Syllabus point 6.1 (6.2, 6.3) (1 Viewer)

[BrotherBread]

Ok so most of this topic is fairly ok to understand. But in making my notes I came across two conflicting sources surrounding point 6.1 (and thus 6.2 6.3 becasue they are closely related)

Specifically relating to anaerobic bacteria and the chemical reactions that relate to corrosion at depths.

One source (text) says that the sulfate ions are reduced to sulfur then to sulfur ions. As shown below.
SO₄^2-_(aq) --> S_(s) --> S^2-_(aq)


_{Now my other source (teachers notes, which make more sense) says that the sulfate ions are oxidised to sulfur then reduced to sulfur. Admittedly seeming more correct, however to where do these electrons from the oxidation go. I know the oxidation of iron leads to the reduction of sulfur but what about the oxidation of sulfate. Does anyone have good notes for all of syllabus point 6? Or at least fully explained chemical equations to explain this.}

_{Any help is very much appreciated.}

EDIT:

I think I have it. Anaerobic bacteria respire creating Sulfur which is then reduced by the iron to sulfur irons and then they react with the Iron to form FeS, or alternately the iron reacts with the hydroxide ions, as two reduction reactions may be operating in non-acidic conditions. Is that right?

EDIT 2:

Wait I have tried to break it down to far thanks to this misleading me.
SO₄^2-_(aq) --> S_(s) --> S^2-_(aq)

 I think what infact happens is that these steps happen all at once in the respiration of the bacteria, and for this to occur there must be electrons coming from the iron. this produces both Sulfur ions, hydroxide ions and Iron ions, that later react to precipitate FeS Iron Hydroxide or hydoxide ions. How's that? I have the equations here in front of me but I don't have the technical ability to write those down.

 

[xiao1985]

IIRC:

Reduction of sulphate:
SO₄^2- + 9H⁺ -> 4H₂O + HS^-

 

[BrotherBread]

I thought 1 of two reductions occurred

SO₄^2- + 4H₂O + 8e- -> HS^- + 8OH<sup>+
and another one similar

Or are you talking about the acid reaction, but even for that I only have 8 hydrogen ions and no hdrigen sulfate produced


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