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

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

8.0 pe as a Master Variable E O,R = E o O,R 2. 303RT log n R O E O, R = E o O, R + 2. 303RT n O log R E R, O = E o R, O 2. 303RT n O log R E R, O = E o R, O 2. 303RT + n log R O 2

pe as a Master Variable G = G o + 2.303RT n log R O 3

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( 9 16 364 1 = = = = = O R n pe pe o log 1 + = R O n pe pe o log 1

pe as a Master Variable G o = 2.303RTlogK = 1.364logK = 23.064nE o o G logk = 1.364 o o G E = 23.604n = = o ne 0.05915 logk 16.9n = = 16.9nE o 0.05915logK n G in Kcal/mol E in volt 5

+ + + = + 6

ph versus pe 7

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

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 = 12.85 pε = pe o log([fe +2 ]/[Fe 3+ ] =1 2.85 log(10-3 /10-5 ) = 12.85-2 = 10.85 9

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 = 41.55 (e) =10-13.45 pε = 13.45 10

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 =10 43.01 K =[Mn 2+ ]/([H + ] 2 [e] 2 ) 43.01 = log([mn 2+ ]/[MnO 2 ]) +4pH + 2pε pε =(43.01/2) (1/2)(log 10-5 ) -16 = 8.0 11

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) = - 0.047 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 =13.75 +(1/4)log (1/P O2 ) = 13.75 + (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

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): =20.78+0.5*LOG(0.21*10^(-4))= 18.44 =20.78+0.5*LOG(0.21*10^(-14))=13.44 13

Log({Fe 3+ }/{Fe 2+ ]) = 18.44-13= 5.4 14

{ Fe } 3+ = { Fe } 3+ 2 15. 5 K 10 = 2+ 2 + 2 0. 5 { Fe } { H } ( P O ) 2 2+ { Fe } = 10 2 ( 15. 5 / 2) + 0. 25 { H }( P O ) = 10 5. 58 {Fe 3+ }+{Fe 2+ } = 10-4 {Fe 3+ }=10-4 {Fe 2+ }=10-4 /10 5.88 =10-9.88 15

+ 2 { H } = 10 =. { Mn }( ) 0. 71 K 2+ P O 2 0 5 ph = 7 10 Mn = = 4. 35x10 10 ( 0. 21) 14 2+ { } 0. 7 0. 5 15 16

Fe 3+ /Fe 2+ 17

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

Water Stability 19

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Example 8.4. Construct a diagram showing the pε dependence of a 10-4 M SO 4 2- -HS - system at ph = 10 and 25 o C. = kj/mol SO4-742 HS 12.6 H2O -237.2-194.2 21

log (H 2 )=-20-2pe log (O 2 ) =-43.1+4pe pe 8(Pe+7) 1+8(Pe+7) log(1+8(pe+7) LOG(HS) LOG(SO4 log O2 Log(H2) -12-40 1 0-4 -44-91.1 4-11 -32 1 0-4 -36-87.1 2-10 -24 1 0-4 -28-83.1 0-9 -16 1 0-4 -20-79.1-2 -8-8 1.00000001 4.34294E-09-4 -12-75.1-4 -7 0 2 0.301029996-4.301-4.301-71.1-6 -6 8 100000001 8.000000004-12 -4-67.1-8 -5 16 1E+16 16-20 -4-63.1-10 -4 24 1E+24 24-28 -4-59.1-12 -3 32 1E+32 32-36 -4-55.1-14 -2 40 1E+40 40-44 -4-51.1-16 -1 48 1E+48 48-52 -4-47.1-18 0 56 1E+56 56-60 -4-43.1-20 1 64 1E+64 64-68 -4-39.1-22 2 72 1E+72 72-76 -4-35.1-24 3 80 1E+80 80-84 -4-31.1-26 4 88 1E+88 88-92 -4-27.1-28 5 96 1E+96 96-100 -4-23.1-30 6 104 1E+104 104-108 -4-19.1-32 7 112 1E+112 112-116 -4-15.1-34 8 120 1E+120 120-124 -4-11.1-36 9 128 1E+128 128-132 -4-7.1-38 10 136 1E+136 136-140 -4-3.1-40 11 144 1E+144 144-148 -4 0.9-42 12 152 1E+152 152-156 -4 4.9-44 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

log (H 2 )=-20-2pe log (O 2 ) =-43.1+4pe 23 Stumm & Morgan

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 ] =10 49.2 [e] 8 [NH 4+ ]=[NH 3 ]10 49.2 [e] 8 N T =[NH 3 ]+[NH 4+ ] N T ]=[NH 3 ](1+10 49.2 [e] 8 ) [NH 3 ]=N T /(1+10 49.2 [e] 8 ) log[nh 3 ]= log N T log(1+10 49.2 [e] 8 ) log[nh 4+ ]=log[nh 3 ]+49.2+log[e] 8 24

