Oxidation-Reduction (Redox) Reactions
Eh, Ph, and the chemistry of natural waters
Important equations and terms for redox reactions:
ΔGR = nFE
Nernst factor at standard state
°
ΔGR = ΔGR + RTlnQ
Eh = E = E° + (2.303RT/nF)logQ = (0.05916/n)logQ
E° = -(0.05916/n)logK
Eh = 1.23 + 0.01479log[O2] - 0.05916pH
R = 1.987 x 10-3 kcal/mol°
T = 298.15°K
F = 23.06 kcal/V
n = number of electrons transferred in the reaction
Q = reaction quotient
E° = standard potential for a half-reaction
Eh = electromotive force (emf) generated between an electrode
in any state and the H2 electrode (the reference
hydrogen scale Eh) in standard state (theoretical voltage)
Zn(s) + Cu 2+ (aq) ←→ Zn 2+ (aq) +Cu(s)
Each metal has a different relative preference for electrons
Electrode Potential
Separate the reaction into half-reactions and add the e-
Zn(s) + Cu 2+ (aq) ←→ Zn 2+ (aq) +Cu(s)
Zn(s) ←→ Zn 2+ (aq) + 2eCu 2+ (aq) + 2e-←→ Cu(s)
There will be Energy gained or lost by the exchange of eUnfortunately, such measurements are not possible
(nor would these reactions occur in the natural environment:
electrons are not given up except to another element or species).
This requires the establishment of an arbitrary reference value
Hydrogen Scale Potential Eh
1
+
H2 (g) ←
→ H (aq ) + e −
2
The potential measured for the entire reaction is then
assigned to the half-cell reaction of interest:
Zn(s) ←→ Zn 2+ (aq) + 2e-
becomes
2+
Zn (aq ) + H 2 ( g ) ←
→ Zn( s ) + 2 H
+
The potential for this reaction = -0.763 V
So the Eh of the reduction of Zn2+ to Zn (s) = -0.763 V
2+
Zn (aq ) + H 2 ( g ) ←
→ Zn( s ) + 2 H
+
(+) Eh means reaction moves left to right
(-) Eh means reaction moves right to left
Eh = E = E° + (2.303RT/nF)logQ = (0.05916/n)logQ
(+) Eh means the metal ion is reduced by H2(g) to the metal
(-) Eh means the metal will be oxidized to the ion and H+ is reduced
Calculating the Eh of Net Reactions
We can calculate Eh values by algebraic combinations of the
reactions and potentials that are listed in Table 14.3 of the text.
There is, however, a “catch”
Calculate the Eh for the reaction:
Fe3+(aq) + 3e- ←→ Fe(s)
Fe3+(aq) + 3e- ←→ Fe(s)
This reaction is the algebraic sum of two reactions listed in
Table 14.3
Fe3+(aq) + e- ←→ Fe2+(aq)
Fe2+(aq) + 2e- ←→ Fe(s)
Since the reactions sum, we might assume that we can simply sum
the Eh values to obtain the Eh of the net reaction.
Doing so, we obtain an Eh of 0.33 V.
However, the true Eh of this reaction is -0.037 V.
What did we do wrong?
We have neglected to take into consideration the number of
electrons exchanged.
Fe3+(aq) + e- ←→ Fe2+(aq)
Fe2+(aq) + 2e- ←→ Fe(s)
In the algebraic combination of Eh values, we need to multiply the
Eh for each component reaction by the number of electrons
exchanged.
We then divide the sum of these values by number of electrons
exchanged in the net reaction to obtain the Eh of the net reaction
1
Eh(net ) =
z net
∑ z Eh
i
i
i
1
Eh(net ) =
z net
∑ z Eh
i
i
i
where the sum is over the component reactions i.
By Hess’s Law, the ΔG of the net reaction must be the simple
sum of the component reaction ΔG’s
…but Eh values are obtained by multiplying ΔG by z.
ΔGR = nFE
1
Eh(net ) =
z net
∑ z Eh
i
i
i
Eh = 1/3 (1*0.77+2*-0.44) = 1/3 (0.77-0.88) = 1/3(-0.11)=-0.037
Fe3+(aq) + e- ←→ Fe2+(aq)
Fe2+(aq) + 2e- ←→ Fe(s)
a
b


RT
(A)
(B)
o

Eh(volts) = E +
ln
c
d 
nF  (C) (D) 
R = gas constant = 0.001987 kcal/molo
RT
[H +]
o Kelvin
T=
temp.
in
Eh
=
E
+
ln
1/2
Eh = E° + (2.303RT/nF)logQ
=
(0.05916/n)logQ
F = 23.061
[PH 2 ] kcal/voltgeq
F = faraday constant
o
oelectrons
n =Enumber
of
= - ΔG R lnF
2.303RT
- factor = 0.05916 volts at 25o C
=
Nernst
aA
F + bB + ne = cC + dD
reduced
E ooxidized
= std. potential of half
- reaction
o
Δ
G
o
f
E =
nF
If an environment is characterized by a certain redox
reaction, it has a characteristic Eh
[O
2]
=1
atm
Oxidizing
Reducing
[O
2]
=1
0 -83.1
From Garrels & Christ (1965)
atm
Similar to Fig. 14.4 on p. 238
Eh = 1.23 + 0.01479log[O2] - 0.05916pH
[O2] conditions for
natural waters
1) write the equation for the reaction
H 2O(l ) ↔ 12 O2 ( g ) + H 2 ( g )
2) “electrode” the equation
H 2O(l ) ↔ 2 H + 12 O2 ( g ) + 2e −
+
O in H2O gives up 2e-
2e-
3)
0.05916
1/ 2
+ 2
Eh = E +
log [O2 ] [ H ]
2
o
(
)
4) get Eo from ΔGro
ΔGro = [2G of ( H + ) + 1 / 2G of (O2 ) + 2G of (e−)] − [G of ( H 2O(l )]
G of H + = G of O2 = G of e− = 0
ΔGro = −G of H 2O = +56.687 kcal
56.687
E =+
= +1.23V
2 x 23.06
o
6)
0.05916
Eh = 1.23 +
log [O2 ]1/ 2 [ H + ]2
2
Eh = 1.23 + 0.01479 log[O2 ] − 0.05916 pH
(
)
This demonstrates the Eh - pH relationship as shown in
figure 14.4 p. 238 and is the equation for the lines limiting
the minimum and maximum [O2] for natural waters
Oxidizing
Reducing
Figure 14.11 on p. 248