Question

calculate the thermodynamic favorability of aerobic Mn(II) oxidation and aerobic Fe(II) oxidation at pH 2 and 7? Would an organism grow from these processes?

Show all work and don't use previous answers similar to this because they aren't correct. Thanks.

Answer 1

Electron donors that occur in soils and participate in chemical redox couples are compounds found in reduced soils. These are as follows;;;;; -

Mn (II) = Mn(IV) + 2e-

Fe (II) = Fe(III) + e-

S2- = SO42- + 8e-

As (III) = As (V) + 2e-

O2 (g), is the sole electron acceptor in aerobic systems.

O2 + 4 e- + 4H+ = 2H2O

When Fe (II) coming in contact with molecular oxygen (O2), following reactions are take place.

H+ + 1/4 O2 + e- = 1/2 H2O 1.229V

Fe2+ = Fe3+ + e- -0.711 V

So Fe2+ + 1/2O2 + H+ = Fe3+ + 1/2 H2O 0.518V

A positive ∆Eh, means that the reaction is thermodynamically favorable.

If a reduced species MnO2 with a low redox potential is present with a species in the oxidized form which has a higher redox potential, then there is a high probability for oxidation of the reduced species and reduction of the oxidized species.

1/8H2S + 1/2 H2O = 1/8 SO42- + 5/4H+ + e- -0.303V

1/2 MnO2 (s) + 2H+ + e- = 1/2 Mn2+ + H2O 1.23V

1/8H2S + 1/2MnO2 (s) + 1/2H2O + 3/4H+ = 1/8SO42- + 1/2 Mn2+ + 1/2H2O 0.927V

A positive ∆Eh, means that the reaction is thermodynamically favorable.

When the pH is 2, H+ consumption in the reduction reactions increases the pH.

MnO2 (s) + 4 H+ + 2e- = Mn2+ + 2H2O

If the pH is 7, then the precipitation of metal ions such as Fe and Mn as hydroxides, carbonates, or sulfides tends to lower the pH.

Fe2+ + 2 H2O = Fe (OH)2 + 2 H+

An organism could get “enough” energy to grow from each of these four processes ​ Fe, Mn, S2 and As cycles.