Solutions to exercises
201
2. a. The graphical determination of the limiting currents gives
2 for platinum
I ℓ = − 47 μA
2 for nickel
I ℓ = − 1.6 μA
b. To obtain z, we have to derive the expression for the overpotential as a
function of current.
We obtain
I
z F
RT I I
1
2
2η = −
−
,
c
m
The inverse gives
I
RT
z F
(I I )
2
2
η = +
− ,
The slope of the line, I
f(I I )
2 2η =
− , , which takes the value
( )
zF RT ,
allow us to determine the value of z.
With a and b denoting two consecutive points of the curve I(η), the line
can be reformulated as
I I
f
2
I I I
b
a
b
a
b
a
η η
−
−
=
− − ,
c
m
I I 2
b
a
−
^
h
is the average I m of the current for which we calculate the
slope I I
b
a
b
a
η η
−
−
^
^
h
h . The values of I
2 2η and of I m − I ℓ are listed
in table 44.
Table 44 – Detail of the calculation of the function I
f (I
I )
m
2 2η
−
=
, .
For the Pt / YSZ interface
10
6
I
2 2η [A V
−1
] 360 280 210 140 90
50
22
11
4
1
10
6
(I m − I ℓ ) [A] 42.5 34.5 25.75 17 11.25 6.5 2.9 1.25 0.5 0.25
For the Ni / YSZ interface
10
6
I
2 2η [A V
−1
] 26.04 16.12 6.86
2.48
0.94
0.23
0.05
0.02
10
6
(I m − I ℓ ) [A] 1.274 0.747 0.374 0.141 0.055 0.02 0.006 0.003
The functions are shown in figure 78 for (a) the Pt / YSZ interface and
(b) the Ni / YSZ interface.