Redox and pE – pH Diagrams OCN 623 – Chemical Oceanography Reading: Libes, Chapter 7
Redox Equations Worth Knowing • ∆G = − nFE a oxid 0 . 0592 ο log • E=E + b n reduc
• •
a oxid ο pE = pE + log b reduc n F pE = Eh 2.3RT
1
Equations are written as REDUCTIONS!!
1
•
pE ο = log K
•
∆G pE = − 2.3nRT
n
This lecture will go briefly over the principles involved in deriving and using these equations. These principles are all based on electrochemical cells.
Electrochemical Cells Consider the following simple electrochemical cell operating at 25°C, used to measure the electrochemical potential of this reaction: Cu2+(aq) → Cu0(s)
ions
We arbitrarily assign a potential of 0 to the reaction in the left cell: 2H+(aq) + 2e- → H2(g)
E°= 0.000 V
Then the potential for the reaction in the right cell is: Cu2+(aq) + 2e- → Cu0(s)
E°= 0.337 V
(always write as a reduction)
The standard potentials for all redox reactions are similarly determined against the standard hydrogen electrode:
= E° = E°H
An electrochemical cell is capable of doing work – by driving electrons across a potential difference. This can be measured as a change in free energy: ∆Gr° = -nFE° where n = number of moles of electrons (equivalents) involved in the reaction F = Faraday constant = 23.1 kcal V-1 equiv-1 E° = the cell potential (V) at standard state For the general case: ∆Gr = -nFE
Nernst Equation We know from a previous class:
{ products }x ∆G = ∆G ° + RT ln {reactants } y
reduced species oxidized species
Substituting ∆G = -nFE, we get the Nernst Equation:
RT { products } x E = E° − ln nF {reactants } y Or:
At 298°K:
y RT {reactants } E = E° + ln nF { products }x y 0.0592 {reactants } E = E° + log n { products } x
Important points: • Geochemists usually use the symbol EH instead of E (indicating the hydrogen scale is being used) • The Nernst Equation relates the EH of a cell to the standard EH and to the activities of reactants and products under given conditions • When at standard state (all activities = 1), EH = E° • We can use EH as an indicator of the state of natural waters:
Example #1: Use Eh values to Calculate Vanadium Speciation V
3+
−
+e ↔V
2+
What species of V dominates in seawater with measured Eh = 0.729 V? 3+ 0 . 059 V ο Eh = Eh + log 2 + 1 V 1. From Table 7.1: Ehº= -0.26 2. Plug into equation:
Definition of pE a(oxidized-species) + ne- → b(reduced-species) From equilibrium theory: b { reduc} K= {oxid }a { e − }n
1
{e }
− n
a { oxid } =K {reduc}b
Take the log :
{ oxid } 1 n log − = log K + log e {reduc}b a
{ }
1 but : log − = − log{e − } ≡ pE {e }
{ 1 1 oxid } ∴ pE = log K + log n n {reduc}b a
but :
pE ο =
1 n
log K
{ 1 oxid } ο ∴ pE = pE + log n {reduc}b a
and divide by n
Example #2: Calculation of pE Assume: • pE in a given environment is controlled by this reaction: Fe3+ + e- → Fe2+
pE = pE
(∴n = 1) • {Fe3+} = 10-5 • {Fe2+} = 10-3
Table 7.1
pE
ο
ο
1 oxid + log n reduc
log (K = n
10 ∴ pE = 13 . 0 + log 10
{ e } = 10 −
) = 13 . 0 −5 −3
− 11
= 11 . 0
