During the discharge of lead storage battery, weight of ${\mathrm{PbSO}}_4$ obtained at cathode involving  of 1F of electricity is(At.Wt of Pb = 208)

Given:

• Reaction at cathode: Pb²⁺ + SO₄^2- → PbSO₄
• 1 mole of PbSO₄ is produced by 2 moles of electrons (i.e., 2 Faradays of electricity).
• Atomic weight of Pb = 208, molecular weight of PbSO₄ = 208 + 32 + 4(16) = 304 g.

For 1 Faraday:

Since 2 Faradays produce 1 mole of PbSO₄, 1 Faraday produces 0.5 moles of PbSO₄.

Weight of PbSO₄ for 1 Faraday:

Weight = 0.5 × Molecular weight of PbSO₄ = 0.5 × 304 = 152 g.

Final Answer

The weight of PbSO₄ obtained at the cathode is 152 g