Fundamentals of Materials Science and Engineering: An Integrated Approach, 3e

GTBL042-16 GTBL042-Callister-v2 September 13, 2007 13:10

Revised Pages

16.4 Prediction of Corrosion Rates • 677

EXAMPLE PROBLEM 16.2

Rate of Oxidation Computation

Zinc experiences corrosion in an acid solution according to the reaction

Zn+2H+→Zn^2 ++H 2

The rates of both oxidation and reduction half-reactions are controlled by ac-

tivation polarization.

(a)Compute the rate of oxidation of Zn (in mol/cm^2 -s) given the following

activation polarization data:

For Zn For Hydrogen

V(Zn/Zn 2 +)=− 0 .763 V V(H+/H 2 )=0V

i 0 = 10 −^7 A/cm^2 i 0 = 10 −^10 A/cm^2

β=+0.09 β=−0.08

(b)Compute the value of the corrosion potential.

Solution

(a)To compute the rate of oxidation for Zn, it is first necessary to establish re-

lationships in the form of Equation 16.25 for the potential of both oxidation

and reduction reactions. Next, these two expressions are set equal to one

another, and then we solve for the value ofithat is the corrosion current

density,iC. Finally, the corrosion rate may be calculated using Equation

16.24. The two potential expressions are as follows: For hydrogen reduc-

tion,

VH=V(H+/H 2 )+βHlog

(

i

i (^0) H

)

and for Zn oxidation,

VZn=V(Zn/Zn 2 +)+βZnlog

(

i

i (^0) Zn

)

Now, settingVH=VZnleads to

V(H+/H 2 )+βHlog

(

i

i (^0) H

)

=V(Zn/Zn 2 +)+βZnlog

(

i

i (^0) Zn

)

And solving for logi(i.e., logiC) leads to

logiC=

(

1

βZn−βH

)

[V(H+/H 2 )−V(Zn/Zn 2 +)−βHlogi (^0) H+βZnlogi (^0) Zn]

=

[

1

0. 09 −(− 0 .08)

]

[0−(− 0 .763)−(− 0 .08)(log 10−^10 )

+(0.09)(log 10−^7 )]

=− 3. 924