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.

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

Answer 2

To determine the correct oxidation and reduction pair for Mn in saturated soils, we need to understand the process of electron transfer in redox reactions.

In this case, we start with the information that in saturated soils, oxygen availability is limited. Therefore, other electron acceptors, such as sulfate (SO42-), are utilized by microbes.

Let's analyze each option and identify the correct oxidation and reduction pairs:

Option A:
Oxidation: Mn --> Mn2+ + 2e-
Reduction: SO4 + 2e- --> SO42-

Option B:
Oxidation: Mn2+ + 2e- --> Mn
Reduction: SO4 + 2e- --> SO42-

Option C:
Oxidation: Mn2+ --> Mn + 2e-
Reduction: SO42+ + 2e- --> SO42-

Option D:
Oxidation: SO42+ + 2e- --> SO4
Reduction: Mn2+ --> Mn + 2e-

We know that in redox reactions, oxidation involves the loss of electrons, while reduction involves the gain of electrons.

From the given information, we can see that option A has Mn being oxidized (losing electrons) and SO4 being reduced (gaining electrons). This suggests that option A is the correct oxidation and reduction pair for Mn in saturated soils.

Therefore, the correct answer is:

A.
Oxidation: Mn --> Mn2+ + 2e-
Reduction: SO4 + 2e- --> SO42-