Physical Chemistry Third Edition

8.5 Thermodynamic Information from Electrochemistry 375

wherenis the number of electrons in the cell reaction equation. For the standard-state

reaction,

−nFE◦∆G◦chem (8.5-2)

The Gibbs energy change in this equation is the same as for the reaction outside of

the cell. The equilibrium constant for the reaction can be calculated using the relations

shown in Eq. (7.1-20) and Eq. (8.5-2)

KQeqe−∆G

◦

chem/RTenFE◦/RT (8.5-3)

EXAMPLE 8.8

a.Find the value ofKfor the reaction of Eq. (8.2-4) at 298.15 K from theE◦value.

b.Find the equilibrium value of the hydrogen pressure whenm(HCl) 0 .500 mol kg−^1 and

γ±(HCl) 0 .757.

Solution

a.

Kexp

(

2(96485 C mol−^1 )(0.2223 V)

(9.3145 J K−^1 mol−^1 )(298.15 K)

)

e^17.^30  3. 28 × 107

b.

3. 28 × 107 

γ^4 ±

(

meq/m◦

) 4

Peq(H 2 )/P◦

Peq(H 2 )P◦

(

(0.757)^4 (0.500)^4

3. 28 × 107

)

(6. 26 × 10 −^10 )P◦

 6. 26 × 10 −^10 bar 4. 69 × 10 −^7 torr

Exercise 8.13

Find the value of the equilibrium constant for the reaction of Exercise 8.3 at 298.15 K.

We can write an expression for the entropy change of a reaction outside of an

electrochemical cell (we now omit the subscript “chem”):

(

∂S

∂ξ

)

T,P

−

(

∂

∂ξ

(

∂G

∂T

)

P

)

T,P

−

(

∂

∂T

(

∂G

∂ξ

)

T,P

)

P

nF

(

∂E

∂T

)

P

(8.5-4)

For the standard-state reaction, this equation becomes

∆S◦−

(

∂∆G◦

∂T

)

P

nF

(

∂E◦

∂T

)

P

(8.5-5)

The rate of enthalpy change per mole of a reaction is given by

(

∂H

∂ξ

)

T,P



(

∂G

∂ξ

)

T,P

+T

(

∂S

∂ξ

)

T,P

−nFE+nFT

(

∂E

∂T

)

P

(8.5-6)