(1) Is seawater normally an oxidizing or reducing medium? Oxidizing

Transcription

1 Lecture 11 Diagenesis, Redx Chemistry Organic C Preservatin (1) Is seawater nrmally an xidizing r reducing medium? Oxidizing Write the xidatin numbers fr all f the atms in the fllwing cmpunds: N 2 ; NO 2 - ; NO 3 - ; Fe 2 O 3 ; FeOOH; MnO; MnO 2 ; S 2 O 3 2-0; 3+, 2-; 5+, 2-; 3+, 2-; 3+, 2-, 1+; 2+, 2-; 4+, 2-; 2+, 2-; Answer the fllwing questins fr: 2Mn 2+ + O 2 + 2H 2 O == 2MnO 2 + 4H + The xidatin # f Mn changes frm 2+ t 4+. The xidatin # fr O changes frm 0 t 2-. The xidizing agent is Mn and the element that gets reduced is O. The half-cell xidatin reactin is Mn H 2 O = MnO 2 +4H + + 2e and the half-cell reductin reactin is 1/4 O 2 + H + + e = ½ H 2 O. If equilibrium is achieved in seawater (ph 8) the E 0 cell fr this reactin wuld be V. Cmpute the equilibrium cnstant fr the reactin. E 0 cell = /n lgk cell lgk = n/ x 0.81= 27.36; K = 2.3 x10 27 (2) What cntrls rganic C cntent in sediments? verlying prductivity mineral surface area water depth seasn sedimentatin rate What des nt supprt the thery that mineral surfaces cntrl n rganic matter preservatin: 1. Crrelatin between clay cntent in the sediment and % rganic matter 2. The empirical relatin between the rganic matter cntent and the ne mlecule deep full cverage n the surface area. 3. the bservatin that when this rganic matter is released frm minerals it is utilized by rganisms 4. the difficulty t mechanically separate the rganic matter frm the assciated sediments. 5. The crrelatin between bttm water xygen cntent and rganic matter preservatin in sediments Why d the deep waters f Saanich Inlet becme anxic? 1. because O 2 released by phtsynthesis in surface waters escapes t the atmsphere 2. because nitrate is utilized as an electrn acceptr 3. because reducing pwer is intrduced by sinking rganic particles 4. because the increased acidity f the deep waters causes a marked decrease in Eh 5. nne f these 6. due t seasnal stratificatin

2 The Oxygen Expsure time depends n: 1. distance frm shre 2. sedimentatin rate 3. bttm water xygen cntent 4. biturbatin f the sediment 5. type f rganic matter in the sediment (3) Uranium is mre sluble in its xidized U +6 frm than in its reduced U +4 frm. Is the cncentratin f disslved U likely t be greater r less in the black sea relative t seawater? Lwer as the water in the black sea is reducing thus U +4 will be the abundant phase and it is less sluble. (4) When diagenesis prceeds in marine sediments, there is a natural sequence f cmpunds used t xidize rganic matter. Describe the sequence f redx pairs and draw a hypthetical vertical prfile f sediment clumn with the apprpriate species fund in pre waters. In what majr way wuld yu expect this sequence t change in fresh water sediments cmpared t marine sediments? Hw wuld yu expect the particulate Mn and Fe prfiles in this sediment t be? The sequence f redx pairs is: Aerbic Respiratin 1/4CH 2 O + 1/4O 2 = 1/4H 2 O + 1/4CO 2 (g) Denitrificatin 1/4CH 2 O + 1/5NO /5H + = 1/4CO 2 (g) + 1/10N 2 (g) +7/20H 2 O Manganese Reductin 1/4CH 2 O + 1/2MnO 2 (s) + H + = 1/4CO 2 (g) + 1/2Mn /4H2O Irn Reductin 1/4CH 2 O + Fe(OH) 3 (s) + 2H + = 1/4CO 2 (g) + Fe /4H 2 O Sulfate Reductin 1/4CH 2 O + 1/8SO /8H + = 1/4CO 2 (g) + 1/8HS - + 1/4H 2 O Methane Fermentatin 1/4CH 2 O = 1/8CO 2 (g) + 1/8CH 4 In fresh water sediments sulfate reductin might nt be prminent as these waters are lw in sulfate. Particulate Fe and Mn will precipitate at the xic anxic bundary where Mn and Fe diffusing up see xygen.

