REDOX (2) pe as a master variable. C. P. Huang University of Delaware CIEG 632

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1 REDOX (2) pe as a master variable C. P. Huang University of Delaware CIEG 632 1

2 8.0 pe as a Master Variable E O,R = E o O,R RT log n R O E O, R = E o O, R RT n O log R E R, O = E o R, O RT n O log R E R, O = E o R, O RT + n log R O 2

3 pe as a Master Variable G = G o RT n log R O 3

4 4 pe as a Master Variable = = + = = = = = + O R log n 1 pe pe O R log n 1 n logk pe pe O R log n 1 n logk log(e) O R log n 1 n logk nlog(e) O R log logk (O)(e) (R) K Red;K ne Ox o n o O R o o E n G n K pe e pe,.. log ) log( = = = = = O R n pe pe o log 1 + = R O n pe pe o log 1

5 pe as a Master Variable G o = 2.303RTlogK = 1.364logK = nE o o G logk = o o G E = n = = o ne logk 16.9n = = 16.9nE o logK n G in Kcal/mol E in volt 5

6 + + + = + 6

7 ph versus pe 7

8 Example 8.2. Calculate the pε values for the following equilibrium systems (25 o C, I =0) 8

9 Example What is the pε of an acid rain containing 10-5 M of Fe 3+ and 10-3 M of Fe 2+? Fe 3+ + e = Fe 2+ ; E o = 0.76 v pε o = (1/n)log K = 16.91E o O,R =16.9x0.76 = pε = pe o log([fe +2 ]/[Fe 3+ ] = log(10-3 /10-5 ) = =

10 Example What is the pε of a natural water in equilibrium with atmospheric air at P O2 = 0.21 atm and ph = 7.5? ½ O 2 (g) + 2H+ + 2e = H 2 O(l); log K = (e) = pε =

11 Example Water in equilibrium with γ-mno 2 and Mn 2+ (10-5 M) at ph = 8. γ-mno 2 +4H + +2e = Mn 2+ +H 2 O; K = K =[Mn 2+ ]/([H + ] 2 [e] 2 ) = log([mn 2+ ]/[MnO 2 ]) +4pH + 2pε pε =(43.01/2) (1/2)(log 10-5 ) -16 =

12 Example Sediment containing FeOOH(s) and FeCO 3 (s) FeOOH(s) +HCO - 3 (10-3 M) + 2H + + e = FeCO 3 (s) + 2H + ; pε o (w) =-0.8 E H = pε/16.9 = pε o /16.9 = (- 0.8/16.9) = v (b) Deep layer of lake; DO =0.03 mg/l; ph=7 K H (O 2 ) = 10-3 M/atm (1/4) O 2 + H + + e = (1/2)H 2 O; pe o (w) =13.75 pε = (1/4)log (1/P O2 ) = (1/4) log(10-3 ) =13.75+(-0.75)=12.0 E H = pε/16.9 = 12.0/16.9 = 0.77 v P O2 = (3x10-5 /32) = 10-6 atm [O 2 )aq)] = 0.03 mg/l = 0.3x10-4 g/l = 3x10-5 (g/l)/32 (g/mol) = 10-6 M P(O 2 ) = [O 2 )aq)] /K H (O 2 ) =10-3 atm 12

13 Example 8.3. calculate the equilibrium composition of the following solutions (25oC, I=0) both in equilibrium with the atmosphere (P(O 2 )= 0.21 atm): = *LOG(0.21*10^(-4))= = *LOG(0.21*10^(-14))=

14 Log({Fe 3+ }/{Fe 2+ ]) = =

15 { Fe } 3+ = { Fe } K 10 = { Fe } { H } ( P O ) 2 2+ { Fe } = 10 2 ( / 2) { H }( P O ) = {Fe 3+ }+{Fe 2+ } = 10-4 {Fe 3+ }=10-4 {Fe 2+ }=10-4 / =

16 + 2 { H } = 10 =. { Mn }( ) K 2+ P O ph = 7 10 Mn = = 4. 35x10 10 ( 0. 21) { }

17 Fe 3+ /Fe 2+ 17

18 Fe 3+ /Fe 2+ log K 18 Stumm & Morgan

19 Water Stability 19

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21 Example 8.4. Construct a diagram showing the pε dependence of a 10-4 M SO HS - system at ph = 10 and 25 o C. = kj/mol SO4-742 HS 12.6 H2O

22 log (H 2 )=-20-2pe log (O 2 ) = pe pe 8(Pe+7) 1+8(Pe+7) log(1+8(pe+7) LOG(HS) LOG(SO4 log O2 Log(H2) E E E E E E E E E E E E E E E E E E E pe=-7+ (1/8)log([SO 4 ]/[HS]) log([so 4 ]/[HS]) = 8(pe+7) [SO 4 ]/[HS]=exp{8(pe+7)} S T =[SO 4 ]+[HS] S T =[HS](1+exp{8(pe+7)} [HS]=S T /(1+exp{8(pe+7)} log(hs) = log S T - log(1+8(pe+7)) log (SO 4 )= log (HS)+8(pe+7) 22

