Question

In saturated soils, oxygen is quickly depleted and microbes must utilize the next most favorable electron acceptor, in this case sulfate (SO42-). Under these conditions, Mn (and Fe) are oxidized to Mn2+ (and Fe2+). Select the correct oxidation and reduction pair for Mn.


A.
Oxidation Mn --> Mn2+ + 2e-

Reduction SO4 + 2e- --> SO42-

B.
Oxidation Mn2+ + 2e- --> M

Reduction SO4 + 2e- --> SO42-

C.
Oxidation Mn2+ --> Mn + 2e-

Reduction SO42+ + 2e- --> SO42-

D.
Oxidation SO42+ + 2e- --> SO4

Reduction Mn2+ --> Mn + 2e-

I would like to know how to derive at an answer and not just know the answer please...thanks. :)

Answer 1

No big problem. Just remember the definitions.
Oxidation is the loss of electrons.
Reduction is the gain of electrons.

In the problem Mn goes to Mn^2+, so
Mn ==> Mn^2+ + 2e is the loss of electrons and is oxidation. A is the only answer that fits for Mn.

I don't see the corresponding reduction for SO4^2- because none of them start with SO4^2-. As far as I know there is no "SO4" without the 2- charge.

Since A is the only answer with the oxidation part of Mn correct then I assume A must fit the key; however, there is no correct answer given in my opinion.
It should be something like
SO4^2- + 2e + 2H^+ --> SO3^2- + H2O