Question

At the interface of anaerobic sediments with oxic water, sharp gradients in the concentrations of oxygen and sulfide can occur. For lake water just above the interface with pH 6.5, 10^-3 M SO₄^2- and 10^-5 M TOTS(-II), what is the pe of the water? What is the value of PO₂ that would be in equilibrium with this water.

Answer 1

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  15-3 R edox Calculati ons T here are two main types of redox calculations The first is the calculation of what controls the pE of the environment. T his is anal ogous to calculating the pH of the environment, for example when it is controlled by the H2C Os system in equi librium with atmospheric Pco2 T he second type of calculati on is to deter mi ne how trace species respond or distri bute themselves with respect to that pE. A gain by analogy when we know the pH we can calculate the pH dependent speciation of trace species, like Fefor ex ample. LW hat is the pE of water at pH 8 i n equilibrium with oxygen (P02-0.20) in the atmosphere. 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 isrelatvely insensitive to Poz but is very dependent on pH 1402 He 1/2 H20 K 1 (Po2)4 (H') (e) 10205 pE pE 1/n log (oxid/ted) E 20.75 log (Poz)1 (H') pE = 20.75 + 1/4 log 0.2 -pH pE 20.75-0.17 8 pE12.58 pE 20.75 Il. W hat is the pE of water at pH 8 in equili brium with the SO4H2S couple? A ssume that HS 105 and SO103 T his is the typi cal of the anoxic end-member condition. pE+4.25 pE pE"-18 log (HS) + 1 ,8 log (SO42-) + 9/8 log (H ') pE4.25+0.63 0.37 - 9/8pH pE4.25+0.63 0.37 -9.0 pE -4.49
  
  III. W hen the pE is externally controleclate the distribution of trace components. For example assume a natural seawater at pH 8 in equilibrium with Po2 0.20. Then pE 12.58 as calculated above. W hat would be the Mn in equilibrium with yMn02? Assume: 1/2 yMnO2(s) 2 H++e1/2 Mn2+ H20 log K pE20.8 E 20.8 log (H/Mn2)1 258- 20.82log (H) - 1/2 log (Mn2) log (Mn2) - 2 (20.8 - 12.58 2 pH) og (Mn2) 2 (-7.78) log (Mn2+)--15.56 or (Mn2.): 10-15 56 In oxygenated surface seawater dissolved Mn21 E quilibrium does not exist. This is a typical case with redox reactions.
  
  K (Red)/(Ox)e)" We can also rearrange the equation to determine the activity of the electron for any redox couple (e) I(Red) /K (Ox) Electron activiti es are usually expressed on ei ther the pE or Eh scales as shown below. pElog (e) 1h [logK - log (Red)AOx)1 or Eh- 2.3RT pE/F The pE provides a nondi mensional scale (like pH) that ex presses the activity of electrons in factors of 10. Eh (called the redox potential) is measured in volts. F is the Faraday which is the electric charge of one mole of electrons(96,500 coulombs). The ratio 2.3 RT.F has a value of 0.059 V at 25°C. W ith these equations you can express the energy available from a reaction in terms of either AG, Eh or pE. We can express the equi li brium constants for half reactions in different forms as well (pE Eho, K, AG°). Remember that AGRT InK. Then: E FEh/2.3 R -1hlog K -1/h AG /2.3R It is most conventi ent to write the various half reactions in terms of one electron although this introduces odd fractions for the species coeffici ents. The species thatloses an electron is the e-donor (thus itis oxidized butis a reductant for other elements) and the species that accepts an electron is the eacceptor (so it is reduced but is an oxidant). A summary of many redox half-reactions that may occur in natural waters (with values for pE#2 log K ) are given in T able 15-1 (from Morel and Hering, 1993). Note that all half- reacti ons 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 equili brium distri buti on of (Red)Ox), this equati on can be used to solve for the pE of the environment Conversely, if you know the pE of the environment you can calculate the equilibrium activity rat o of the reduced and oxidized forms. Because the half reactions have different pH dependencies a common approach is to compare the various half reactions by calculating the equili brium constant at pH7. T his constant is called pE°(w). For example, consi der the SO4HS haf reaction: E-4.25 we can write: = 4.25-18 log (HS)/(SOJ + 9,8 log (H') at pH 7 we calculate pE (w) pE (w) pE 9/8 log (H+) E°(w) 4.25-9/8 (7)-3.63 Values of pE(w) for the main half-reactions are summarized in T able 15-2
  
  Table 15-1 Redox Half Reactions (from Morel and Hering, 1993) TADL.E TI alfRedox Reactions pe" log K il ydrragen 35.1 20 75 0.0 13.0 +14.15 (Note also NO IINOlo K 3.351 +-16.15 4-18.9 +21.05 +14.9 Suifur 一1.65 [Note also SC), 't H +-IISC½ ; log K-7) +4.85 +6.03 +5.40 +5.88 (Note also +11 . _ HSs : log K =0.1 and 1 15, . Ι.1'-H.ss; log K-3.51 +5.12 (Noue also sa t ll'-lis, i log K 115, + 11°-115%: log K-3.81 7-0 and 5.13 Note also H Sg) H2Siaq og KH.0, and otie acil basc, oordin and precipitaion reactionsi Carba Inorganic Carbon monoxide 1.74 Graphite 3.50
  
  T able 15-1, cont. TABLE 7.i (Contine ranic-C. Fommato 5.2) Formaliclyde 1.20 Melhan: +2.80 OanieC.2 10.7 Acealdchyec Ethanol Ethane Ethvienc + 1.52 +-241 1.34 0.8 OrganiC.3 yrualc 400.68 Cilyceral +0.31 Orga C4 Succinat -113 Organie C.b Giucose -0.20
  
  Table 15-2 TABI ET 2 Eficerive-Equilibrium Constants of Aquatic Redo、Couples +23.41 1373 + 13 65 +9.8 20,75 +21เห i14.15 14.9 15.2 11,0 +4.5 +525 +4.25 -3.6 -4.1 3.3 +3.2 -0.8 14.65 4.3 7.0 0.2 +2.3 -8.3 8.7 "I" lhe reacb帕of ruductive dissolution of "telal oxides. Inc concenirations urthe dissolves! 111elals bisullile ion HS
  

  

III. W hen the pE is externally controleclate the distribution of trace components. For example assume a natural seawater at pH 8 in equilibrium with Po2 0.20. Then pE 12.58 as calculated above. W hat would be the Mn in equilibrium with yMn02? Assume: 1/2 yMnO2(s) 2 H++e1/2 Mn2+ H20 log K pE20.8 E 20.8 log (H/Mn2)1 258- 20.82log (H) - 1/2 log (Mn2) log (Mn2) - 2 (20.8 - 12.58 2 pH) og (Mn2) 2 (-7.78) log (Mn2+)--15.56 or (Mn2.): 10-15 56 In oxygenated surface seawater dissolved Mn21 E quilibrium does not exist. This is a typical case with redox reactions.