23 log (H 2 )=-20-2pe log (O 2 ) = pe 23 Stumm & Morgan

24 Example 8.6. reduction of NO 3 - to NH 4 + at ph =7 N T =10-4 M At ph = 7 [NH 4+ ]/[NH 3 ] = [e] 8 [NH 4+ ]=[NH 3 ] [e] 8 N T =[NH 3 ]+[NH 4+ ] N T ]=[NH 3 ]( [e] 8 ) [NH 3 ]=N T /( [e] 8 ) log[nh 3 ]= log N T log( [e] 8 ) log[nh 4+ ]=log[nh 3 ]+49.2+log[e] 8 24

25 pe e 1+10^(49.2) NH3 NH4 O2 H2 e E E E E E E E E E E E E E E

26 Example 8.7. Redox equilibrium of chlorine How do Cl 2 (aq), HOCl, OCl - and Cl - depend on pε in a solution in which TOTAL = 10-5 M? 26

27 Cl T = [HOCl] + [Cl - ] [Cl - ] = [HOCl] [H + ][e] 2 ph =2 Cl T = [HOCl]( [e] 2 ) Cl T = [HOCl]( [e] 2 ) log [HOCl] = -5 + log( [e] 2 ) log[ocl - ] = ph + log[hocl] log[cl - ] = pε + log[hocl] log[cl 2 (aq)] = pH - 2pε + 2 log[hocl] log[cl 2 (aq)] = pε + 2 log[hocl] log[cl 2 (aq)] = pε + 2 log[cl - ] ph =5 Cl T = [HOCl]( [e] 2 ) Cl T = [HOCl]( [e] 2 ) log [HOCl] = -5 + log( [e] 2 ) log[ocl - ] = ph + log[hocl] log[cl - ] = pε + log[hocl] log[cl 2 (aq)] = pε + 2 log[hocl] 27 log[cl 2 (aq)] = pε + 2 log[cl - ]

28 ph =8 Cl T = [HOCl]( [e] 2 ) Cl T = [HOCl]( [e] 2 ) log [HOCl] = -5 + log( [e] 2 ) log[ocl - ] = ph + log[hocl] log[cl - ] = pε + log[hocl] log[cl 2 (aq)] = pH - 2pε + 2 log[hocl] log[cl 2 (aq)] = pε + 2 log[hocl] log[cl 2 (aq)] = pε + 2 log[cl - ] 28

29 ph = 2 pe e 1+10^(48.8)e2 [HOCl] [Cl] [Cl2(aq)] [OCl] 18 1E E E E E E E E E E E E

30 pe e 1+10^(45.8)e2 [HOCl] [Cl] [Cl2(aq)] [OCl] 18 1E E E E E E E E E E E

31 pe e 1+10^(42.8)e2 [HOCl] [Cl] [Cl2(aq)] [OCl] 18 1E E E E E E E E E E E

32 Electrochemical potential revisit 32

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36 pε vs ph diagrams 36

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42 Fe(III)/CO 2 /H 2 O (1) Fe 3+ + e = Fe 2+ ; E o = ; K = (2) Fe 3+ + OH - = Fe(OH) 2+ ; K = (3) Fe(OH) 2+ + e = Fe 3+ + OH-; E o = 0.086; K = (4) Fe(OH) 3 (s) + e = Fe OH - ; E o = ; K = (5) FeCO 3 (s) = Fe 2+ + CO 2-3 ; K = (6) Fe(OH) 3 (s) + e + CO 2-3 = FeCO 3 (s) + 3OH - ; K = (7) FeCO 3 (s) + 2OH - = Fe(OH) 2 + CO 2-3 ; K = (8) Fe(OH) 3 (s) + CO e = FeCO 3 (s) + 3OH - ; E o = ; K = (9) Fe(OH) 2 (s) + OH - = Fe(OH) 3- ; K = (10) Fe(OH) 3 (s) + e = Fe(OH) 3- ; E o = -0.56; K =

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45 45 Stumm & Morgan

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49 2 + SO 4 + 8H + 6e = S(s) + 4H2O;logK = S (s) + 2H + 2e = H2 S(aq);logK = HS = S(s) + H + 2 e;logk = SO 4 + 3H 2S(aq) + 2H = 4S(s) + 4H2O;logK = Stumm & Morgan

50 S T = 10-2 M; ph =4 i) H 2 S-S S(s) + 2H + + 2e = H 2 S(aq); K = = [ H S( aq) ] [ S( s) ][ H ] [ e] 4.8 = log {[H 2 S(aq]/[S(s)]} +2pH +2e log {[H 2 S(aq]/[S(s)]} +2pE =4.8-2pH log {[H 2 S(aq]/[S(s)]} +2pE =-3.2 at ph =4 at log {[H 2 S(aq]/[S(s)]} = 0 pe = -0.6 pe < -0.6; [H 2 S(aq)] =S T =10-2 M pe >-0.6; log[h 2 S(aq)] =-3.2-2pE 50

