Question

Please answer the following question and provide explanation as to how the answer was concluded, please answer all questions asked.

Given two reduction half-reactions: Acetaldehyde + 2 H ions + 2 eletrons ---------> Ethanol, where the standard reduction potential is -0.20 V, and Pyruvate + 2 H ions + 2 electrons --------> Lactate, where the standard reduction potential is -0.18 V.

Consider the following reaction at 25 degrees celsius, Pyruvate + Ethanol ----> Acetalaldehyde + Lactate

a). Using the data provided above, calculate the standard free energy change for the reaction in kJ/mol.

b). Is the reaction spontaneous as written?

c). If the concentration of concentrations of ethanol, pyruvate, and lactate are each 1.00 mM, what does the concentration of acetaldehyde need to be to make the reaction spontaneous?

Answer 1

(a) Delta G^o = -nFE^ocell

Calculate E^ocell from two half cells

E^ocell = Ecathode - Enaode

          = -0.18 -(-0.20)

          = 0.02 V

n = 2

F = 96485 C.mol-1

Feed this value,

Delta G^o = -nFE^ocell

               = -2 x 96485 x 0.02

               = -3859.4 J = -3.86 kJ

thus, the standard free energy change for the reaction is -3.86 kJ

(b) The -ve sign of standard free energy calculated for the reaction in (a) shows the reaction will be spontaneous as written.

(c) We will use the relation,

Delta G^o = -RT lnK

R = 8.314 J.mol.C-1             

T = 25 oC = 25 + 273 = 298 K

K = [acetaldehyde][lactate] / [pyruvate][ethanol]

Feed the values,

-3859.4 = -8.314 x 298 ln [acetaldehyde]

[acetaldehyde] = 4.75 mM

Thus, 4.75 mM concentration of acetaldehyde is needed to make the reaction spontaneous.