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

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

Revised Pages

668 • Chapter 16 / Corrosion and Degradation of Materials

Concept Check 16.2

Modify Equation 16.19 for the case in which metals M 1 and M 2 are alloys.

[The answer may be found at http://www.wiley.com/college/callister (Student Companion Site).]

EXAMPLE PROBLEM 16.1

Determination of Electrochemical Cell Characteristics

One-half of an electrochemical cell consists of a pure nickel electrode in a

solution of Ni^2 +ions; the other half is a cadmium electrode immersed in a Cd^2 +

solution.

(a)If the cell is a standard one, write the spontaneous overall reaction and

calculate the voltage that is generated.

(b)Compute the cell potential at 25◦C if the Cd^2 +and Ni^2 +concentrations are

0.5 and 10−^3 M, respectively. Is the spontaneous reaction direction still the

same as for the standard cell?

Solution

(a)The cadmium electrode will be oxidized and nickel reduced because cad-

mium is lower in the emf series; thus, the spontaneous reactions will be

Cd→Cd^2 ++ 2 e−

Ni^2 ++ 2 e−→Ni

Ni^2 ++Cd→Ni+Cd^2 + (16.21)

From Table 16.1, the half-cell potentials for cadmium and nickel are,

respectively,−0.403 and−0.250 V. Therefore, from Equation 16.18,

V=VNi^0 −VCd^0 =− 0 .250 V−(− 0 .403 V)=+ 0 .153 V

(b)For this portion of the problem, Equation 16.20 must be utilized, since

the half-cell solution concentrations are no longer 1M. At this point it is

necessary to make a calculated guess as to which metal species is oxidized

(or reduced). This choice will either be affirmed or refuted on the basis

of the sign ofVat the conclusion of the computation. For the sake of

argument, let us assume that in contrast to part (a), nickel is oxidized and

cadmium reduced according to

Cd^2 ++Ni→Cd+Ni^2 + (16.22)

Thus,

V=

(

VCd^0 −VNi^0

)

−

RT

nf

ln

[Ni^2 +]

[Cd^2 +]

=− 0 .403 V−(− 0 .250 V)−

0. 0592

2

log

(

10 −^3

0. 50

)

=− 0 .073 V