Physical Chemistry Third Edition

13.5 Nonequilibrium Electrochemistry 605

nFk◦exp

[

( 1 −α)nF

(

φeq−φ◦

)

RT

]

[R]bnFk◦exp

[

−

αnF

(

φeq−φ◦

)

RT

]

[O]b

(13.5-15)

Each side of Eq. (13.5-15) is equal to the exchange current density (the exchange current

per unit area). When equal factors on the two sides of this equation are canceled, the

symmetry factorαdisappears from the equation:

exp

[

nF

(

φeq−φ◦

)

RT

]



[O]b

[R]b

(13.5-16)

This equation is equivalent to the Nernst equation, which means that our treatment is

consistent with equilibrium electrochemistry.

Exercise 13.22

Carry out the algebraic steps to put Eq. (13.5-16) into the standard form of the Nernst equation.

If both sides of Eq. (13.5-16) are raised to the−αpower, we obtain

exp

[

−αnF

(

φeq−φ◦

)

RT

]



[

[O]b

[R]b

]−α

(13.5-17)

When Eq. (13.5-17) is substituted into Eq. (13.5-15), we obtain an expression for the

exchange current density,j 0 :

j 0 nFk◦[R]^1 b−α[O]αb (13.5-18)

If the voltage is changed from the equilibrium value, the resulting current density is

jja−|jc|

nFk◦exp

[

( 1 −α)nF(φ−φ◦)

RT

]

[R]s

−nFk◦exp

[

−αnF(φ−φ◦)

RT

]

[O]s (13.5-19)

where we count the current as positive if the electrode half-reaction proceeds in the

oxidation direction. Equation (13.5-19) gives the dependence of the current on the

potential for any value of the potential. We now express this dependence in terms of

the overpotential,η, defined by

ηφ−φeq (13.5-20)

We divide the first term on the right-hand side of Eq. (13.5-19) by the left-hand side of

Eq. (13.5-15) and divide the second term by the right-hand side of Eq. (13.5-15) and

obtain

j

j 0



[R]s

[R]b

exp

[

( 1 −α)nF η

RT

]

−

[O]s

[O]b

exp

[

−αnF η

RT

]

(13.5-21)