Page 221 - Handbook of Battery Materials

190 6 Lead Oxides

Table 6.7 Density and volume ratio of corrosion products related to lead.

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Density (g cm ) Volume ratio relative to Pb
Pb 11.34 1
PbO (red) 9.64 1.26
α-PbO 2 9.87 1.32
β-PbO 2 9.3 1.40
PbSO 4 6.29 2.64

active material is represented by the area on the left; the grid is shown on the
right. Underneath the porous lead dioxide that constitutes the active material, a
dense layer, also of lead dioxide, covers the grid surface. This layer is formed
by corrosion and protects the grid. On account of acid depletion a rather stable
oxide layer (mainly of α-PbO 2 ) is formed [28]. However, lead dioxide and lead
cannot exist beside each other for thermodynamic reasons, and a thin layer of
less-oxidized material is always formed between the grid and the lead dioxide
(PbO x in Figure 6.8) [29]. The existence of lead oxide (PbO) in this layer has been
determined; the existence of higher oxidized species is assumed (PbO x phases: cf.
Ref. [5], p. 18), but their structure is not yet known exactly [30].
The PbO 2 /PbO x border slowly penetrates into the metal, but only at a very slow
rate as a solid-state reaction. Cracks are formed when the oxide layer exceeds a
given thickness, on account of the growth in volume when lead becomes converted
into lead dioxide (Table 6.7). Underneath the cracks the corrosion process starts
again and again. As a whole, the corrosion proceeds at a fairly constant rate. It
never comes to a standstill, and a continually flowing anodic current, the corrosion
current, is required to re-establish the corrosion layer.
When the grid material (Pb) is converted into lead dioxide (PbO 2 ), the basic
electrochemical reaction is
Pb → Pb 4+ + 4e − (6.31)

Twice the amount of electricity is required compared with the discharge reaction
at the negative electrode according to Equation 6.18, since corrosion involves four
valences, which means 4F = 107.21 Ah per multiple of Equation 6.31. Consequently,
for the corrosion reaction according to Equation 6.31 the equivalent values are:
207.19
= 1.9326 g Pb/Ah or 517.4Ah/kgPb (6.32)
107.21
3
This value means that 1 cm of lead (11.34 g; cf. Table 6.2) is equivalent to 5.89 Ah.
A current of 1 µAcm −2 means 8760 µAh cm −2 per year. Referred to
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5.89 Ah cm , a penetration rate of 1.49 × 10 −3 cm/year results, assuming that the
corrosion attack occurs uniformly and progresses at a constant rate. Thus the value
in Figure 6.8 means that a corrosion current of 2µAcm −2 implies a penetration
rate of about 0.03 mm/year.