[Iamyourfather]

I'm doing some work so I can prepare for me exams and I need help with these questions. I don't know what to do.

Standard Reduction Potentials The standard reduction potential, E'°, of any redox pair is defined for the half-cell reaction:
Oxidizing agent + n electrons → reducing agent.
The E'° values for the NAD+/NADH and pyruvate/lactate conjugate redox pairs are - 0.32 V and -0.19V, respectively.
a) Which redox pair has the greater tendency to lose electrons? Explain.
b) Which pair is the stronger oxidizing agent? Explain
c) Beginning with 1 M concentrations of each reactant and product at 25°C and pH 7 in
which direction will the following reaction proceed?
Pyruvate + NADH + H   lactate + NAD+
d) What is the standard free-energy change (ΔG'°) for the conversion of pyruvate to lactate?
e) What is the equilibrium constant (K' eq) for this reaction?
Constants: R = 8.31 J mol-1 k-1
F = 96,500 J mol-1 V-1

[Dan]

Think about the reduction potentials. What corresponds to a more favourable process - a more negative or more positive number?

http://en.wikipedia.org/wiki/Reduction_potential

[Babcock_Hall]

I would start this problem by writing out the two half reactions with their half cell potentials.  Each one involves 2 electrons, and it is important to know which species is picking up two electrons.  Between pyruvate and lactate, which one gains two electrons (plus two protons) to become the other species.  Between NAD and NADH, which one gains two electrons (plus one proton) to become the other one.

[Dan]

I would start this problem by writing out the two half reactions with their half cell potentials.

Go on then. You have to show you are trying, it is a forum rule.

First hit on google: http://www.wiley.com/college/pratt/0471393878/student/review/redox/4_reduction_potential.html

[Babcock_Hall]

Iamyourfather,

What I suggested is that you write out the two half-reactions.  You may use these two as a model (from Dan's link):

(1)   fumarate + 2H+ + 2 e–    succinate   E°' = 0.030 V
(2)   FAD + 2H+ + 2 e–    FADH2             E°' = -0.180 V

Here is a quote from the same link:  "remember the rule that a higher E°' value indicates a stronger tendency for the compound to gain electrons."