Glycerol 3-phosphate dehydrogenase catalyzes the reversible reaction:
G3P + NAD+ --> NADH + H+ + DHAP
If in the cell, the concentrations of G3P, NAD+, NADH, and DHAP are 2 mM, 4 mM, 6 mM, and 8 mM respectively. Determine if the reaction spontaneous as written?
DHAP + 2e– + 2H+ -->G3P DE'° = –0.29 V
NAD+ + H+ + 2e– --> NADH DE'° = –0.32 V
To determine if the reaction is spontaneous as written, we can calculate the overall standard cell potential (E'°) of the reaction.
The standard cell potential of a reaction can be calculated by summing the standard reduction potentials of the half-reactions involved. The half-reactions given are:
DHAP + 2e– + 2H+ --> G3P DE'° = –0.29 V
NAD+ + H+ + 2e– --> NADH DE'° = –0.32 V
The reaction equation is:
G3P + NAD+ --> NADH + H+ + DHAP
Now, let's calculate the overall standard cell potential (E'°) of the reaction:
E'° = E'°(cathode) - E'°(anode)
= E'°(NADH) - E'°(G3P)
E'°(NADH) = -0.32 V (given)
E'°(G3P) = -(-0.29 V) = 0.29 V (since the reduction potential is given for DHAP to G3P, we need to use the opposite sign for the oxidation potential)
E'° = -0.32 V - 0.29 V
= -0.61 V
Since the overall standard cell potential (E'°) is negative (-0.61 V), we can conclude that the reaction is spontaneous as written.
To determine if a reaction is spontaneous, we can calculate the standard Gibbs free energy change (ΔG°) using the equation:
ΔG° = -nFΔE°
where ΔE° is the standard cell potential (also known as the standard reduction potential) and n is the number of moles of electrons transferred. In this case, we need to consider the reduction potentials for both half-reactions involved in the reaction:
DHAP + 2e– + 2H+ --> G3P ΔE'° = –0.29 V (reduction potential for DHAP/G3P)
NAD+ + H+ + 2e– --> NADH ΔE'° = –0.32 V (reduction potential for NAD+/NADH)
Let's calculate the standard Gibbs free energy change for each half-reaction:
ΔG'°(1) = -nFΔE'°(1)
= -2 × 96485 C/mol × (–0.29 V) [since 2 moles of electrons are transferred]
= 56.078 J/mol
ΔG'°(2) = -nFΔE'°(2)
= -2 × 96485 C/mol × (–0.32 V) [since 2 moles of electrons are transferred]
= 61.6448 J/mol
Now, let's calculate the overall ΔG'° for the reaction:
ΔG'°(reaction) = ΣΔG'°(products) - ΣΔG'°(reactants)
= (0 J/mol + 0 J/mol + 0 J/mol) - (56.078 J/mol + 61.6448 J/mol)
= -117.7228 J/mol
Since the overall ΔG'° for the reaction is negative (-117.7228 J/mol), the reaction is spontaneous as written.