Solutions to exercises
265
E
0.378 7.73 10 T [V]
T
4
Δ
= −
−
−
#
°
b. Use of the thermodynamic data gives
E
0.378 7.73 10
773
0.220 V
500 °C
4
Δ
= −
−
= −
−
#
#
°
^
h
Exploitation of the experimental curve in figure 88 leads to
E
0.217 V
500 °C
Δ
= −
°
The difference is close to 1.4%. We find that the thermodynamic data
satisfactorily account for the sensor response. This agreement validates
the choice of equilibria considered in question 1.
c. The characteristic slope of the experimental curve is s expt = 0.078 V per
decade. The slope of the equation describing the thermodynamic data is
given by
s
2F
RT
ln 10
calc =
.
ln
s
2 96 480
8 314 773
10
calc =
#
#
s calc = 0.077 V per decade
The difference between the two slopes is of the order of 1.28%. These
results indicate that the sensor response is nernstian and confirms the
good agreement between the thermodynamic data and experiment.
Solution 5.9 – Sulfur oxide sensor
1. The potential ΔE between the terminals of the electrochemical chain is
ΔE = φ Mes − φ Ref
where φ Mes and φ Ref are the potentials of the measurement and reference
electrodes. ΔE may equally well be written in the form
ΔE = φ Mes − φ Ag + φ Ag − φ Ref
where φ Ag is the potential of the metallic silver in the chain.
2 Let us first calculate φ Mes − φ Ag . For this, we write
• the equilibrium at the measurement electrode,
SO 3(g) + 2Ag
+ + 2e Pt + 2
1
O 2(g) m Ag 2 SO 4(s)
which gives