Concepts from Chemical Reaction Kinetics - Elements of Environmental Engineering: Thermodynamics and Kinetics - page 238

Formally, there is an analogy between the transfer of e − in a redox process and the
transfer of H + in an acid-base reaction. It is useful to first understand how the electron
activity in solutions is represented. A redox reaction can be generally represented as
the sum of two half-cell reactions , one where the e − is accepted (reduction) and the
other where it is donated (oxidation). For example, a redox reaction involving the
oxidation of Zn by Cu 2 + can be represented as

Cu 2 + ( aq ) 2 e − −→ + Cu ( s ) Reduction ) ,

Zn 2 + ( aq ) 2 e −

(5.165)

Zn ( s ) −→ + ( Oxidation ) ,

Cu 2 + ( aq ) Zn 2 + ( aq ) Redox ) . + Zn ( s ) −→ Cu ( s ) +

It is useful to represent each as a reduction reaction and then the overall process is
the difference between the two:

Cu 2 + ( aq ) 2 e − → + Cu ( s ) Reduction ) ,

Zn 2 + ( aq ) 2 e − → + Zn ( s ) Reduction ) , (5.166)

Cu 2 + ( aq ) Zn 2 + ( aq ) Redox ) . + Zn ( s ) → Cu ( s ) +

+ ne − →

Each half-reaction is denoted as Ox/Red and is represented by Ox Red.

][ e − ]

n .

The equilibrium quotient for the reaction is K =[ Red ] / [ Ox

Analogous to the definition of H + activity, pH H + ]

=− log [ , we can define the

electron activity [ e − ] using pe e − ] =− log [ . Thus for the above redox reaction,

n log [ n log [

,

1 Red ] 1 Ox ]

pe 0 pe 0

pe = − = + (5.167)

[ Ox ] [ Red ]

where pe 0

( 1 /n) log K is the electron activity at unit activities of [Red] and [Ox]
species. The above definition is generally applicable to any reaction involving e −

=

such as i ν i A i +

ne − = 0, where v i is the stoichiometric coefficient (positive for

reactants, negative for products) and for which we have

n log
i

(A i ) v i .

1

pe = pe 0 + (5.168)

E XAMPLE 5.17 pe OF R AINWATER

Neglecting all other ions, we have the following equation driving the pe of rain-
water: ( 1 / 2 ) O 2 ( g ) + 2H + + 2 e − → H 2 O ( l ) . Since the activity of pure water is 1,
we can write pe = pe 0

O 2 [ H + ]

+ ( 1 / 2 ) log [ P 0.5

2

] , where pe 0 = ( 1 / 2 ) log K with K =

1 /P 0.5 = 10 41 . If atmospheric partial pressure of O 2 ( = 0.21 atm) is used

and a pH of 5.6 in rainwater is considered, pe = 14.7.

O 2 [ H + ]

2

[ e − ]

2