You are employed for a consulting company as a consultant working for a project

Question

You are employed for a consulting company as a consultant working for a project related with the biodegration of organic waste after graduating KU. Microbial process is dominant but you are not sure how efficient the process should be Microbial analysis indicated that there are mixed bacterial communities that utilize both oxygen and nitrate as the terminal electron acceptor. Nitrate concentration in the of the waste is 4.0 times 10^-4 mol L^-1 and pH was 8.0. Now, you want to estimate partial pressure of oxygen below which nitrate becomes more energetically efficient. Answer the following questions. Standard redox potentials relevant for this analysis are found in Appendix B.5. O_2 + 4H^+ + 4e^- righatrrow 2H_2O E degree = +1.229 V pE degree = +2.08 2NO_2 + 12 H^+ + 10e righatrrow N_2 + 6H_2O E degree = +1.25 V pE degree = +21.1 What is pE for the reduction of nitrate to nitrogen gas under this condition (pH 8 and [NO_2 +] 4.0 times 10^4 mol L^-1)? Assume that the partial pressure of nitrogen is 1 atm. Faraday constant is 96, 485 C mol^-1, temperature is 298 K, and gas constant is 8.314 J K^-1 mol^-1 Describe pE for the reduction of oxygen as a function of the partial pressure of oxygen. Determine the partial pressure of oxygen below which nitrate becomes more efficient terminal electron acceptor than molecular oxygen.

Explanation / Answer

pE = pE0 - 1/n log Q

E = E0 - 0.0592 / n log Q

For redcution of nitrate

Q = pN2 / [NO3-]^2 [ [H+]12

Given pH = 8 therefore [H+] = 10^-8

Q = 1/ (4 X 10^-4 )^2 (10^-8)^12

E0 = 1.25

E = 1.25 - 0.0592 / 10 log(1/ (4 X 10^-4 )^2 (10^-8)^12)

E =1.869 V

pE = E / 2.303RT / F

F= 96485

R = 8.314

T = 298

pE = 0.1.869 / 0.0592 = 0.619 / 0.0592 = 31.57

b) pE for reduction of oxygen as a function of the partial pressure of oxygen

E = E0 -0.0592 / n log Q

pE = pE0 - 1/n log Q

Q = 1 / [H+]^4 X partial pressure of oxygen

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