CBSE Class 12 Chemistry 2025: question 23(a)

Hints

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Step-by-step solution

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1. The cell generally used in inverters is the lead storage battery.

   1 mark

2. Anode (oxidation): $\mathrm{Pb}(\mathrm{s})+\mathrm{S}{\mathrm{O}}_4^{2-}(\mathrm{aq})\to \mathrm{PbS}{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{e}}^{-}$

   1 mark

3. Cathode (reduction): $\mathrm{Pb}{\mathrm{O}}_2(\mathrm{s})+\mathrm{S}{\mathrm{O}}_4^{2-}(\mathrm{aq})+4{\mathrm{H}}^{+}(\mathrm{aq})+2{\mathrm{e}}^{-}\to \mathrm{PbS}{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

   Overall reaction: $\mathrm{Pb}+\mathrm{Pb}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to 2PbS{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: Lead storage battery; anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻; cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O

Revise the chapter

• Chapter: Electrochemistry (notes, formulas, where marks are lost and every past question)