Page 56 - Soil and water contamination, 2nd edition

Basic environmental chemistry 43

changes if the reaction is written as an oxidation reaction. Consequently, the sign of the
redox potential is negative if the system is reducing, but positive if the system is oxidising.
0
For example, the E for half reaction 2.57 is -0.77 Volt and for half reaction 2.58 is +1.23
0
Volt. The E for the complete redox reaction (Equation 2.59) is obtained by adding both
values, which results in a standard potential of +0.46 Volt. The positive sign implies that the
reaction (2.59) proceeds spontaneously to the right if all activities are equal to one.
In theory, the redox potential of an aqueous solution could be measured using an
electrode system as shown in Figure 2.2. Unfortunately, in practice it appears that results
from measurement using a redox probe are not unambiguous and may contain considerable
errors. The major causes of these erroneous results are the lack of equilibrium and analytical
difficulties using a platinum electrode due to, for example, insensitivity to some redox
couples and contamination of the electrode (Appelo and Postma, 1996).

Example 2.15 Calculation of redox speciation using the Nernst equation
2+
3+
A groundwater sample with a pH of 6.0 contains [Fe ] = 10 -3.22 and [Fe ] = 10 -5.69 at
2+
25 °C. Calculate the Mn activity if the sample were in equilibrium with MnO . The
2
standard redox potential s for the reactions involved are
0
Fe 3 + + e Fe 2 + E = +0.77 V
0
MnO 2 + 4 H + + e 2 Mn 2+ + 2 H 2 O E = –1.23 V
Solution
First, calculate the Eh using the Nernst equation (Equation 2.66) and the activities of the
Fe species:
[Fe 2 ] 10 . 3 22
Eh . 0 77 . 0 0592 log . 0 77 . 0 0592 log
[Fe 3 ] 10 . 5 69

. 0 77 . 0 0592 . 2 47 . 0 624 V

The Nernst equation for the second reaction involving the Mn species is
. 0 0592 [Mn 2 ]
Eh . 1 23 log
2 [H + ] 4

Note that MnO and H O are not considered in this equation, since they have unit
2 2
+
activity by definition. This equation can be rewritten as (note that pH = –log[H ])
2
2
Eh . 1 23 . 0 030 (log[ Mn ] 4 pH) . 1 23 . 0 030 log[ Mn ] . 0 118 pH
Thus,
. 1 23 . 0 030 log[Mn 2 ] . 0 118pH . 0 624 V

. 0 624 . 1 23 . 0 118 0 . 6
log[Mn 2 ] . 3 41
. 0 030
[Mn 2 ] 10 . 3 41 mol l -1

10/1/2013 6:44:18 PM
Soil and Water.indd 55 10/1/2013 6:44:18 PM
Soil and Water.indd 55