Physical Chemistry , 1st ed.

Example 8.6

Consider the following formation reaction for H 2 O ():

2H 2 (g) O 2 (g) →2H 2 O ()

IfH°571.66 kJ at 25°C and n4 for this reaction, determine the tem-

perature coefficient of the standard potential E°.

Solution

We can use equation 8.30 to determine E°/ T, which is the temperature co-

efficient of interest. Using Table 8.2, we can determine that E° for the reac-

tion is 1.23 V. Substituting into the equation for the known quantities:

571,660 J (4 mol e^ )(96,500 C/mol e^ )1.23 V (298 K)^

E

T

°

The “mol e^ ” units cancel, and when we divide by Faraday’s constant we get

J/C as a unit, which equals a volt. We get

1.481 V1.23 V (298 K)

E

T

°



0.25 V(298 K)^

E

T

°



E

T

°

8.4 10

4 V/K

The final units are appropriate for a temperature coefficient of electromotive

force.

Changes in Eversus pressure aren’t normally considered, since the expression



G

p



T

V

implies that



(

p

G°)



T


nF

E

p

°



T



V

and rearranging:



E

p

°



T





nF

V

 (8.31)

Since most voltaic cells are based in some condensed phase (that is, liquid or
solid), the change in volume of this condensed phase is very small unless pres-
sure changes are very, very high. Since Vvalues are typically very small and
Fis numerically very large, we can ignore the pressure effects on E°. However,
partialpressure variations of gaseous products or reactants involved in the
electrochemical reaction can have a large effect on E°. These effects are usually
handled with the Nernst equation, since the partial pressure of a reactant or
product contributes to the value of the reaction quotient Q.
Finally, the relationship between the equilibrium constant and the EMF of
a reaction should be considered. This relationship is commonly used to make
measurements on various systems, by measuring the voltage across some con-
trived electrochemical cell. Using the relationships

G°

nFE°

G°

RTln K

8.5 Nonstandard Potentials and Equilibrium Constants 221