Chapter 15 Organic Matter Diagenesis Jim Murray(5/09/01) Univ. Washington15-1 Oxidation-reduction reactionsMany elements in the periodic table can exist in more than one oxidation state. Oxidationstates are indicated by Roman numerials in parantheses (e.g. (+I), (-II), etc). The oxidationstate represents the "electron content" of an element which can be expressed as the excessor deficiency of electrons relative to the elemental state.If you have a compound and you want the oxidation state of an element, assign oxygen (O)to be (-II) and hydrogen (H) to be (+I) and calculate the oxidation state of the element ofinterest, taking into account the charge on the compound.For example:Element Oxidation State SpeciesNitrogen N (+V) NO3-N (+III) NO2-N (O) N2N (-III) NH3, NH4+Sulfur S (+VI) SO42-S (+II) S2O32-S (O) S°S(-II) H2S, HS-, S2-Iron Fe (+III) Fe3+Fe (+II) Fe2+Manganese Mn (+VI) MnO42-Mn (+IV) MnO2 (s)Mn (+III) MnOOH (s)Mn (+II) Mn2+15-2 Redox half-reactionsRedox reactions are written as half-reactions which are in the form of reductions (whichmeans an element is transformed from a higher oxidation state (e.g. +II) to a loweroxidation state (e.g. +I)):Ox + ne- = Red; ∆G°;Kwhere the more oxidized form of an element is on the left and the reduced form is on theright. n is the number of electrons transferred. We can write an equilibrium constant for thisreaction as we can any other reaction. Formally the concentrations should be expressed asactivities. Thus:K = (Red) / (Ox)(e-)nWe can also rearrange the equation to determine the activity of the electron for any redoxcouple:(e-) = [ (Red) / K (Ox) ] 1/n Electron activities are usually expressed on either the pE or Eh scales as shown below.pE = - log (e-) = 1/n [logK - log (Red)/(Ox) ]orEh = 2.3 RT pE / FThe pE provides a nondimensional scale (like pH) that expresses the activity of electrons infactors of 10. Eh (called the redox potential) is measured in volts. F is the Faraday which isthe electric charge of one mole of electrons (96,500 coulombs). The ratio 2.3 RT.F has avalue of 0.059 V at 25°C. With these equations you can express the energy available from areaction in terms of either ∆G, Eh or pE.We can express the equilibrium constants for half reactions in different forms as well (pE°,Eh°, K, ∆G°). Remember that ∆G° = - RT ln K. Then:pE° = F Eh° / 2.3 RT = 1/n log K = - 1/n ∆G°/2.3RTIt is most conventient to write the various half reactions in terms of one electron althoughthis introduces odd fractions for the species coefficients. The species that loses an electronis the e- donor (thus it is oxidized but is a reductant for other elements) and the species thataccepts an electron is the e- acceptor (so it is reduced but is an oxidant).A summary of many redox half-reactions that may occur in natural waters (with values forpE° = log K) are given in Table 15-1 (from Morel and Hering, 1993). Note that all half-reactions are written in terms of one electron.The pE equation can be expressed as:pE = pE° - log (Red) / (Ox)where pE° can be calculated from log K.If you know the equilibrium distribution of (Red)/(Ox), this equation can be used tosolve for the pE of the environment. Conversely, if you know the pE of the environmentyou can calculate the equilibrium activity ratio of the reduced and oxidized forms.Because the half reactions have different pH dependencies a common approach is tocompare the various half reactions by calculating the equilibrium constant at pH = 7. Thisconstant is called pE°(w). For example, consider the SO4/HS half reaction: 1/8 SO42- + 9/8 H+ + e- = 1/8 HS- + 1/2 H2OpE° = 4.25we can write:pE = pE° - log (HS-)1/8 / (SO42-)1/8 (H+)9/8 = 4.25 - 1/8 log (HS-)/(SO4) + 9/8 log (H+)at pH = 7 we calculate pE(w)pE°(w) = pE° + 9/8 log (H+)pE°(w) = 4.25 - 9/8 (7) = -3.63Values of pE(w) for the main half-reactions are summarized in Table 15-2.Table 15-1 Redox Half Reactions (from Morel and Hering, 1993)Table 15-1, cont.Table 15-1, cont.Table 15-215-3 Redox CalculationsThere are two main types of redox calculations. The first is the calculation of what controlsthe pE of the environment. This is analogous to calculating the pH of the environment, forexample when it is controlled by the H2CO3 system in equilibrium with atmospheric PCO2.The second type of calculation is to determine how trace species respond or distributethemselves with respect to that pE. Again by analogy when we know the pH we cancalculate the pH dependent speciation of trace species, like Fe for example.I.What is the pE of water at pH 8 in equilibrium with oxygen (PO2 = 0.20) in theatmosphere. This is the typical pE for oxic conditions.We assume that the pE is controlled by the oxygen-water half reaction from Table 15-1, 15-2. Note that the pE is relatively insensitive to PO2 but is very dependent on pH.1/4O2 + H+ + e- = 1/2 H2OpE° = 20.75K = 1 / (PO2)1/4 (H+) (e-) = 1020.75pE = pE° + 1/n log (oxid/red)pE = 20.75 + log (PO2)1/4 (H+)pE = 20.75 + 1/4 log 0.2 - pHpE = 20.75 - 0.17 - 8pE = 12.58II. What is the pE of water at pH = 8 in equilibrium with the SO4/H2S couple?Assume that HS- = 10-5 and SO42- = 10-3.This is the typical of the anoxic end-member condition.1/8SO42- + 9/8H+ + e- = 1/8HS- + 1/2H2O pE° = +4.25pE = pE° - log (HS-)1/8 / (SO42-)1/8 (H+)9/8pE = pE° - 1/8 log (HS-) + 1/8 log (SO42-) + 9/8 log (H+)pE = 4.25 + 0.63 - 0.37 - 9/8pHpE = 4.25 + 0.63 - 0.37 - 9.0pE = -4.49III. When the pE is externally controlled calculate the distribution of trace components. Forexample, assume a natural seawater at pH = 8 in equilibrium with PO2 = 0.20. Then pE =12.58 as calculated above. What would be the Mn2+ in equilibrium with γMnO2?Assume:1/2 γMnO2(s) + 2 H+ + 1e- = 1/2 Mn2+ + H2O log K = pE° = 20.8pE = 20.8 + log (H+)2 / (Mn2+)1/212.58 = 20.8 + 2 log (H+) - 1/2 log (Mn2+)log (Mn2+) = 2 (20.8 - 12.58 - 2 pH)log (Mn2+) = 2 ( -7.78 )log (Mn2+) = -15.56 or (Mn2+) = 10-15.56In oxygenated surface seawater dissolved Mn2+ = 10-9.Equilibrium does not exist. This is a typical case with redox reactions.15-4 Balanced Redox ReactionsA balanced reaction has an electron passed from an electron donor to an electron acceptor.Thus:Ox1 + Red2 = Red1 + Ox2In this case Red2 is the electron donor, passing electrons to Ox1 which is the electronacceptor. Thus Red2 is oxidized to Ox2 and