Chemical Thermodynamics of Tin - Volume 12 - OECD Nuclear ...

B.1 2BThe specific ion interaction equations
447
is:
For this reaction
UO + H (g) + 2H U + 2H O(l) . (B.15)
2+ + 4+
2 2 2
2
4+
ο
U HO 2
10
K ⎜
10 2
⎜a 2+ ⋅a + ⋅ f
UO
H
2 H 2
log = log
⎛
⎝
a
⋅ a
⎞
⎟.
⎟
⎠
log K = log K + log γ −log γ −2log γ −log
γ
ο
10 10 10 4+ 2+ +
U 10 UO
10 10 ,H
2
H
f 2
+ 2log
a
10 H2O
,
f
H 2
≈ p
H 2
at reasonably low partial pressure of H 2 (g),
approximation for I m ≤ 3), and
Hence,
log γ = − 16 D + ε(U ,ClO ) m −
4+ −
10 4+
U
4
a
HO
≈ 1 (which is a reasonable
2
ClO4
log γ = − 4 D + ε(UO ,ClO ) m −
2+ −
10 2+
UO
2 4
2 ClO4
log γ = − D + ε(H ,ClO ) m −
+ −
10 +
H
4
log K = log K −10
D
ο
10 10
ClO4
+ ( ε(U ,ClO ) −ε(UO ,ClO ) −2 ε(H ,ClO )) m −
4+ − 2+ − + −
4 2 4 4 ClO4
The relationship between the equilibrium constant and the redox potential is:
ln K = nF E
RT
K
nF
E
RT
ο
ο
ln = .
(B.16)
(B.17)
(B.18)
ο
E is the redox potential in a medium of ionic strength I, E is the corresponding
standard potential at I = 0, and n is the number of transferred electrons in the reaction
considered. Combining Eqs. (B.16), (B.17) and (B.18) and rearranging them leads to
Eq. (B.19):
⎛ RT ln(10)
RT ln(10)
⎜ ⎞ ⎟ ε
⎛ ⎜ ⎞
⎟
⎝ nF ⎠ ⎝ nF ⎠
ο
E− 10D = E −Δ m −
ClO4
For n = 2 in the present example and T = 298.15 K, Eq. (B.19) becomes:
(B.19)
where
ο
E/mV −295.8 D = E /mV −29.58Δε m −
ClO4
Δε = ε(U ,ClO ) −ε(UO ,ClO ) − 2 ε(H ,ClO ).
4+ − 2+ − + −
4 2 4 4
CHEMICAL THERMODYNAMICS OF TIN, ISBN 978-92-64-99206-1, © OECD 2012