In: Chemistry
Calculate the thermodynamic favorability of aerobic Mn(II) oxidation and aerobic Fe(II) oxidation at pH 2 and 7. Is it likely that an organism could get “enough” energy to grow from each of these four processes?
Electron donors that occur in soils and participate in chemical redox couples are compounds found in reduced soils. These ares -
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 cycle.