Question

Calculate the electromotive force (in volts) registered by an electrode immersed in a solution containing

the following mixtures of NAD+

and NADH at pH 7.0 and 25oC, with reference to a half-cell of E’

o

0.00 V.

(a) 1.0 mM NAD+

and 10 mM NADH

(b) 1.0 mM NAD+

and 1.0 mM NADH

(c) 10 mM NAD+

and 1.0 mM NADH​

Answer 1

Equation used:

E = $E^{0}$ + $\frac{RT}{nF} ln \frac{NAD^{+} }{NADH}$

At 25 $\frac{RT}{nF}$ simplfies to 0.026 $\frac{V}{n}$

Explanation

• NAD/NADH redox pair is -0.320 V. Because two electrons

        are transferred, n = 2. Thus,

         E = (-0.320 V) + (0.026 V/2) ln $\frac{1* 10^{-3} }{10 * 10^{-3} }$

            = -0.320 V + (-0.03V) = -0.35V

• The conditions specified here are “standard conditions,” so we expect that E = $E^{0}$. As proof, we know that ln 1 = 0, so under standard conditions the term [(0.026 V/n) ln 1] = 0, and E = $E^{0}$ = -0.320 V.

• Here the concentration of NAD (the electron acceptor) is 10 times that of NADH (the electron donor). This affects the value of E:

         E = (-0.320 V) + (0.026/2 V) ln $\frac{1* 10^{-3} }{10 * 10^{-3} }$

            = -0.320 V + 0.03V = -0.29V