Compute the free energy that is released upon transfer of a pair of electrons be

Question

Compute the free energy that is released upon transfer of a pair of electrons between the following donor/acceptor pairs. Unless specified otherwise, assume standard reaction conditions; Standard reduction potentials are listed below. Provide answers in kJ/mol. Please explain the process, equations used, etc.

i. Computte the protonmotive force for this gradient.

ii. Given the energy yield you computed above for respiration (v), how many protons could be pumped for every two electrons transferred?

C. You study this bisulfide(HS^-) using cell further and find that it used some of the electrons from bisulfide to enable CO2 reduction, in a process that requires NADH. Thus, some electrons must be driven 'uphill' from bisulfide into NAD+. The energy to drive this normally unfavorable electron transport is obtained from protons flowing down the gradient described in part B. How many protons must be consumed to form each NADH in this 'reverse electron transfer' process? (Note that NADH formation requires two electrons.)

Please explain your process and equations used. My teacher does not go over examples in class and my book is very brief on covering these topics. Your help will be greatly appreciated!

Explanation / Answer

I am supose to answer only 4 subparts of 1 question

1.
i)
Eo = Eo(1/2O2 /H2O) - Eo(NO3-NO2-)
      = 0.82 - 0.43
      =0.39 V

delta Go = -F*Eo
                   = -96500 *0.39
                   = - 37635J/mol
                   = - 37.6 KJ/mol

iI)
Eo = Eo(NO3-NO2-) Eo(SO4- /HSO3-)
      = 0.43 - (-0.52)
      =0.95 V

delta Go = -F*Eo
                   = -96500 *0.95
                   = - 91675 J/mol
                   = - 91.7 KJ/mol

iii)
Eo = Eo(NO3-NO2-) - Eo(2H+/H2)
      = 0.43 - (-0.41)
      =0.84 V

delta Go = -F*Eo
                   = -96500 *0.84
                   = - 81060 J/mol
                   = - 81.1 KJ/mol
iV)
Eo = Eo(N2O-N2) - Eo(2H+/H2)
      = 1.36 - (-0.41)
      =1.77 V

delta Go = -F*Eo
                   = -96500 *1.77
                   = - 170805 J/mol
                   = - 170.8 KJ/mol

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