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664 19 PROCESSES AT SOLID SURFACES
E19D.8(b) Because the standard potential of Zn2+/Zn is −0.76 V, under standard condi-
tions Zn metal will only be deposited when the applied potential is more nega-
tive than−0.76V.�e current density is given by [19D.2–848], j = j0(e(1−α) f η−
e−α f η), but under these conditions only the second term (the cathodic current)
is signi�cant. Using the data given for H+, assuming α = 0.5, and recalling that,
at 298.15 K, f = 38.921V−1
jH+ = − j0e−α f η
= −(0.79 × 10−3 A cm−2) e−0.5×(38.921 V
−1)×(−0.76 V)
= −2.1 × 103 A cm−2
It is usually considered that themetal can be deposited if the current density for
discharge of H+ is less than about 1 mA cm−2. In this case, the current density
for discharge of H+ is vastly in excess of this criterion, which means that zinc
metal will not be deposited, and all that will happen is the evolution of H2.
Solutions to problems
P19D.2 Simultaneous deposition is expected if the two potentials are the same, and this
is acheived by altering the ratio of the concentrations of Pb2+ and Sn2+. Using
the Nernst equation [6C.4–221], the potentials of the two half cells are
E(Pb2+/Pb) = E(Pb2+/Pb)−○ + RT
2F
ln aPb2+
E(Sn2+/Sn) = E(Sn2+/Sn)−○ + RT
2F
ln aSn2+
�ese will be equal when
E(Pb2+/Pb)−○ + RT
2F
ln aPb2+ = E(Sn2+/Sn)−○ +
RT
2F
ln aSn2+
From which it follows that
ln(aSn2+/aPb2+) =
2F
RT
[E(Pb2+/Pb)−○ − E(Sn2+/Sn)−○]
= 2 × (96485Cmol−1)
(8.3145 JK−1mol−1)×(298 K)
[(−0.126 V) − (−0.136 V)]
= 0.778...
�erefore aSn2+/aPb2+ = e0.778... = 2.18 . By making the concentration of Sn2+
a bit over twice that of Pb2+, simultaneous deposition can be achieved.
P19D.4 �e overpotential is η = E′ − E, where E is the equilibrium potential; in this
case η = E′ + (0.388 V).�e data given correspond to positive overpotentials,
so the anodic current will dominate and hence ln j = ln j0 + (1 − α) f η. For
the In3+/In electrode three electrons are transferred, therefore the relationship
needs to be modi�ed to ln j = ln j0 + 3(1 − α) f η. A plot of ln j against η will
have slope 3(1 − α) f and intercept ln j0. Such a plot is shown in Fig. 19.7.