Question

Describe the charging and discharging mechanism of lead storage battery with the help of chemical reactions. (1½+1½=3)

Hint:

Anode Pb, cathode PbO₂, electrolyte H₂SO₄. Overall discharge: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O; charging is the reverse.

Answer 1

Concept: A lead storage battery (car battery) is a secondary (rechargeable) cell. It has a lead (Pb) anode and a grid of lead packed with lead dioxide (PbO₂) as the cathode, both dipped in about 38% sulphuric acid (H₂SO₄) as the electrolyte.

Step 1: Discharging (working as a cell). While supplying current, oxidation occurs at the Pb anode and reduction at the PbO₂ cathode.
At anode (oxidation):
\[ Pb + SO_4^{2-} \rightarrow PbSO_4 + 2e^- \]
At cathode (reduction):
\[ PbO_2 + 4H^+ + SO_4^{2-} + 2e^- \rightarrow PbSO_4 + 2H_2O \]
Overall discharge reaction:
\[ Pb + PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O \]
Both electrodes get coated with insoluble white lead sulphate (PbSO₄) and sulphuric acid is used up, so the density of the acid falls.

Step 2: Charging (recharging). When an external current is passed in the opposite direction, the electrode reactions reverse; the cell now acts as an electrolytic cell and the PbSO₄ is converted back to Pb and PbO₂.
\[ 2PbSO_4 + 2H_2O \rightarrow Pb + PbO_2 + 2H_2SO_4 \]
The sulphuric acid is regenerated, so its density rises again, restoring the battery.

Result: On discharge, Pb and PbO₂ both turn into PbSO₄ and acid is consumed; on charging this is exactly reversed, regenerating Pb, PbO₂ and H₂SO₄.