Page 139 - Basic physical chemistry for the atmospheric sciences

Oxidation-reduction reactions 1 2 S

conditions and � e ll· The reader was asked to derive this relationship
in Exercise 2 . 2 9 (the solution is given in Appendix VII) ; applied to
standard states the relationship is

AG° = - nF �ell (6 .20)
where F is the Faraday constant, which was defined in Exercise 6 . 5
s
and is equal t o 96,489 coulomb . The value o f n t o b e used in Eq.
(6 .20) for any particular redox reaction is determined by express­
ing the reaction as the sum of two half-reactions, in which elec­
trons appear in equal numbers. The number of electrons in each half­
. For example, for the redox
reaction is then equal to the value of n
reaction (6. 1 8 ) , n = 2, because there are two electrons in Eqs. (6. 1 6 )
)
and (6 . 17 .
Exercise 6.8. Using values of E° given i n Table 6 . 2 , determine the
change in the Gibbs free energy under standard conditions for the
redox reaction

2Cr(s) + 6H + (aq)- 2Cr 3 + (aq) + 3Hz(g)
Is this a spontaneous reaction?
Solution. The half-reactions are

2Cr(s)- 2Cr3 + (aq) + 6e - f!x = - ( - 0 .74 V) = 0.74 V
6H + (aq) + 6e - - 3 H z(g) med = O V

Net: 2Cr(s) + 6 H + (aq)-
3
2Cr + (aq) + 3H2(g)
Note that the oxidation half-reaction involving the Cr(s) - CrH (aq)
couple given in Table 6.2 has been multiplied by 2, and the reduction
half-reaction involving the H + (aq) - H2(g) couple given in Table 6 . 2
has been multiplied by 3 , in order t o make the number of electrons
released by the oxidation half-reaction equal to those consumed by the
n
reduction half-reactio .
Since s i x electrons are involved in each o f the balanced half­
reactions, n = 6 . Also, �ell = 0 . 7 4 V and F= 96,489 coulombs. There­
fore, from Eq. (6.20)

- 6(96,489)(0. 74) = - 4 .3 X 1 0 5 J
AG0 = - n F�ell =
i
Since AG0 s negative (and �ell is positive), the redox reaction is spon­
taneous .
In the above exercise w e derived AG0 from � ell· Conversely, we