Discharging and Charging Reactions of a Lead Acid Battery

Lead-acid batteries are the most commonly used battery bank systems in substations. This is a popular and robust rechargeable battery system. Before diving into the details of a lead-acid battery, we should first understand its basic charging and discharging cycles.

Discharging Reaction of Lead Acid Battery

During Discharging at Negative Pb Plates

Diluted sulfuric acid has H⁺ and SO₄^2- ions in the water medium. Lead (Pb), as a metal, readily releases its loosely bonded outermost valence electron. As a result, it becomes positive and easily combines with SO₄^2- ions to form PbSO₄. Then the electrons released by Pb reach the PbO₂ plate via the external load circuit.

$Pb + 2H^+ + SO_4^{2-}\rightarrow PbSO_4+2H^++2e$

We refer to Pb plates as negative plates because they donate electrons during discharging.

During Discharging at Positive PbO₂ Plates

Electrons reach the PbO₂ plates through the load. In the presence of electrons (e), PbO₂ reacts with dilute sulfuric acid and forms PbSO₄. Electrons from the Pb plates are absorbed by the PbO₂ plates to form PbSO₄. This is because the oxidation state of Pb changes from +4 in PbO₂ to +2 in PbSO₄.

$PbO_2+2H^++SO_4^{2-}+2e+2H^+\text{(from previous reaction)}\rightarrow PbSO_4+2H_2O$

We refer to PbO₂ plates as positive plates because they absorb electrons during discharging.

Overall Discharge Reaction of Lead Acid Battery

$Pb +PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4+2H_2O$

Charging Reaction of Lead Acid Battery

Now, let us connect the lead-acid battery with an external DC source with the same polarity. Hence, the source injects electrons into the negative plates.

During Charging at Negative Plates

After discharging, the negative plates now contain lead sulfate (PbSO₄). After receiving electrons from the source, the lead sulfate (PbSO₄) forms pure lead (Pb) and removes sulfate ions (SO₄^2-) into the electrolyte solution.

$PbSO_4+2e\rightarrow Pb+SO_4^{2-}$

During Charging at Positive Plates

On the other side, during charging, one PbSO₄^2- releases an SO₄^2- ion and two electrons (2e), forming a positive Pb⁴⁺ ion. Then the Pb⁴⁺ ion attaches with oxygen to form PbO₂. So, the water breaks and donates H+ ions to the electrolyte solution. The released electrons flow into the source through the positive terminal.

$PbSO_4+2H_2O\rightarrow Pb^{4+} +SO_4^{2-}+2e+2H_2O\\\\\rightarrow PbO_2+4H^++SO_4^{2-}+2e$

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