Question

1.) Calculate the thermodynamic favorability of aerobic Mn(II) oxidation and aerobic Fe(II) oxidation at pH 2 and 7.

2.) Is it likely that an organism could get “enough” energy to grow from each of these four processes?

Answer 1

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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-

    S^2- = SO₄^2- + 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 SO₄^2- + 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⁺ = 1/8SO₄^2- + 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.