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

Answer:

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Answer 2

Answer:

The equation for calculating E for this problem is

E = E' + $\frac{RT}{nτ}$ ln $\frac{NAD+}{NADH}$

Explanation:

At 25 degree C, the RT/n term simplifies to 0.026 V/n.

(a) From Table 13–7, E' for the 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 (1 X 10⁻³) /(10 X 10⁻³)

  = - 0.320 V + (- 0.03 V) = - 0.35 V

(b) The conditions specified here are “standard conditions,” so we expect that E = E'. As proof, we know that ln 1 = 0, so under standard conditions the term [(0.026 V/n) ln 1] =n0, and E = E' = -0.320 V.

(c) 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 (10 X 10⁻³)/(1 X 10⁻³)

 = -0.320 V + 0.03 V = -0.29 V