Agriculture Reference
In-Depth Information
Equilibrium constants of reduction half-reactions at pH 7 and 25
◦
C
pe
o
Table 4.2
pe
o
∗
1
4
O
2
(
aq
)
H
+
+
e
−
1
2
H
2
O
+
=
20.75
13.75
1
5
NO
3
−
+
6
5
H
+
+
e
−
1
10
N
2
(
g
)
3
5
H
2
O
=
+
21.05
12.65
1
4
NO
3
−
+
5
4
H
+
+
1
8
N
2
O(g)
5
8
H
2
O
e
−
=
+
18.81
10.06
1
2
MnO
2
(
s
)
+
2H
+
+
1
2
Mn
2
+
+
e
−
=
2H
2
O
21.82
9.67
a
1
2
Mn
3
O
4
(
s
)
4H
+
+
e
−
3
2
Mn
2
+
+
+
=
2H
2
O
30.79
8.33
3H
+
+
e
−
Mn
2
+
+
8.02
a
MnOOH(s)
+
=
2H
2
O
25.33
1
2
NO
3
−
+
H
+
+
1
2
NO
2
−
+
1
2
H
2
O
e
−
=
14.15
7.15
1
8
NO
3
−
+
5
4
H
+
+
e
−
1
8
NH
4
+
+
3
8
H
2
O
=
14.90
6.15
1
6
NO
2
−
+
4
3
H
+
+
e
−
1
6
NH
4
+
+
1
3
H
2
O
=
15.14
5.82
1
4
CH
2
O
H
+
+
e
−
1
4
CH
4
(
g
)
1
4
H
2
O
+
=
+
6.94
−
0.06
3H
+
+
e
−
Fe
2
+
+
Fe(OH)
3
(
s
)
+
=
3H
2
O
16.54
−
1.46
a
1
2
CH
2
O
H
+
+
e
−
1
2
CH
3
OH
+
=
3.99
−
3.01
8
SO
4
2
−
+
1
5
4
H
+
+
e
−
1
8
H
2
S(g)
1
2
H
2
O
=
+
5.25
−
3.50
1
8
SO
4
2
−
+
9
8
H
+
+
1
8
HS
−
+
1
2
H
2
O
e
−
=
4.25
−
3.63
1
8
CO
2
(
g
)
+
H
+
+
1
8
CH
4
(
g
)
+
1
4
H
2
O
e
−
=
2.87
−
4.13
1
6
N
2
+
4
3
H
+
+
e
−
1
3
NH
4
+
=
4.63
−
4.70
3H
+
+
e
−
Fe
2
+
+
6.69
a
α
-FeOOH(s)
+
=
2H
2
O
11.31
−
1
2
H
2
(
g
)
H
+
+
e
−
=
−
0.00
7.00
1
4
CO
2
(
g
)
H
+
+
e
−
1
24
(
glucose
)
1
4
H
2
O
+
=
+
−
0.20
−
7.20
1
4
CO
2
(
g
)
H
+
+
e
−
1
4
CH
2
O
1
2
H
2
O
+
=
+
−
1.20
−
8.20
a
For reductive dissolution of Mn and Fe oxides pe
o
∗
values are calculated with
(
Mn
2
+
)
=
0
.
2mM and
(
Fe
2
+
)
1 mM to represent conditions in submerged soil solutions; in other couples reactants are given
unit activities.
=
Consider the reaction
Ox
1
+
Red
2
=
Red
1
+
Ox
2
for which the reduction half reactions are:
Ox
1
+
n
e
−
=
Red
1
Ox
2
+
n
e
−
=
Red
2
and
K
2
, .e.pe
1
=
pe
2
=
with
equilibrium
constants
K
1
1
/n
log
K
1
and
1
/n
log
K
2
. Therefore
=−
2
.
303
RT
log
K
=−
2
.
303
RT
log
K
1
G
o
K
2
=−
2
.
303
RT n(
pe
1
−
pe
2
)
(
4
.
23
)
