Page 210 - Handbook of Battery Materials

6.3 The Thermodynamic Situation 179

1) Oxygen evolution according to Equation 6.5 is possible above 1.23 V and forms
the couple with the discharge reaction (the reversal of Equation 6.21):
Discharge of the positive electrode
−
+
PbO 2 + H 2 SO 4 + 2H + 2e → PbSO 4 + 2H 2 O
and oxygen evolution
+
H 2 O → 1/2O 2 + 2H + 2e −
with the result
PbO 2 + H 2 SO 4 → 1/2O 2 + PbSO 4 + H 2 O (6.25)

2) Hydrogen oxidation according to Equation 6.5 is possible above 0 V. If
hydrogen evolution occurs at the negative electrode and the H 2 evolved reaches
the positive electrode, from the thermodynamic situation the reaction, that is,
to be expected is:
Discharge of the positive electrode
−
+
PbO 2 + H 2 SO 4 + 2H + 2e → PbSO 4 + 2H 2 O
and hydrogen oxidation
+
H 2 → 2H + 2e −
with the result
+
PbO 2 + H 2 SO 4 + 2H → PbSO 4 + 2H 2 O (6.26)
In practice, however, this reaction can be neglected since its rate is extremely
low at the PbO 2 surface.
3) Corrosion of lead starts at the equilibrium potential of the negative electrode.
It induces self-discharge of the positive electrode on account of the following
couple of reactions:
Discharge of the positive electrode
−
+
PbO 2 + H 2 SO 4 + 2H + 2e → PbSO 4 + 2H 2 O
and grid corrosion
+
Pb + H 2 O → PbO + 2H + 2e −
with the result

+
−
PbO 2 + H 2 SO 4 + 2H + 2e → PbSO 4 + 2H 2 O (6.27)
Above the potential given by line F in Figure 6.1, the formation of PbO is possible,
but this is stable only in an alkaline environment. As soon as it comes to contact
with sulfuric acid, it is converted into lead sulfate. For this reason PbO is in
parentheses in Figure 6.2. The final result of these reactions is:
PbO 2 + Pb + 2H 2 SO 4 → 2PbSO 4 + 2H 2 O (6.28)
Above the equilibrium potential of the positive electrode, lead is oxidized to PbO 2 .