Nicotinamide-adenine dinucleotide (in its two redox forms NAD+ and NADH) is an i

Question

Nicotinamide-adenine dinucleotide (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 What is the value of this potential versus Ag/AgCl electrode: repeat same question but versus the SCE reference electrode 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 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 1/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 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

Ans

(a)

Ag/AgCl electrode has reduction potential of 0.23 V against SHE , so value of this potential vs Ag/AgCl electrode = -(0.230 + 0.1128) = -0.3428 V

SCE electrode has reduction potential of 0.244 V against SHE , so value of this potential vs SCE electrode = -(0.244 + 0.1128) = -0.3568 V

(b)

Potential of this couple vs itself is 0.

Potential of SCE vs NAD+/NADH is +0.3568 V

Potential of SCE vs NAD+/NADH is +0.3428 V

(c)

Nernst eqn :

E = Eo - 0.0591/n log Q , here Q = reaction qoutient = [NADH] / [NAD+][H+]

Thus using these relations, since the ratio NAD+/NADH will be known to us, and we can measure the potential E via potentiometer, and Eo is known to us, so only [H+] is unknown which can thus be calculated. Thus pH can be determined.

(d)

A = e*l*c

Assuming path length of 1,

At 260 nm, A260 = 0.4 = A1 + A2 = 18000*1*c1 + 14400*1*c2

At 340 nm, A340 = 0.2 = A2 = 14400*1*c2

Here subscript 1 denotes NAD+

Thus, c2 = 1.38*10-5 and c1 = 2.22*10-5

Since pH = 7 , so [H+] = 10-7

Putting these values, E = -0.1128 - 0.0591/2 log((1.38*10-5)/(2.22*10-5)*(10-7)) = 0.200 V

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