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DM Sherman, University of Bristol
Oxidation-Reduction
Equilibria
Geochemistry, DM Sherman University of Bristol
Oxidation States of Atoms and Ions
- The oxidation state of an atom in an elemental form is 0.
- When bonded to something else, oxygen is in oxidation state -2 and hydrogen is in oxidation state of +1 (except for peroxide and superoxide).
- The oxidation state of a single-atom ion is the charge on the ion. For Fe2+, Fe is in +2 oxidation state. In CO 3 2-, O is in -2 state, C is in +4 state. In O 2 , O is in the 0 oxidation state. The oxidation state of an atom is defined with the following convention:
DM Sherman, University of Bristol
Oxidation-Reduction Reactions
- Fe2+^ is being oxidized to Fe3+^ (as Fe(OH) 3 )
- Mn4+^ (as MnO 2 ) is being reduced to Mn2+ We can express the overall reaction as two half- reactions: 2Fe2+^ + 6H 2 O = 2Fe(OH) 3 + 6H+^ + 2e- MnO 2 + 4H+^ + 2e-^ = Mn2+^ + 2H 2 O
2Fe2+^ + MnO 2 + 4H 2 O = 2Fe(OH) 3 + 2H+^ + Mn2+ Consider the reaction 2Fe2+^ + MnO 2 + 4H 2 O = 2Fe(OH) 3 + 2H+^ + Mn2+
K for half-reactions
K =
[ B]
[ A][ e"^ ] For each half-reaction, !
A + e"^ = B
We can define an equilibrium constant Where [e-] is the activity of electrons. ( This does not mean that bare electrons are floating around in solution!!)
DM Sherman, University of Bristol
The pe concept..
For convenience, take -log (= p) of the K expression to get By analogy with pH, the pe (= -log[e-]) can be used to characterize the redox state of a system.! pK = pB " pA " pe Knowing the pe, we can predict the oxidation state of the species in the system.
Speciation and pe
K =
[ Cr^3 +^ ][ H 2 O] [ HCrO 4 ][ H+^ ]^7 [ e"^ ]^3 Example: A soil has pH = 6.0 and pe= 5.0 (Eh = 0.3 V). Calculate speciation of Cr given pK = -60.6 for the reaction HCrO 4 -^ + 7H+^ + 3e-^ = Cr3+^ + 4H 2 O. ! p[ Cr] " p[ HCrO 4 " ] = " 3. 6 or [ Cr^3 +^ ] [ HCrO 4 " ]
Solution: pK = (p[Cr] - p[HCrO 4 ]) - 7pH - 3pe Since pH = 6.0 and pe 5.
DM Sherman, University of Bristol
Calculating pe from Concentrations
Example: Given that pK = -16.5 for the half-reaction Fe(OH) 3 (s) + 3H+^ + e- = Fe2+^ + H 2 O calculate the pe of groundwater in which [Fe2+] = 10-4^ M, pH = 7 and the solution is saturated in Fe(OH) 3. Solution: !
K =
[ Fe^2 +^ ]
[ H+^ ]^3 [ e"^ ]
pK = p[Fe] - 3pH - pe -16.5 = 4 - 21 - pe Hence, pe = -0.5.
Predicting Stability of Species
Would (NO 3 )-^ be stable in this groundwater? pK = -104. for the reaction NO 3 -^ + 6H+^ + 5e-^ = 1/2N 2 (g) + 3H 2 O !
K =
(PN 2 (g) )1/ [NO 3
][H+^ ]^6 [ e"^ ]^5
pK = -(1/2)log(PN2) -p[NO 3 - ] - 6pH - 5pe Since, PN2 = 0.8, pe = -0.5 and pH = 7, we get -104.6 = -(1/2)(-0.097) -p[NO 3 - ] - 42.0 +2. p[NO 3 - ] = 65. (Note: P = partial pressure)
DM Sherman, University of Bristol
Example: Speciation of Fe
Range of pe of Aqueous Solutions
By convention, ΔG^0 = 0.0 for the reaction H+(aq) + e-^ = 1/2H 2 (g) The most reducing condition that is possible at the Earth’s surface will have pH2 = 1 bar. Hence, pH + pe = 0 or
K =
( pH 2 )^1 /^2
[ H+^ ][ e"^ ]
pH = -pe
DM Sherman, University of Bristol
Range of pe of Aqueous Solutions (cont.)
The most oxidizing condition under which an aqueous solution can exist is buffered by the half-reaction Under the most oxidizing condition, pO 2 =1. Since [H 2 O] = 1, we have pH + pe = 20.75 or ! K = ( pO 2 )^1 /^4 [ H+][ e"] [ H 2 O]^1 /^2 = 10 "20.^75 1/2H 2 O = e-^ +1/4O 2 + H+ pe = 20.75-pH
pe-pH Environments
DM Sherman, University of Bristol
Cu-H 2 O System
Acidic solutions are necessary to mobilize Cu for supergene enrichment.
Summary
- Be able to decompose an oxidation-reduction reaction into half-reactions.
- Plot pe-pH diagrams..
- Know pe-pH environments in nature.
- Calculate pe from concentrations.
- Prediction of stable oxidation states.
