Chemical Forms of Sulfur (S) in Seawater (Pourbaix Diagram)

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• Select section Conditions under which sulfate ions are reduced to form hydrogen sulfide ions

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

  • Select activity 　Sulfate ions (SO42-) are abundant in seawater as ...

    　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.

    
    

  • Select activity 　Now, let us proceed with the calculation.The stan...

    　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)}

     

    As a rule of physical chemistry, let [H₂O] = 1.
    

     

    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)

    
    

    For example, if we substitute seawater pH = 8, the boundary between the greater and lesser abundance ratios of [SO₄^2-] and [HS^-] is E = -0.28 (V).
    

     

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

    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.
    

     

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• Select section Pourbaix diagram of sulfur (s)

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  Pourbaix diagram of sulfur (s)

  
  

  • Select activity E = 0.25&nbsp;－&nbsp;0.06648×pH&nbsp;&nbsp;&nbsp;&amp;...

    E = 0.25 － 0.06648×pH                    (式1)
    

    
    　Equation 1 is shown in the graph below with pH on the horizontal axis and E(V) on the vertical axis. Note that the SO₄^2-⇔HS^- line (black thick line) in the figure is the boundary condition obtained above.

    
    

    　If we measure the oxidation-reduction potential of oxygen-rich seawater (pH 8), it is about +0.3 to +0.6 (V). (If only a sufficient amount of oxygen is dissolved in pure water, the oxidation-reduction potential of that water will be +0.8 (V), but since natural water also contains reducing substances (e.g., organic matter), it will be less than +0.6 (V) in sea water.)

    　From this figure, it can be determined that sulfur (S) is stable in oxygenated seawater (E = 0.3-0.6 V) as SO₄^2-.

    　When oxygen is reduced to zero in the sediment, the redox potential suddenly drops to a low level. Furthermore, when oxidants such as nitric acid are also eliminated, the redox potential falls below the SO₄^2-⇔HS^- line. Sulfate-reducing bacteria, which breathe sulfuric acid (SO₄^2-) as an oxidant, proliferate in the sediment, converting SO₄^2- in the pore water to hydrogen sulfide (H₂S). In the sediment, carbonic acid is produced as a result of organic matter decomposition, resulting in a slight acidification of the sediment to a pH of about 7. The redox potential and pH decrease in the sediment are indicated by the green dashed arrows in the figure.
    

  • Select activity Earlier, we plotted the conditional equation [SO42...

    Earlier, we plotted the conditional equation [SO₄^2-] / [HS^-] = 1 (Eh = 0.25 - 0.06648 x pH) on a pourbaix diagram. What would happen with a different ratio.

    Under the condition [SO₄^2-] / [HS^-] = 1000, Eh = 0.27 - 0.06648 x pH. The potential has shifted by +0.02. The figure is almost the same as the previous one. This means that the ratio of [SO₄^2-] to [HS^-] changes abruptly near this plot.

    

  • Select activity &nbsp;In addition, hydrogen sulfide ion (HS-) is i...

     In addition, hydrogen sulfide ion (HS^-) is in dissociative equilibrium with hydrogen sulfide (H₂S).

    Reaction equation for the dissociation equilibrium of hydrogen sulfide and hydrogen sulfide ion

    H₂S (aq)   = H⁺  + HS^- 　　；K = 10^－7（K is the dissociation equilibrium constant）

     

    　This is an acid-base reaction, since the oxidation number of S does not change in this reaction at -2 valence. Therefore, the redox potential is irrelevant in this dissociation equilibrium condition. The relationship between the equilibrium constant and pH is expressed by the following equation.

    
    

    Equilibrium constant = {concentration product of the product form} / {concentration product of the original form} so,
    

    　　 [HS^-][H⁺] / [H₂S]} = 10^－7

     

    Take both sides Log and express the hydrogen ion concentration in terms of pH.
    

    　　Log{ [HS^-] / [H₂S] } － pH = －7

     

    　The equilibrium condition where HS- and H₂S coexist in the same ratio: [HS^-] / [H₂S] = 1 is at pH = 7.

    　Below pH 7, most will exist as H₂S, and above pH 7, they will exist as HS^-. I have rewritten the previous figure to add this information.
    

    
    

    　This is called the "Pourbaix diagram," which shows how the chemical form of a substance changes with pH and redox potential. The "water oxidative decomposition region" and the "water reduction decomposition region" are regions where H₂O decomposes, and therefore cannot exist in an aqueous environment. 

    　Again, the boundary conditions noted in the Pourbaix diagram are calculated from thermodynamic constants, and it has been confirmed that sulfate reduction can proceed even in more oxidizing environments (from about -0.1 (V)) due to microbial action.
    
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