pe e 1+10^(49.2) NH3 NH4 O2 H2 e -2 100 1.5849E+65-69.2-4 -63.1-10 2-1 10 1.5849E+57-61.2-4 -59.1-12 1 0 1 1.5849E+49-53.2-4 -55.1-14 0 1 0.1 1.5849E+41-45.2-4 -51.1-16 -1 2 0.01 1.5849E+33-37.2-4 -47.1-18 -2 3 0.001 1.5849E+25-29.2-4 -43.1-20 -3 4 0.0001 1.5849E+17-21.2-4 -39.1-22 -4 5 0.00001 1584893193-13.2-4 -35.1-24 -5 6 1E-06 16.8489319-5.22657238-4.02657238-31.1-26 -6 7 1E-07 1.00000016-4.00000007-10.8000001-27.1-28 -7 8 1E-08 1-4 -18.8-23.1-30 -8 9 1E-09 1-4 -26.8-19.1-32 -9 10 1E-10 1-4 -34.8-15.1-34 -10 11 1E-11 1-4 -42.8-11.1-36 -11 12 1E-12 1-4 -50.8-7.1-38 -12 25

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

Cl T = [HOCl] + [Cl - ] [Cl - ] = [HOCl]10 50.8 [H + ][e] 2 ph =2 Cl T = [HOCl](1 + 10 48.8 [e] 2 ) Cl T = [HOCl](1 + 10 48.8 [e] 2 ) log [HOCl] = -5 + log(1+10 48.8 [e] 2 ) log[ocl - ] = -7.3 + ph + log[hocl] log[cl - ] = 48.8-2pε + log[hocl] log[cl 2 (aq)] = 53.8-2pH - 2pε + 2 log[hocl] log[cl 2 (aq)] = 49.8-2pε + 2 log[hocl] log[cl 2 (aq)] = -47.2 + 2pε + 2 log[cl - ] ph =5 Cl T = [HOCl](1 + 10 45.8 [e] 2 ) Cl T = [HOCl](1 + 10 45.8 [e] 2 ) log [HOCl] = -5 + log(1+10 45.8 [e] 2 ) log[ocl - ] = -7.3 + ph + log[hocl] log[cl - ] = 45.8-2pε + log[hocl] log[cl 2 (aq)] = 43.8-2pε + 2 log[hocl] 27 log[cl 2 (aq)] = -47.2 + 2pε + 2 log[cl - ]

ph =8 Cl T = [HOCl](1 + 10 43.8 [e] 2 ) Cl T = [HOCl](1 + 10 43.8 [e] 2 ) log [HOCl] = -5 + log(1+10 43.8 [e] 2 ) log[ocl - ] = -7.3 + ph + log[hocl] log[cl - ] = 43.8-2pε + log[hocl] log[cl 2 (aq)] = 53.8-2pH - 2pε + 2 log[hocl] log[cl 2 (aq)] = 37.8-2pε + 2 log[hocl] log[cl 2 (aq)] = -47.2 + 2pε + 2 log[cl - ] 28

ph = 2 pe e 1+10^(48.8)e2 [HOCl] [Cl] [Cl2(aq)] [OCl] 18 1E-18 6.30957E+12-17.8-5 -21.8-21.2-23.1 19 1E-19 63095734449-15.8-5 -19.8-19.2-21.1 20 1E-20 630957345.5-13.8-5 -17.8-17.2-19.1 21 1E-21 6309574.445-11.8000001-5 -15.8-15.2-17.1 22 1E-22 63096.73445-9.80000688-5 -13.8-13.2-15.1 23 1E-23 631.9573445-7.80068777-5.0007-11.801-11.201-13.101 24 1E-24 7.309573445-5.86389203-5.0639-9.9278-9.3278-11.164 25 1E-25 1.063095734-5.02657238-6.2266-10.253-9.6531-10.327 26 1E-26 1.000630957-5.00027393-8.2003-12.201-11.601-10.3 27 1E-27 1.00000631-5.00000274-10.2-14.2-13.6-10.3 28 1E-28 1.000000063-5.00000003-12.2-16.2-15.6-10.3 29

pe e 1+10^(45.8)e2 [HOCl] [Cl] [Cl2(aq)] [OCl] 18 1E-18 6309573446-14.8-5 -21.8-21.2-17.1 19 1E-19 63095735.45-12.8-5 -19.8-19.2-15.1 20 1E-20 630958.3445-10.8000007-5 -17.8-17.2-13.1 21 1E-21 6310.573445-8.80006883-5.0001-15.8-15.2-11.1 22 1E-22 64.09573445-6.80682913-5.0068-13.814-13.214-9.1068 23 1E-23 1.630957344-5.2124426-5.4124-12.625-12.025-7.5124 24 1E-24 1.006309573-5.0027316-7.2027-14.205-13.605-7.3027 25 1E-25 1.000063096-5.0000274-9.2-16.2-15.6-7.3 26 1E-26 1.000000631-5.00000027-11.2-18.2-17.6-7.3 27 1E-27 1.000000006-5 -13.2-20.2-19.6-7.3 28 1E-28 1-5 -15.2-22.2-21.6-7.3 30

