Page 57 - Soil and water contamination, 2nd edition

44 Soil and Water Contamination

2.10.4 Redox reactions and pe
The equilibrium constant for the half-reaction reaction 2.53 is:

[Fe 3+ ][ e ] 13 . 05
K 2+ 10 (2.68)
[Fe ]
-
In contrast to the Nernst equation (Equation 2.66), the activity of the electrons, [e ], appears
explicitly in Equation (2.68). Analogous to the pH, the pe is defined as:

pe log[e ] (2.69)
The advantage of using the pe approach is that the calculations of redox reactions are similar
to other equilibrium calculations. A disadvantage of the pe is that it cannot be measured
directly, but fortunately the pe is directly related to the redox potential :
. 2 303 RT
Eh pe (2.70)
F
At 25 °C this is equal to

Eh . 0 0592 pe (2.71)

Example 2.16 Calculation of redox speciation using the pe concept

2+
3+
A groundwater sample contains [Fe ] = 10 -3.22 and [Fe ] = 10 -5.69 at 25 °C (see Example
2.15). Calculate the pe and the Eh given the equilibrium constant for the half reaction
Fe 3 + + e Fe 2 + pK = 13.05

Solution
The mass action expression for the half reaction is
[Fe 2+ ] 13 . 05
K 10
[Fe 3+ ][e ]

Note that this equilibrium constant is the reciprocal of Equation (2.68), since the above
half reaction has been written as a reduction reaction and the half reaction corresponding
to Equation (2.68) has been written as an oxidation reaction (Equation 2.57).
The above mass action expression can be rewritten as
2

3
log K log[ Fe ] log[ Fe ] pe
pe log K log[Fe 3 ] log[Fe 2 ]
Filling in the values gives
pe 13 . 05 . 5 69 . 3 22 10 . 58
The Eh is calculated using Equation (2.67):
Eh . 0 0592 pe . 0 626 V
Which is approximately the same as calculated in Example 2.15.

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