using the Nernst Equation)

Click For Summary
SUMMARY

The redox midpoint potential (Eo’) of NAD+/NADH at pH 7.0 is established at -320 mV. Given a total concentration of 1.5 mM for {[NAD+] + [NADH]}, the cellular redox potential is approximately -200 mV. Under oxidative stress, this potential increases to -100 mV, affecting the concentrations of NAD+ and NADH. The Nernst equation, Eh = Eo' - (RT/nF)ln(reduced/oxidized), is crucial for calculating these concentrations, where "reduced" refers to NADH and "oxidized" refers to NAD+.

PREREQUISITES
  • Understanding of the Nernst equation and its application in electrochemistry.
  • Knowledge of redox reactions and the roles of NAD+ and NADH.
  • Familiarity with the concepts of oxidation and reduction in biochemical contexts.
  • Basic grasp of thermodynamic principles, including the constants R and F.
NEXT STEPS
  • Study the application of the Nernst equation in biochemical systems.
  • Explore the impact of oxidative stress on cellular metabolism and redox states.
  • Learn about the calculation of redox potentials in different pH environments.
  • Investigate the significance of NAD+/NADH ratios in cellular respiration and metabolism.
USEFUL FOR

Biochemists, molecular biologists, and students studying cellular metabolism and redox biology will benefit from this discussion, particularly those focused on the roles of NAD+ and NADH in metabolic pathways.

Gannon
Messages
73
Reaction score
0
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!
 
Physics news on Phys.org
"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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 2 ·
Replies
2
Views
11K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
4
Views
6K
  • · Replies 1 ·
Replies
1
Views
10K
Replies
3
Views
10K
Replies
2
Views
6K