The Foundations of Chemistry

EE^0 

0.0

n

592

log 

[

[

R

O

e

x

d

]

]

x

y

 (reduction half-reaction)

For the familiar half-reaction involving metallic zinc and zinc ions,

Zn^2  2 e88nZn E^0 0.763 V

the corresponding Nernst equation is

EE^0 

0.0

2

592

log

[Zn

1

 2 ] (for reduction)

We substitute the E^0 value into the equation to obtain

E0.763 V

0.0

2

592

log

[Zn

1

 2 ]

EXAMPLE 21-5 The Nernst Equation

Calculate the potential, E,for the Fe^3 /Fe^2 electrode when the concentration of Fe^2 is

exactly five times that of Fe^3 .

Plan

The Nernst equation lets us calculate potentials for concentrations other than one molar. The

tabulation of standard reduction potentials gives us the value of E^0 for the reduction half-

reaction. We use the balanced half-reaction and the given concentration ratio to calculate the

value of Q.Then we substitute this into the Nernst equation with nequal to the number of

moles of electrons involved in the half-reaction.

Solution

The reduction half-reaction is

Fe^3 e88nFe^2  E^0 0.771 V

We are told that the concentration of Fe^2 is five times that of Fe^3 , or [Fe^2 ] = 5[Fe^3 ].

Calculating the value of Q,

Q

[

[

R

O

e

x

d

]

]

x

y



[

[

F

F

e

e

2

3





]

]



5

[

[

F

F

e

e

3

3





]

]

 5

The balanced half-reaction shows one mole of electrons, or n1. Putting values into the

Nernst equation,

EE^0 

0.0

n

592

log Q0.771

0.0

1

592

log 5(0.7710.041) V

0.730 V

You should now work Exercise 78.

The Nernst equation can be applied to balanced equations for redox reactions. One

approach is to correct the reduction potential for each half-reaction to take into account

the nonstandard concentrations or pressures.

878 CHAPTER 21: Electrochemistry

Metallic Zn is a pure solid, so its
concentration does not appear in Q.