Agriculture Reference
In-Depth Information
Writing
(
1
/n)
log
K
=
pe
o
,
1
n
log
(
Red
)
pe
=
pe
0
−
(
4
.
7
)
(
Ox
)
and since
G
=−
2
.
303
RT
log
K
,
G
o
2
.
303
nRT
pe
o
=−
(
4
.
8
)
pe
o
is the electron activity with all the components of the redox couple at unit
activity, with unit activity of gases taken as a partial pressure of 1 atm. Table 4.1
gives values of pe
o
for important redox couples in natural systems expressed in
terms of unit electron transfer (i.e.
n
=
1 and pe
o
log
K
).
The pe of a solution indicates its relative tendency to accept or donate electrons
in the same way that pH indicates the tendency to accept or donate protons. In a
strongly reducing solution the tendency to donate protons and the corresponding
electron activity are large, and the pe is low. Likewise an acid solution has a
low pH. Table 4.1 shows that the pe values and hence equilibrium constants of
many redox reactions are very large or very small, which means that the reactions
proceed to completion in one direction or the other and the free energy changes
involved are large.
The values of pe
o
and
G
o
for any particular redox half reaction can be
calculated from the values for complete redox reactions or combinations of reac-
tions having the half reaction in common. For example, the oxidation of glucose
by oxygen,
=
1
6
glucose
+
O
2
(
g
)
=
CO
2
(
g
)
+
H
2
O
(
4
.
9
)
is equivalent to the two half reactions
1
2
O
2
(
g
)
+
2H
+
+
2e
−
=
H
2
O
(
4.9a
)
and
1
6
glucose
+
H
2
O
=
CO
2
(
g
)
+
4H
+
+
4e
−
(
4.9b
)
The change in free energy for the glucose — CO
2
half reaction can be found from
G
o
for the full reaction less that for the O
2
-H
2
O half reaction
=−
477
.
7
−
2
×
(
−
236
.
6
)
=−
4
.
5kJmol
−
1
.
The values of pe
o
and
G
o
can also be calculated from the standard free
energies of formation,
G
f
, of each of the reactants and products in a redox half
reaction. For example, for the reduction of ferrihydrite [amorphous Fe(OH)
3
]the
half reaction is
3H
+
+
2e
−
=
Fe
2
+
+
Fe(OH)
3
(
s, amorph
)
+
3H
2
O
(
4
.
10
)
and the values of
G
f
712 kJ mol
−
1
are between
−
699 and
−
for amorphous
−
78
.
87 kJ mol
−
1
for Fe
2
+
and
−
711
.
54 kJ mol
−
1
Fe(OH)
3
,
for H
2
O. Therefore
=
G
f
products
−
G
f
reactants
=−
91 to
−
78 kJ mol
−
1
, and pe
o
G
r
=
(
log
K)/n
=−
G
r
/
2
.
303
nRT
of
G
f
=
16
.
0
to
13.7.
Values
and
other