3 (5) What wuld be the Mn 2+ in equilibrium with MnO 2 in natural seawater at ph=8 in equilibrium with P O2 = 0.20 thus, pe = Hw wuld yu answer change fr the Black Sea with ph= 8 and pe = -4? S under anxic cnditins wuld yu expect manganese t be in the frm f Mn 2+ r MnO 2? Assume: MnO 2 (s) + 4H + + 2e - = Mn H 2 O lgk = 2p = 41.6 K = [Mn 2+ ]/ (H + ) 4 (e) 2 lg K = lg [Mn 2+ ] 4lg (H + ) 2lg (e) lg (Mn 2+ ) = lg K - 4pH - 2pE fr seawater lg (Mn 2+ ) = r (Mn 2+ ) = r 2.7 x nt abundant fr the black sea (Mn 2+ ) = 3.98 x s Mn 2+ is a lt mre abundant in anxic cnditins b) In xygenated surface seawater disslved Mn 2+ = This thermdynamic calculatin des nt appear t crrectly describe the distributin f Mn species in seawater since the cncentratin f disslved Mn ranges frm nm in seawater. Why nt? (a) Equilibrium des nt exist in this case. (b) Ligands and cmplexatin f Mn 2+ in slutin increases it ttal cncentratin ver that expected frm equilibrium cnsideratins in ideal slutins. (6) In anxic basins dminated by sulfate reductin what will the dminant Idine species be? IO H + + 6e- == I - + 3H 2 O lgk = 110.1; pe = -4.5? Cmpare this with xygenated seawater (pe =13), assume ph is 8 at bth cases lgk = lg {[I - ]/[IO 3 - ]} + 6pH +6pE = lg {[I - ]/[IO 3 - ]} = 89.1 fr anxic and 15.9 fr seawater; s [I - ]/[IO 3 - ] = 1.26 x e.g. I - dminant in anxic cnditins and 1.26 x thus the dminant frm will be idate IO 3 - in seawater (7) Please answer the fllwing questins with the use f the pe/ph plt belw fr varius species f lead. In this figure, slid phases are indicated by dashes entending alng phase bundaries tward the center f the stability field. As with all such stability field applicatins, assume that the system is at equilibrium: (a) What is the expected majr species f lead under surface seawater cnditins? PbOH +, PbO 2 (s) (b) Under what ph range is PbO 2(s) stable under xic cnditins? Frm ph 7-14 (c) Over what pe range can tw sluble lead species cexist? 6 t 12 (d) What chemical prduct may frm if yu drp metalic lead, Pb (s), int acidic (ph = 2) water? Pb 2+ (e) What tw chemical species can PbO(s) directly cnvert t withut any net transfer f electrns? Pb(OH) 3 -, PbOH +

4 (8) Has the sediment cmpsitin in the prfiles presented here changes much with depth? Explain. Where is the xic anxic bundary in these prfiles? Why d bth disslved Fe and Mn increase dwnward frm the transitin? Hw wuld yu expect the disslved U in this prfile t behave? What will the sulfate pre water prfile lk like? In the upper 20 centimeters the majr change in cmpsitin wuld be rganic mater xidatin as indicated by the lss f xygen and nitrate. Belw that xides in the sediment are als lst releasing Mn 2+ and Fe 2+ int slutin. The majr aluminumsilicates have nt changed.

5 (9) OXIDATION - REDUCTION OF IRON "RUST" Which xidatin state f Fe (+2 r +3) is expected t be mre abundant in seawater (pe = and which in the deep waters f the black sea (pe = -3.5)? Cnsider this reactin: Fe /2 H 2 (g) <==> Fe +2 + H + Which is cmpsed f tw half reactins: reductin Fe +3 + e - <==> Fe +2 lgk2 = 13.0 xidatin 1 / 2 H2(g) <==> H + + e - lgk3 = 10 By cnventin, K3 = 1 ( G = 0) K1 = K2 x K3 ==> K 1 K 2 (Fe 2 ) (Fe 3 )(e ) lg K 1 lg(e ) lg (Fe 2 ) (Fe 3 ) Fr seawater: lg (Fe 2+ /Fe 3+ ) = = -0.75; (Fe 2+ /Fe 3+ ) =0.177 Fe 3+ abundant Fr the black sea: lg (Fe 2+ /Fe 3+ ) = = 16.5; (Fe 2+ /Fe 3+ ) =3.16 X10 16 Fe 2+ abundant Nte analgy t ph, and acid-base reactins. Recall that ph= pk + lg (A - )/(HA) By cnventin, the reduced species is in the numeratr, xidized in the denminatr.

