Page 199 - Basic physical chemistry for the atmospheric sciences
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Answers, hints, and solutions to exercises 1 8 S
Solution: From Table 6.2 we see that the standard potential
for the FeH (aq) - Fe2 + ( aq) couple is 0.77 1 V . Since the
seawater has a redox potential of 0.600 V, it provides a
2
better reducin g environment than the FeH (aq) - Fe + (aq)
couple. Therefore , this couple will be reduced ; that is, it
will be driven from left to right.
�ed = 0 . 7 7 1 V
The seawater will be involved in the oxidation half-cell
reduction (i. e . , it acts as the reductant) with E0x = - (0.600
V) = - 0.600 V . When the seawater is in equilibrium with
the iron system Ecell = Ere d + E0x = 0 or Ered = 0 . 6 00 V. For
nonstandard concentrations at 298K , the total cell potential
developed by the iron system when paired with the hydro
gen half-cell is given by Eq. (6.26). However, the hydrogen
half-cell generates zero electrode potential . Therefore, the
electrode potential developed by the iron system is, from
Eq. (6 . 26)
2
0.059 1 [ F [Fe + (aq)]
£ d = �e d - - 1 - log eH (aq)]
re
or,
2
[Fe + (aq)]
0 . 6 00 = 0 . 77 1 - 0.059 1 l o g
[ FeH (aq)]
Therefor ,
e
2
Fe + ( aq)
782
FeH (aq)
6.28. 1 equiv. of H2S(aq) is 1 7 g, and 1 equiv . of HN03(aq) is 1
2
6
. Hint: From Exercise . 1 1 we have
g
1 mole of the species
Oxidation number of a species = 1 . ·
eqmv. o t
f h e species
0
6 . 2 9. Normality is . 5 7 N .
CHAPTER 7
7 . 7 . Ratios of energies are I : 0 . 08: 0 . 8 : 0.004.
7.8 . 4 x 1 0 5 .
J