pe e 1+10^(42.8)e2 [HOCl] [Cl] [Cl2(aq)] [OCl] 18 1E-18 6309574.445-11.8000001-5 -21.8-21.2-17.1 19 1E-19 63096.73445-9.80000688-5 -19.8-19.2-15.1 20 1E-20 631.9573445-7.80068777-5.0007-17.801-17.201-13.101 21 1E-21 7.309573445-5.86389203-5.0639-15.928-15.328-11.164 22 1E-22 1.063095734-5.02657238-6.2266-16.253-15.653-10.327 23 1E-23 1.000630957-5.00027393-8.2003-18.201-17.601-10.3 24 1E-24 1.00000631-5.00000274-10.2-20.2-19.6-10.3 25 1E-25 1.000000063-5.00000003-12.2-22.2-21.6-10.3 26 1E-26 1.000000001-5 -14.2-24.2-23.6-10.3 27 1E-27 1-5 -16.2-26.2-25.6-10.3 28 1E-28 1-5 -18.2-28.2-27.6-10.3 31

Electrochemical potential revisit 32

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pε vs ph diagrams 36

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Fe(III)/CO 2 /H 2 O (1) Fe 3+ + e = Fe 2+ ; E o = 0.7817; K = 10 13.029 (2) Fe 3+ + OH - = Fe(OH) 2+ ; K = 10 11.57 (3) Fe(OH) 2+ + e = Fe 3+ + OH-; E o = 0.086; K = 10 1.45 (4) Fe(OH) 3 (s) + e = Fe 2+ + 3OH - ; E o = -1.4267; K = 10-24.12 (5) FeCO 3 (s) = Fe 2+ + CO 2-3 ; K = 10-13.45 (6) Fe(OH) 3 (s) + e + CO 2-3 = FeCO 3 (s) + 3OH - ; K = 10-13.46 (7) FeCO 3 (s) + 2OH - = Fe(OH) 2 + CO 2-3 ; K = 10 4.05 (8) Fe(OH) 3 (s) + CO 2-3 + e = FeCO 3 (s) + 3OH - ; E o = -0.796; K = 10-13.46 (9) Fe(OH) 2 (s) + OH - = Fe(OH) 3- ; K =10 8.44 (10) Fe(OH) 3 (s) + e = Fe(OH) 3- ; E o = -0.56; K =10-9.46 42

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2 + SO 4 + 8H + 6e = S(s) + 4H2O;logK = 36. 2 + S (s) + 2H + 2e = H2 S(aq);logK = 4. 8 + HS = S(s) + H + 2 e;logk = 2. 2 2 + SO 4 + 3H 2S(aq) + 2H = 4S(s) + 4H2O;logK = 21. 8 49 Stumm & Morgan

S T = 10-2 M; ph =4 i) H 2 S-S S(s) + 2H + + 2e = H 2 S(aq); K = 104.8 10 4. 8 = [ H S( aq) ] 2 + 2 2 [ 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

ii) S(s)-SO 4 2- SO 4 2- + 8H + + 6e =S(s) +4H 2 O; K=10 36.2 10 36. 2 = [ S( s) ] 2 + 8 [ SO ][ H ] [ e] 6 4 36.2 = 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 -4.2 51

iii) H 2 S(aq)/SO 4 2- SO 4 2- + 10H + +8e = H 2 S(aq)+ 4H 2 O; K=10 41 41 10 = [ H S( aq) ] [ ][ ] 2 + 10 SO H [ e] 8 4 2 41 = log{[h 2 S(aq)]/[SO 4 2- ] +10 ph + 8 pe log{[h 2 S(aq)]/[SO 4 2- ] }+8 pe = 41-10 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 = 0.125 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- ] = 10-2 52

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Reference electrodes AgCl/Ag electrode Calomel electrode 54

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(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 =1x96530.47x1.13569/1000=109.612 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)=1.13596+0.22234= 1.3580V G o = FE= -96530.47X1.3580 = -131.068 kj 62

(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 = 1.28958-(4.31562x10-1 )T-(2.7922x10-7 )T 2 de cell /dt=-4.31562x10-1 -2x2.7922x10-7 T At 25 o C, de cell /dt=-0.43173 (mv/deg) E cell = -0.12742 V H o = 96500(-0.12742-0.00043173*298.19)/1000 = -24.463 kj/mol S o = 96500*(0.00043173)=41.6618 J/deg/mol 63

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

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 ] ) 4 2 4 ] 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

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 4 2-66

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This, for a set of assumed actual conditions, such as p(co 2 )=10-3.5 atm, [CH 2 O]=10-6 M, [SO 4 2- ]=10-3 M, p(h 2 S) =10-2 atm, Q =10 2.35 Q/K=10-37.6+2.35 =10-35.25 < 0 69

Redox Intensity and the biochemical cycle 70

Microbial mediation 71

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Microbial Mediated Redox Reactions Oxygen reduction C organic + O 2 == CO 2 Nitrate reduction (nitrate reducing bacteria) 5C organic + 4NO 3 + 4H + == NH 4 + + 5CO 2 + 2H 2 O Fe 3+ reduction C organic + 4Fe(OH) 3 + 8H + == 4Fe 2+ + CO 2 + 10H 2 O Fe 3+ is insoluble, Fe 2+ is soluble Sulfate reduction SO 4 2- + 2C 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

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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Succession of Redox Environments 85

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