Question

Bacteria are not limited to organic matter as a source of electrons or energy. Autotrophic denitrification occurs when NO₃^- is used as an electron acceptor and converted to N₂ gas and H₂ gas is used a source of electrons.

Below are the two half reactions we will be working with.

2NO₃^- + 10e^- -> N₂

2H⁺ + 2e^- -> H₂

Which reaction needs to be reversed?  Enter the letter A or the letter B

Balance the half reaction for the conversion of nitrate to dinitrogen gas. Input the reaction co-efficient for each component of the half reaction in the blank provided. The co-efficient might be 1 in some cases.

2NO₃^- + e^- + H⁺ -> N₂ + H₂O

Balance the half reaction for the conversion of hydrogen gas to hydrogen ions. Questions 12-13 are asking for the reaction co-efficients for each piece of the reaction. The co-efficient might be 1 in some cases.

H₂ + -> 2e^- + H⁺

Combine and balance both equations.

NO₃^- + H⁺ + H₂ -> N₂ + H₂O

How many electrons were transferred?

What is the redox potential of this reactions in volts? E⁰_cell = ?

Using the Nernst Equation, calculate the Gibbs Free Energy to deduce how many kilojoules are available to do work based on the stoichiometry you determined? ΔG = -nFΔE, where F = 96.48 kJ (mol e^-)  (Report your answer to 3 decimal places)

Answer 1

H2 gas is used a source of electrons.

H2 gas produces electrons

The two half cell reactions are

Reduction reaction

2NO3- + 10e- - - - - > N2

Ered = 0.74 V

Oxidation reaction (The reverse reaction)

H2 - - - - - > 2H+ + 2e-     

Eox = 0.42 V

The balanced reaction

unbalanced equation

NO3- + H+ + H2 = N2 + H2O

oxidation numbers for each atom within brackets

N(+5)O(-2)3- + H(+1)+ + H(0)2 = N(0)2 + H(+1)2O(-2)

The two half cell reactions are

Oxidation reaction

H2 = H2O

H is being oxidized

Reduction reaction

NO3- = N2

N is bring oxidized

Balance all atoms except H and O.

Oxidation reaction

H2 = H2O

Reduction reaction

2NO3- = N2

Balance the oxygen atoms by adding water molecules.

O:

H2 + H2O = H2O

R:

2NO3- = N2 + 6H2O

Balance the hydrogen atoms by adding protons (H+).

O:

H2 + H2O = H2O + 2H+

R:

2NO3- + 12H+ = N2 + 6H2O

Balance the charge.

O:

H2 + H2O = H2O + 2H+ + 2e-

R:

2NO3- + 12H+ + 10e- = N2 + 6H2O

Make electron balance

O:

H2 + H2O = H2O + 2H+ + 2e-

Multiply by 5

R:

2NO3- + 12H+ + 10e- = N2 + 6H2O

Multiply by 1

O:

5H2 + 5H2O = 5H2O + 10H+ + 10e-

R:

2NO3- + 12H+ + 10e- = N2 + 6H2O

Add the half-reactions together.

5H2 + 2NO3- + 5H2O + 12H+ + 10e- → N2 + 10H+ + 11H2O + 10e-

Simplify

5H2 + 2NO3- + 2H+ = 1N2 + 6H2O

10 electrons were transfered

E°cell = Eox + Ered

= 0.74 + 0.42

= 0.98 V

Free Energy change

ΔG = -nFΔE

= 10 x 96480 Coulomb x 0.980 V

= 945504 x 1kJ/1000 J

= 945.504 kJ