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Discharging Lead/Acid Battery
Resistance
Negative plate = Anode
Pb → PbSO4
Sponge Lead → Lead Sulfate
E0 = -0.356V
Pb + H2SO4

[2e-] + 2H+ + PbSO4
Reaction at -terminal
(Oxidation)
 
1. By removing two electrons [2e-] the lead oxidizes and gets a positive charge:
Pb → [2e-] + Pb2+
 
2. The positively charged lead atom reacts with the sulfuric acid and makes lead sulfate, and two hydrogen ions are released to the electrolyte:
Pb2+ + H2SO4 → PbSO4 + 2H+
 
Electrolyte = Sulfuric Acid
H2SO4 → H2O
Sulfuric Acid → Water
Lead Electrodes
Grid Plate
Lead Sulfate
Lead Dioxide
Sponge Lead
Electrolyte
Net reaction at discharging
Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O
Sponge Lead+Lead Dioxide+2 Sulfuric Acid→2 Lead Sulfate+2 Water
(chemical energy → electrical energy)
E = 2.041V
Positive plate = Cathode
PbO2 → PbSO4
Lead Dioxide → Lead Sulfate
E0 = +1.685V
H2SO4 + PbO2 + [2e-] + 2H+

2H2O + PbSO4
Reaction at +terminal
(Reduction)
 
3. The two added electrons [2e-] reduces the lead dioxide that splits into one lead ion(II) and two oxygen ions:
[2e-] + PbO2 → Pb2+ + 2O2-
 
4. The lead ion from [3] reacts with the sulfuric acid and makes lead sulfate, and another two hydrogen ions are being released:
Pb2+ + H2SO4 → PbSO4 + 2H+
 
5. The hydrogen ions from [2] and [4] joins with the oxygen ions from [3] and makes two water molecules:
2H+ + 2H+ + 2O2- → 2H2O