Page 81 - Corrosion Engineering Principles and Practice

60 C h a p t e r 4 C o r r o s i o n T h e r m o d y n a m i c s 61

Although these equations appear slightly overwhelming, they can
be computed relatively simply with the use of a modern spreadsheet.
Step One: Calculate G for Each Species
For species O, the free energy of 1 mol can be obtained from G
0
with Eq. (4.24):
G o T) = G 0 o T) + . 2 303 RT log 10 a (4.24)
(
(
O
For x mol of species O the free energy is expressed by Eq. (4.25):
O)
xG o T) = x G ( o T) + . 2 303 RT log 10 a (4.25)
0
(
(
For pure substances such as solids, a is equal to 1. For a gas, a is
O
O
equal to its partial pressure (p ), as a fraction of 1 atm. For soluble
O
species, the activity of species O (a ), is the product of the activity
O
coefficient of that species (g ) with its molar concentration ([O]) (i.e.,
O
a = g O[O]). The activity coefficient of a chemical species in solution
O
is close to 1 at infinite dilution when there is no interference from
other chemical species. For most other situations the activity
coefficient is a complex function that varies with the concentration of
the species and with the concentration of other species in solution.
For the sake of simplicity the activity coefficient will be assumed to be
of value 1; hence Eq. (4.25) can be rewritten as a function of [O]:
xG o T) = x G ( 0 o T) + . 2 303 RT log [ O]) (4.26)
10
(
(
Taking the global reaction for the Al-O system expressed in
2
0
Eq. (4.19) and the G values calculated for 60°C in Table 4.3 and
Table 4.4, one can obtain thermodynamic values for the products
and reactants, as is done in Table 4.6.
Step Two: Calculate Cell DG
The ∆G of a cell can be calculated by subtracting the G values of the
reactants from the G values of the products in Table 4.6. Keeping the
example of the global reaction at 60°C in mind, one would obtain
,
(
,
,
,
,
∆G = G products − G reactants = −3 846 087 − −670 6115) = − 3 175 472 J (4.27)
The ∆G energy can be converted into potential:
−∆ G 3 188 818
,
,
E = nF = 12 × 96,485 = 2 74 V (4.28)
.
where n = 12 because each of the four Al in Eq. (4.19) gives off 3 e .
–
Step Three: Calculate the Specific Capacity (Ah kg )
—1
The specific capacity relates the weight of active materials with the
charge that can be produced, that is, a number of coulombs or ampere-
hours (Ah). Because 1 A = 1 C s , 1 Ah = 3600 C, and because 1 mol of
–1
e = 96,485 C (Faraday), 1 mol of e = 26.80 Ah.
–
–