Environmental Engineering Reference
In-Depth Information
The
pe
valuecanbeobtainedexperimentallyfromthe
electrodepotentialofaredox
reaction, E
H
(in volts). The subscript H denotes that it is measured on a hydrogen
scale. Consider the redox reaction, Ox
ne
−
→
+
Red. The Gibbs function for the
reaction is given by
nFE
H
, which is the work required to move
n
electrons
from the anode to the cathode of an electrochemical cell.
E
H
is the electrode potential
(Volts) on a hydrogen scale, that is, assuming the reduction of H
+
to H
2
has a zero
reduction potential.
F
is the Faraday constant (96,485 Coulombs/mole). The standard
Gibbs energy,
Δ
G
=−
G
0
is defined as equal to
nFE
H
. Since we know that
pe
pe
0
Δ
−
=
+
(
1
/n)
log
(
[
Ox
]
/
[
Red
]
)
, and
pe
0
=
(
1
/n)
log
K
=−Δ
G
0
/
2.303
RT
,wehave
2.303
RT
n
log
[
Ox
]
G
0
−Δ
G
=−Δ
+
(5.169)
[
Red
]
or
2.303
RT
n
log
[
Ox
]
E
H
+
E
H
=
,
(5.170)
[
Red
]
which is called the
Nernst equation
for electrochemical cells. It is also apparent from
the above that
FE
H
2.303
RT
.
pe
=
(5.171)
Thus,
pe
is obtained from
E
H
, which is easily measured using electrochemical
cells. Some examples of
E
H
values for redox reactions of environmental interest are
given in Table 5.6.
For redox equilibria in natural waters, it is convenient to assign activities of H
+
and OH
−
values that are applicable to neutral water. The
pe
0
values relative to
E
H
are
now designated
pe
0
(w) relative to
E
H
(w). The two are related through the ion product
of water
pe
0
(
w
)
pe
0
=
+
(n
H
/
2
)
log
K
w
, where
n
H
denotes the number of protons
exchanged per mole of electrons. This type of characterization of
pe
0
allows one to
grade the oxidizing capacity of ions at a specified pH (namely, that of neutral water,
7). Thus a compound of higher
pe
0
(w) will oxidize one with a lower
pe
0
(w).
E
XAMPLE
5.18 C
ALCULATION OF
E
0
H
AND
pe
0
(
w
)
FOR A
H
ALF
-
CELL
R
EACTION
Consider the half-cell reaction, SO
2
−
4
+
9H
+
+
8
e
−
HS
−
+
4H
2
O. The standard
= Δ
G
H
=
G
f
(
HS
−
,aq
)
+
4
G
f
(
H
2
O
)
−
G
f
(
SO
2
4
,
aq
)
, since both H
+
and
e
−
have zero
G
f
by convention. Therefore
Δ
G
H
=
12
+
4
(
−
237
)
−
(
−
744
)
=−
192 kJ/mol. Hence
E
H
=−Δ
G
H
/nF
=
(
192
×
1000
)/(
8
×
96485
)
=+
0.24V.To obtain
E
H
at any other pH (say 7), we can use the Nernst equation
to obtain
E
H
atapHof7,
E
H
=
E
H
+
(
0.059
/
8
)
log
(
[
free energy of the reaction is
Δ
G
0
SO
2
−
4
6
)
.
If all species are at their standard states of unit activities, except [H
+
]=
10
−
7
M,
we have,
E
H
(
w
)
=+
0.24
+
(
0.0074
)
log
[
10
−
7
H
+
]
9
/
[
HS
−
][
][
H
2
O
]
9
=−
0.22V. Hence
pe
0
(
w
)
=
]
E
H
(
w
)/
0.059
=−
3.73.
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