Batscan - 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⁺ + 2O^2-
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⁺ + 2O^2- → 2H₂O