6 (10) In cean waters with very lw xygen cncentratins such as the xygen minimum zne ff Peru (O 2 < 5 ml/l), denitrificatin reactins are f imprtance: NO rgc == CO 2 + ½N 2 (a) If bth NO 3 - and N 2 are present what wuld yu expect the pe t be if this redx cuple NO 3 - / N 2 ) wuld buffer the pe level? (b) What will the partial pressure f xygen be at this pe? (c) What Idine species wuld be dminant in these cnditins (IO 3 -, I - r I 2 ). The fllwing infrmatin might be useful: (assume 25ºC, ph 8, 1atm, pn 2 = 0.8 atm, NO - 3 = 40 M) NO H + + 5e = ½ N 2 + 3H 2 O lg K = O 2 + 4H + + 4e = 2 H 2 O lg K = 83.1 IO H + + 5e = I - + 3H 2 O lg K = ½I 2 + e = I - lg K = 10.5 (a) K = [pn 2 ] ½. [NO 3 - ] [H + ] 6 (e) 5 lg K = ½ lg [N 2 ] lg [NO 3 - ] 6 lg [H + ] 5 lg (e) = ½ lg [0.8] lg [40 x 10-6 ] + 6 ph + 5 pe 5 pe = ½ lg [0.8] + lg [40 x 10-6 ] - 6 ph 5 pe = ( ) + (-4.398) 48 =52.95 pe = (b) K = [H 2 O] 2. [po 2 ] [H + ] 4 (e) 4 lg K = -lg [po 2 ] 4 lg [H+] 4 lg (e) lg [po 2 ] = - lg K + 4pH + 4pE lg [po 2 ] = = po 2 =1.8 x 10-9 Very little xygen!!! (c) IO H + + 5e = I - + 3H 2 O slve this fr ph = 8 and pe = lg K = lg {[I - ]/[IO 3 - ]} + 6pH + 6pE = lg {[I - ]/[IO 3 - ]} = S: [I - ]/[IO 3 - ] = The dminant frm will be IO 3 - (Idate) but bth are present and if the pe was 10 I - wuld have been dminant. Similarly slve fr ½I 2 + e = I - abundant. yu wuld find that [I - ]/[I 2 ] ½ = 0.8 e.g I 2 mre

7 (11) The Black Sea is the wrld s largest permanently anxic water bdy. It is a semienclsed marginal sea with a physical and chemical structure that is determined by its hydrlgical balance. Seawater flws in thrugh the Bsprus t the deep layer f the basin. Freshwater inflw frm Eurpean rivers keeps the salinity lw in the surface layer. As a result the water clumn is strngly stratified with respect t density. A cnsequence is that the surface layer (abut 0 t 50m) is well xygenated while the deep layer (100m t 2000m) has n xygen and high sulfide cncentratin. At the bundary between the xic surface and anxic deep layers there is a subxic zne (frm apprximately 50m t 100m depth) where O 2 and HS - are essentially bth absent. The distributins f nutrients (e.g. NO 3 -, NH4 +, PO 4 ) and metals (e.g. Mn, Fe) and the absence f O 2 and HS suggest the ccurrence f unique r unusual metablic prcesses within the subxic zne (e.g., anaerbic Mn(II), Fe(II), NH 4 +, HS, and S(0) xidatin and the pssibility fr unusual phtsynthetic reactins). Amy C. has prpsed that nitrate reacts with ammnia t prduce N 2 (g). Using the half-reactins in Table 7.3 f the lecture ntes d the fllwing: a) Write a balanced reactin fr NO 3 + NH 4 + t prduce N 2 (g) 0.8 NH NO 3 - == 0.05 H N H 2 O b) The equilibrium cnstant fr this reactin? E 0 = = 0.39 E 0 = /n lgk lgk = (n/0.0592) x E 0 = 6.58; K =3.8 x 10 6