Example #3: Effect of Atmospheric O2 Assume: Natural water at pH 7.5 in equilibrium with the atmosphere PO2 = 0.21atm 1 4
∴{ O2 } = 0.21
O2 ( g ) + H + (aq ) + e − ↔
1 2
H 2O(l )
pE ο = 20.75
ο
pE = pE +
(Table 7.3, Libes)
{ O2 } {H + } log 1 n {H 2O}2 1
1
pE = 13.08
1 4
{H } = 10 +
−7.5
{ }
∴ e − = 10−13.08
Conclusion: This environment has lower electron activity than Example #2, and is thus more oxidizing
Redox Reactions Have Characteristic pE Values pE =
F EH 2.3RT
pE =
− ∆G 2.3nRT
Thus, if an environment is characterized by a certain redox reaction, it has a characteristic pE
¼ O2 + H+ + e- ← ½ H2O
H2O + e- → ½ H2 + OH-
Garrels & Christ (1965)
pE-pH Diagrams pE-pH stability field diagrams show in a comprehensive way how protons (pH) and electrons (pE) simultaneously shift equilibria of reactions under various conditions These diagrams also indicate which species predominate under any given condition of pE and pH Two equations are used to produce the diagrams:
{ 1 oxid } pE = pE ° + log n {reduc}r o
1 pE ° = log K n
Oxidizing limit of diagrams: ¼ O2 + H+ + e- → ½ H2O pE° = +20.75 (Table 7.3) n=1 pH = -log{H+} Set limit: {O2} = 1
{ }
{ 1 O2 } 4 H+ pE = pE ° + log 1 n {H 2O} 2 1
pE = 20.75 − pH
Reducing limit of diagrams: H2O + e- → ½ H2 + OHOH- + H+ → H2O H+ + e- → ½ H2 pE° = 0.0 (Table 7.3) n=1 pH = -log{H+} Set limit: {H2} = 1
{ }
1 H+ pE = pE ° + log 1 n {H 2 } 2 pE = − pH
Oxidizing limit of diagrams: O2 + H+ + e- → ½ H2O
pE = 20.75 − pH
Reducing limit of diagrams: H+ + e- → ½ H2 pE = − pH
• Phase-boundary lines on a pE-pH diagram indicate stability field boundaries – defined as lines where activities of both adjacent dominant species are equal. • Lines are defined by reactions between adjacent dominant species • Reactions must have known log K or pE° values.
Construction of pE-pH Diagrams Acid-base reactions with no pE dependency −
6. HSO 4 ↔ SO 4
2−
+ H+
logK = − 2.0
{ SO }{ H } K = { HSO } { SO } log K = − 2.0 = log − pH { HSO } 2−
+
4
−
4
2−
4
−
4
{
When SO4
2−
}= { HSO }, −
4
pH = 2.0
Redox reactions of dissolved species 2−
5. SO 4 + 9H + + 8e − ↔ HS− + 4H 2 O
{
2−
log K = 34.0
}{ }
9
+ 1 SO H 4 pE = pE ο + log 4 8 HS − {H 2O}
{
pE ο =
}
1 34 log K = 8 n
{
2−
34 1 SO4 pE = + log 8 8 HS −
{
{
When SO4
2−
=1
} = { HS }, −
}−
}
pE =
9 pH 8 34 9 − pH 8 8
Redox reactions of dissolved and solid species 2−
1. SO 4 + 8H + + 6e − ↔ S(s ) + 4H 2 O
{
2−
log K = 36.2
}{ }
8
36.2 1 SO4 H+ pE = + log 6 6 { S(s )}{ H 2O }4
{
=1
}
36.2 1 8 2− pE = + log SO4 − pH 6 6 6
{
When SO4
2−
} = 10
−2
,
pE = 5.70 −
Activity of dissolved species must be given
8 pH 6
SI
Homework – Due Thurs, Jan. 24, 2013 1) Write a balanced equation for the nitrification of NH4+ to NO3- (oxidation by O2). 2) Calculate the Eh at 25°C for this reaction in offshore Hawaiian seawater at 100 m depth. Assume [NH4+] = 1 µM, [NO3- ] = 150 nM, and the partial pressure of O2 = 0.21 atm; assume the activity coefficients for NH4+ and NO3- are 0.7 . (Hint: calculate ∆G° and then Eh.)
3) In which direction would the reaction proceed in an environment with an Eh of -200 mV? Why?
