The Iron Mystery If: The Fe +2 /Fe +3 boundary is at a pe o and - PDF document

CEE 680 Lecture #48 4/29/2020 Print version Updated: 29 April 2020 Lecture #48 Redox Chemistry: Log C vs pe Diagrams (Stumm & Morgan, Chapt.8 ) Benjamin; Chapter 9 David Reckhow CEE 680 #48 1 The Iron Mystery  If:  The Fe +2 /Fe +3 boundary is at a pe o and pe(w) of 13.03  Oxygen saturated water should have a pe(w) of 13.6, but Pankow says the effective pe(w) of surface water is more like 12.6  Will reduced iron spontaneously oxidize to ferric in surface waters? Yes 1. No 2. Then why is it so hard to keep reduced iron (ferrous) from oxidizing to the ferric form? David Reckhow CEE 680 #48 2 1

CEE 680 Lecture #48 4/29/2020 Redox and pH effects  Often oxidation of metals results in a more hydrolyzed species  Acidity of oxidized species is higher, resulting in release of protons  Speciation changes and affects the overall reaction  A good example is the oxidation of ferrous iron to ferric  Fe +3 + e ‐ ↔ Fe +2  pe o = 13.03  This is very close to the theoretical pe o defined by saturated O 2 in water (13.6), or the effective pe o (e.g., 12.6)  But this is deceptive, because Fe +3 isn’t the dominant species at neutral pH David Reckhow CEE 680 #48 3 Ferrous Hydroxides: α diagram  Fe +2 dominates at pH 7.0 neutral pH Fe +2  3 1e+0  0 1e-1 1e-2  2  1 1e-3 1e-4  FeOH + 1e-5 1e-6 - 1e-7 Fe(OH) 3 o 1e-8 Fe(OH) 2 1e-9 1e-10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH David Reckhow CEE 680 #48 4 2

CEE 680 Lecture #48 4/29/2020 Ferric Hydroxides: α diagram  Fe +3 is a minor pH 7.0 species at neutral pH  2  4 1e+1  0  1  3 1e+0 1e-1 o Fe(OH) 3 Fe +3 1e-2 1e-3 1e-4  FeOH +2 1e-5 - Fe(OH) 2 1e-6 1e-7 - Fe(OH) 4 1e-8 1e-9 1e-10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH David Reckhow CEE 680 #48 5 Stumm & Morgan, 1996; Iron redox diagram Fig. 8.1, pg. 435 Similar to: Benjamin, 2002 Fig 9-3, pg.486  2 [ Fe ]   Fe  Analogous to log C vs pH diagram K { e }  3 [ ] 𝐿 � 10 �� . �� David Reckhow CEE 680 #47 6 3

CEE 680 Lecture #48 4/29/2020 Iron: considering speciation 4.70 13.03 David Reckhow CEE 680 #48 7 HOCl and H 2 S example  Neutral pH (~7.0)  0.1 mM HOCl, 1 mM Cl ‐ ‐ 2  1 mM H 2 S and SO 4 David Reckhow CEE 680 #48 8 4

CEE 680 Lecture #48 4/29/2020 Summary David Reckhow CEE 680 #48 9 1  pe o log K n Hypochlorite  5 x 10 ‐ 4 M Cl T  Where: Cl T = [HOCl] + [OCl ‐ ] + [Cl ‐ ]        HOCl H e Cl H O 1 1 1 1 2 2 2 2 2  { Cl } 0 . 5    25 . 1 K 10 0 . 5  0 . 5  { HOCl } { H } { e }   1 [Re d ]       p p o log   n [ Ox ]      0 . 5 { Cl }     25 . 1 log    { HOCl } 0 . 5 { H } 0 . 5      { Cl }      25 . 1 1 log 1 pH   2 2 David Reckhow { CEE 680 #48 HOCl } 10   5

CEE 680 Lecture #48 4/29/2020 Determining Equilibrium Concentrations  Graphical solution analogous to acid/base problems  Create LogC vs pe diagram  Determine location on graph using electron balance  Analogous to proton balance in acid/base problems  Example: HOCl and NaHS  Reduced species: Cl ‐ which is 2e ‐ poor ‐ 2 which is 8e ‐ rich  Oxidized species: SO 4 David Reckhow CEE 680 #48 11 HOCl & HS ‐  10 ‐ 4 M S T  5 x 10 ‐ 4 Cl T David Reckhow CEE 680 #48 12 6

CEE 680 Lecture #48 4/29/2020 Electron Balance         Oxidation HOCl H 2 e Cl H O 2      Reduction 2     SO 9 H 8 e HS 4 H O 4 2  e ‐ Balance: HS- HOCl -2 ] = 2[Cl - ] 8[SO 4 David Reckhow CEE 680 #48 13 Constants  Reference reaction     H e 1 H ( g ) 2 2  Where {e ‐ }=1, if all chemical species activities are also unity 0 . 5 { H ( g )}   K 2 1 . 0   { H }{ e } David Reckhow CEE 680 #47 14 7

CEE 680 Lecture #48 4/29/2020 pE bounds for water I  Oxygen and Hydrogen half cell reactions { H } 0 . 5 2  ( g )  K 1 . 0   { H }{ e }    0 . 5 { H }{ e } { H } 2 ( g )     log{ H } log{ e } 0 . 5 log{ H } 2 ( g )     log P 2 pH 2 p H 2 Stumm & Morgan, 1996; Fig. 8.2, pg. 437 David Reckhow CEE 680 #48 15 pE bounds for water II David Reckhow CEE 680 #48 16 8

CEE 680 Lecture #48 4/29/2020 Electron Balance for HOCl & HS ‐        HOCl H 2 e Cl H O  Oxidation 2      O 4 H 4 e 2 H O  Reduction 2 2     2     SO 9 H 8 e HS 4 H O 4 2     4 H 4 e 2 H  e ‐ Balance: 2 H 2 O HS- HOCl 4[O 2 (aq) ] + 8[SO 4 -2 ] = 2[Cl - ] + 2[H 2 (aq) ] 17 David Reckhow CEE 680 #48 HOCl and HS ‐ David Reckhow CEE 680 #48 18 9

CEE 680 Lecture #48 4/29/2020 Close ‐ up of electron balance David Reckhow CEE 680 #48 19 Nitrogen David Reckhow CEE 680 #48 20 10

CEE 680 Lecture #48 4/29/2020 Nitrogen & Chlorine David Reckhow CEE 680 #48 21 Redox Predominance for N David Reckhow CEE 680 #48 22 11

CEE 680 Lecture #48 4/29/2020  To next lecture DAR David Reckhow CEE 680 #48 23 12