Chemistry - A Molecular Science

Chapter 11 Electron Transfer and Electrochemistry

The mechanism shown in Figure 11.8 accounts for each of the preceding observations.

The pits form at the anodic region as metallic

iron is oxidized to iron(II), which passes into

solution in the water drop. The electrons released by the iron oxidation are used to reduce O^2

at the cathode, where the importance of

acid in the corrosion process can be seen. The

two half-reactions and their standard reduction potentials are:

Cathode

: O

+ 4H 2

1+ + 4e

1-

→

2H

O 2

Anode: 2Fe

→

2Fe

2+ + 4e

1-^

Reaction I:

O^2

+ 4H

1+ + 2Fe

→

2H

O + 2Fe 2

2+^

oE

=

oE

cathode

-^ E

oanode

= 1.23 - (-0.44) = +1.67 V*

The Fe

2+ ions in solution are further oxidized to Fe

3+ by oxygen, either in solution or at

the surface. The iron(III) precipitates as Fe

O 2

(rust). Oxygen from the atmosphere is again 3

the oxidizing agent. The following anode ha

lf-reaction has been multiplied by four to

make the electrons released by Fe

2+ equal to the number gained by O

. The 2

electrochemical reactions are

* The standard reduction potential is us

ed here rather than the reduction

potential of pure water because we

are calculating the standard

Fe potential.

O

1.5

O

+2Fe

23

2

¬

2+

rust

cathode

O^2 O^2

(g) (aq)

4 H

+O

+4

e

1+

1-

2H^2

O 2

anode

2Fe

2Fe

2 ++4e

1-

Figure 11.8 Corrosion of iron The electrochemical reactions that occur when a slightly acidic water drop sits on top of a piece of iron

Cathode: O

+ 4H 2

1+ + 4e

1-^

→

2H

O 2

Anode:

4Fe

2+

→

4Fe

3+ + 4e

1-^

Reaction II: O

+ 4H 2

1+ + 4Fe

2+^

→

4Fe

3+ + 2H

O 2

oE

=

oE

cathode

-^ E

oanode

= 1.23 - 0.77 = +0.46V*

The overall reaction for the corrosion of iron can be obtained by combining Reaction I

and Reaction II. However, Reaction I must fi

rst be multiplied by two in order to deliver

the four Fe

2+ ions required in Reaction II. The Fe

2+ ions then cancel in the summation:

Reaction I: 2O

+ 8H 2

1+ + 4Fe

→

4H

O + 4Fe 2

2+^

Reaction II

: O

+ 4H 2

1+ + 4Fe

2+^

→

4Fe

3+ + 2H

O 2

Corrosion: 3O

+ 12H 2

1+ + 4Fe

→

4Fe

3+ + 6H

O 2

The six H

O molecules can be viewed as 12H 2

1+ + 6O

2-, in which case, the 12H

1+ on each

side cancel. Finally, 4Fe

3+ + 6O

2- is equivalent to 2Fe

O 2

, which allows us to write the 3

common chemical equation for the rusting of iron:

3O

+ 4Fe 2

→

2Fe

O 2

(^3)