Potentiometry and Spectroscopy: Nicotinamide-adenine dinuclcotide (in its two re

Question

Potentiometry and Spectroscopy: Nicotinamide-adenine dinuclcotide (in its two redox forms NAD+ and NADH) is an important biological molecule and is key in a number of electron transfer reactions needed for physiological functions. Consider the corresponding redox couple NAD^+/NADH NAD^+ + H^+ + 2e^- doubleheadarrow NADH (Couple NAD+/NADH) The formal potential of this half reaction is -0.1128 V vs. SHE a. What is the value of this potential versus Ag/AgCl electrode; repeat same question but versus the SCE reference electrode b. Under certain circumstances, this redox couple can be used as a reference to report other potential values. What is the potential of couple NAD+/NADH versus this same reference? What is the potential of SCE and Ag/AgCl vesus NAD+/NADH c. Potentiometry and spectrophotometry can be combined on special optical fibers that can also work as electrodes; NAD+ absorb light at 260 nm but not at 340 whereas NADH absorbs at both wavelengths; with fairly easy manipulation an experimentalist, using the optical fiber coated with known amounts of NADH, can access the concentration ratio of NAD+ to NADH. Explain/describe how this modified optical fiber can be used to potentiometrically determine the pH d. The optical fiber reports the following absorbance values at 260 nm and 340 nm A(@260) = 0.4 and A(@340) = 0.2 for the same path length (same fiber). The pH of the bathing solution is 7.00; Calculate the measured single-electrode potential for the NAD+/NADH. The molar absorptivity values for NAD+ and NADH are as follows:

Explanation / Answer

1)

Ag/AgCl electrode has a potential of 0.230 V against SHE, so the value of NAD+/NADH potential vs Ag/AgCl is :

-(0.230 + 0.1128) = -0.3428 V

SCE electrode has a potential of 0.244 V against SHE, so the value of NAD+/NADH potential vs Ag/AgCl is :

-(0.244 + 0.1128) = -0.3568 V

(2)

Potential of NAD+/NADH vs itself is zero.

Potential of NAD+/NADH vs Ag/AgCl is +0.3428

Potential of NAD+/NADH vs SCE is +0.3568

(3)

Nernst equation :

Ecell = Eocell - 0.0591/n log Q

Here n = 2 and Q = [NADH]/( [NAD+]*[H+] ) for this half cell

Since [NAD+]/[NADH] is also known to us, so the only unknown left in the above equation is [H+], which can be easily solved.

From this we get the pH as -log[H+]

(4)

A260 = e1*l*c1 + e2(260 nm)*l*c2 ---(1)

A340 = e2(340 nm)*l*c2 ---(2)

Here subscript 2 denotes NADH, and path length l = 1 cm

Putting the values known in the above two equations and solving we get :

c2 = [NADH] = 0.2 / 6220 = 3.21*10-5 M

Putting this in (1) we get :

c1 = [NAD+] = -3.5*10-6 M

Since conc. cannot be -ve, the absorbance reading at 260 nm has to be greater than 0.4

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