Question

Consider the following two half-cells. One half-cell (half-cell A) initially has 1.0 M each of 3PG and GAP; the other half-cell (half-cell B) initially has 1 M each of NAD+ and NADH. What is the value of ΔG°’ (in kcal/mol to the nearest tenths) when the two half-cells are connected by a salt bridge. To answer this question, you will need the standard reduction potential of NAD+ (which is given in your text), and you will need the standard reduction potential of 3PG (i.e., 3PG + 2H+ + 2e- --> GAP + H2O), which is not listed in your text but which I will provide for you. This value is -0.55 V.

Standard reduction potential for NAD+ -> NADH is -0.32 V.  

Answer 1

The half reactions are given as

3-PG + 2 H⁺ + 2 e^- --------> GAP + H₂O; E⁰_red = -0.55 V

NAD⁺ + 2 H⁺ + 2 e^- ---------> NADH + H⁺; E⁰_red = -0.32 V

Since NAD⁺/NADH has a lower reduction potential than PG/GAP couple, hence, NAD⁺ is preferentially reduced while 3-PG/GAP is oxidized.

Add the oxidation and the reduction half reactions to get the redox process as

NAD⁺ + 2 H⁺ + GAP + H₂O ---------> NADH + H⁺ + 3-PG + 2 H⁺

Cancel out the common terms and obtain

NAD⁺ + GAP + H₂O --------> NADH + 3-PG + H⁺

E⁰_cell = E⁰_red(NAD⁺/NADH) – E⁰_red(3-PG/GAP)

= (-0.32 V) – (-0.55 V)

= -0.32 V + 0.55 V

= 0.23 V

The standard free energy change for the reaction is given as

ΔG⁰’ = -n*F*E⁰cell where n = number of electrons transferred = 2 and F = 96485 J/V.mol

= -(2)*(96485 J/V.mol)*(0.23 V)

= -44383.1 J/mol

= -(44383.1 J/mol)*(1 kcal)/(4184 J) [1 kcal = 4184 J]

= -10.6078 kcal/mol

≈ -10.6 kcal/mol (ans).