When a lead storage battery is discharged — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Components of a Lead Storage Battery A lead storage battery is a secondary cell (rechargeable battery) commonly used in automobiles. It consists of: * **Anode (Negative Electrode):** A sponge lead plate ($\ce{Pb(s)}$). * **Cathode (Positive Electrode):** A grid of lead packed with lead dioxide ($\ce{PbO2(s)}$). * **Electrolyte:** An aqueous solution of sulphuric acid ($\ce{H2SO4(aq)}$) containing approximately $38\%$ of acid by mass with a specific gravity of about $1.215$ to $1.300\text{ g mL}^{-1}$. Step 2 - Write the Electrode Half-Reactions During Discharge When the battery is discharging (delivering electric current to an external circuit), spontaneous oxidation and reduction reactions occur at the electrodes: 1. **Oxidation at the Anode:** Solid lead ($\ce{Pb}$) is oxidized to lead(II) ions ($\ce{Pb^{2+}}$), which immediately react with sulphate ions ($\ce{SO4^{2-}}$) in the electrolyte to form insoluble solid lead sulphate ($\ce{PbSO4}$): $$\ce{Pb(s) + SO4^{2-}(aq) -> PbSO4(s) + 2e^-}$$ 2. **Reduction at the Cathode:** Lead dioxide ($\ce{PbO2}$) is reduced from oxidation state $+4$ to $+2$ ($\ce{Pb^{2+}}$) in the presence of hydrogen ions ($\ce{H^+}$), forming water and depositing insoluble lead sulphate ($\ce{PbSO4}$) on the cathode grid: $$\ce{PbO2(s) + SO4^{2-}(aq) + 4H^+(aq) + 2e^- -> PbSO4(s) + 2H2O(l)}$$ Step 3 - Determine the Overall Net Discharging Reaction By adding the anode and cathode half-reactions together, we obtain the net chemical reaction that takes place during discharge: $$\ce{Pb(s) + PbO2(s) + 2SO4^{2-}(aq) + 4H^+(aq) -> 2PbSO4(s) + 2H2O(l)}$$ Since $\ce{2SO4^{2-}(aq)}$ and $\ce{4H^+(aq)}$ associate to form two moles of sulphuric acid ($\ce{2H2SO4(aq)}$), the reaction can be written more simply as: $$\ce{Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)}$$ From this balanced chemical equation, we can observe that: * Sulphuric acid ($\ce{H2SO4}$) is **consumed** during discharge. * Water ($\ce{H2O}$) is **produced**, which dilutes the electrolyte, causing the density and concentration of $\ce{H2SO4}$ to drop. * Insoluble lead sulphate ($\ce{PbSO4(s)}$) is **produced** and deposited on both electrodes. Step 4 - Evaluate and Explain the Options * **Option (A) is incorrect:** No sulphur dioxide ($\ce{SO2}$) gas is evolved during the discharging process. The sulphate ions are converted into solid lead sulphate ($\ce{PbSO4}$). * **Option (B) is incorrect:** Lead sulphate ($\ce{PbSO4}$) is a product of the discharging reaction, meaning it is **formed/produced**, not consumed. It is only consumed during the recharging process of the battery. * **Option (C) is incorrect:** Metallic lead ($\ce{Pb}$) is a reactant at the anode, meaning it is **consumed** during discharge to form $\ce{PbSO4}$. It is not formed. * **Option (D) is correct:** As shown in the balanced overall cell equation, sulphuric acid ($\ce{H2SO4}$) is consumed, which leads to a decrease in the acid's concentration and specific gravity (density) during discharging. $$\text{Correct Option: } \boxed{\text{D}}$$