51 ii) S(s)-SO 4 2- SO H + + 6e =S(s) +4H 2 O; K= = [ S( s) ] [ SO ][ H ] [ e] = log {[S(s)/[SO 2-4 ]} + 8pH + 6 pe log {[S(s)/[SO 2-4 ]} + 6 pe =36.2 8pH at ph =4 log {[S(s)/[SO 2-4 ]} + 6 pe =36.2 8pH = 4.2 at log {[S(s)/[SO 2-4 ]} =0, pe = 4.2/6 = 0.4 pe > 0.4; [SO 4 2- ]] = 10-2 pe <0.4; log [SO 4 2- ] = 6pE

52 iii) H 2 S(aq)/SO 4 2- SO H + +8e = H 2 S(aq)+ 4H 2 O; K= = [ H S( aq) ] [ ][ ] SO H [ e] = log{[h 2 S(aq)]/[SO 4 2- ] +10 ph + 8 pe log{[h 2 S(aq)]/[SO 4 2- ] }+8 pe = ph at ph = 4; log{[h 2 S(aq)]/[SO 4 2- ] } + 8pE = 1 at [H 2 S(aq)]/[SO 4 2- ] = 1; pe =1/8 = pe < 0.125; log[so 4 2- ] = 8 pe -1; [H 2 S(aq)] = 10-2 pe > 0.125; log[h 2 S(aq)] = 1-8 pe; [SO 4 2- ] =

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54 Reference electrodes AgCl/Ag electrode Calomel electrode 54

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62 (a) The chemical reaction taking place in cell (i) is composed of the following half reactions: E cell (i) is a direct measure of G o =-nfe cell (i) G o =1x x /1000= kj (b) The standard potential of the Cl 2 /Cl electrode, 0.5Cl 2 (g) + e = Cl - is given by: E(Cl 2,Cl)=E cell (i)+e(agcl,cl)= = V G o = FE= X = kj 62

63 (c) S o and H o for the cell Ag(s) +0.5Cl 2 (g)=agcl(s) S = nf(de cell /dt) H =-nfe + nft(de cell /dt) E cell = ( x10-1 )T-(2.7922x10-7 )T 2 de cell /dt= x x2.7922x10-7 T At 25 o C, de cell /dt= (mv/deg) E cell = V H o = 96500( *298.19)/1000 = kj/mol S o = 96500*( )= J/deg/mol 63

64 Effect of complex formation on EMF I = 0 I >0 = 64

65 E Zn RT 2F 1 ln [ Zn 2+ = E + 2 o, Zn( s) 2 Zn, Zn( s) + ] Zn( II) = = [ Zn 2+ ] + [ ZnCl 2+ 4 [ Zn ]( 1 + β [ Cl ] ) ] E Zn Zn( II) = 1 ] 2+ [ ] + β4[ Cl Zn 2+, Zn( s) o = E + Zn 2 4, Zn( s) RT 2F ln 4 ( 1+ β [ Cl ] ) 4 [ Zn( II)] E Zn 4 RT ( 1+ [ Cl ] ) + ln ([ Zn( )]) o RT = E 2+ ln II, Zn( s) Zn, Zn( s) 4 2F 2F 2+ β 65

66 Redox conditions in natural water pε o (w) =pε o +(n H /2) log K w Table 8.6a at ph = 7, NO 3 - can oxidize HS - to SO

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69 This, for a set of assumed actual conditions, such as p(co 2 )= atm, [CH 2 O]=10-6 M, [SO 4 2- ]=10-3 M, p(h 2 S) =10-2 atm, Q = Q/K= = < 0 69

70 Redox Intensity and the biochemical cycle 70

71 Microbial mediation 71

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76 Microbial Mediated Redox Reactions Oxygen reduction C organic + O 2 == CO 2 Nitrate reduction (nitrate reducing bacteria) 5C organic + 4NO 3 + 4H + == NH CO 2 + 2H 2 O Fe 3+ reduction C organic + 4Fe(OH) 3 + 8H + == 4Fe 2+ + CO H 2 O Fe 3+ is insoluble, Fe 2+ is soluble Sulfate reduction SO C organic + 2H 2 O == H 2 S + 2HCO 3 Pyrite versus siderite If Fe 2+ is present: Fe 2+ + H 2 S + S = FeS 2 (pyrite) + 2H + Pyrite (and other sulfide minerals are highly insoluble. If iron is not present, but HCO 3 is (i.e., ph > 7): Fe 2+ + HCO 3 == FeCO 3 (siderite) + H + 76

77 Microbially Mediated Redox Reactions Methanogenesis CO 2 reduction Aceticlasis Fermentation CO 2 + 4H 2 = CH 4 + 2H 2 O CH 3 COO - + H + = CO 2 + CH 4 No outside electron donors organic compounds serve as electron donor and acceptor Products included acetate and H 2 77

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85 Succession of Redox Environments 85

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