Conditions under which sulfate ions are reduced to form hydrogen sulfide ions

　Sulfate ions (SO₄^2-) are abundant in seawater as salt components. Under an oxidative environment where oxygen is abundant, SO₄^2- is the most stable of the sulfur compounds, so most of the sulfur in seawater exists as SO₄^2-. On the other hand, under a reducing seafloor, an environment is created in which several types of sulfur compounds are present. 

　The oxidation number of sulfur (S) in SO₄^2- is +6. If we add eight electrons (e^-) and one hydrogen ion (H⁺) to this S and remove O, we get hydrogen sulfide ion (HS^-) with the oxidation number of S being -2. Conversely, when HS^- is present in water, take away 8 electrons from S and do the opposite and you get SO₄^2-. At a certain condition, it is determined whether most of the S exists as SO₄^2- or HS^-. At that boundary, SO₄^2- and HS- exist in equal proportions. Let us find the boundary condition (redox potential).

　First, the half-reactions in which SO₄^2- and HS^- exchange electrons are described below.

half-reaction(④’)        SO₄^2- + 9H⁺ + 8e^-  =  HS^-  + 4H₂O

　As explained in the previous course, a redox reaction does not proceed by a single half-reaction. If the reaction is sulfuric acid respiration, it will be combined with the half-reaction of organic matter oxidation (carbon in organic matter releases electrons, carbon in carbon dioxide receives electrons) to form a redox reaction. Which direction the reaction proceeds in, however, depends on the conditions.

　In other words, when HS^- and SO₄^2- are mixed, if an electron is passed from another half-reaction to the above half-reaction (④'), the reaction proceeds in the direction of HS^- formation, and if another half-reaction takes an electron away from the half-reaction (④'), the reaction proceeds in the direction of SO₄^2- formation.

　Now, let us proceed with the calculation.

The standard Gibbs energy of formation for each substance in the half reaction (④') is noted below, and the total difference between the product form and the original form ⊿G_f⁰ is obtained.

                                original form                                      product form

half reaction(S)         SO₄^2-   +   9H⁺   +   8e^-      =      HS^-   +     4H₂O    

G_f⁰ (kJ/mol)　　    －744.5    　　0   　　     　　    12.08        －237.2            

Total difference between the standard Gibbs energy of formation of the substance in its original and product forms.：⊿G_f⁰ 

⊿G_f⁰ (J) = 12.08×10³ + 4(－237.2×10³) － (－744.5 ×10³) = －192220

From the condition that the energies (including electrons) of the original and the product forms are equal, we obtain the standard electrode potential E₀.

E₀ = －⊿G_f⁰ /(n・F) = －(－192220) / (8・96485) =  0.249 (V)（≒0.25）

Applying Nernst equation,

E = 0.25 － 0.003208・Ln {([HS^－][H₂O]⁴) / ([SO₄^2-][H⁺]⁹)}

E = 0.25 + 0.003208・Ln([SO₄^2-] / [HS^－]) + 0.003208・9・Ln [H⁺]

  = 0.25 + 0.003208・Ln([SO₄^2-] / [HS^－]) ＋ 0.003208・9・Log[H⁺] / Log(e)

　= 0.25 + 0.003208・Ln([SO₄^2-] / [HS^－]) － 0.06648×pH

Hydrogen ion concentration rewritten in pH（ pH = -Log₁₀[H⁺] )

If we attach a boundary condition ([SO₄^2-] / [HS^-]=1) where SO₄^2- and HS^- are present in the same ratio, E becomes

E = 0.25 － 0.06648×pH.                    (equation 1)

the ratio of [SO₄^2-] << [HS^-] is reached when the redox potential of the ambient water is E <-0.28 (V).

It is important to note that this is the boundary condition obtained from the thermodynamic constant, and it has been confirmed that sulfate reduction (HS- generation) occurs at E<-0.1 (V) in actual sediments. Due to the action of organisms (local reducing environment within the organisms and enzymes), sulfate reduction occurs in a more oxidizing environment than the boundary conditions in the thermodynamic calculations. The rate of sulfate reduction is also believed to be much faster due to microbial action.