Lead storage cell discharged: changes from 40% (density 1.260 g/ml) to 28% in 1 L original volume. I — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Write and Balance the Discharging Reactions of the Lead Storage Cell A lead storage battery operates as a galvanic cell during discharge. The individual half-reactions occurring at the electrodes are: * **Anode (Oxidation half-reaction):** $$\ce{Pb(s) + SO4^{2-}(aq) -> PbSO4(s) + 2e^-}$$ * **Cathode (Reduction half-reaction):** $$\ce{PbO2(s) + SO4^{2-}(aq) + 4H^+(aq) + 2e^- -> PbSO4(s) + 2H2O(l)}$$ Combining these two half-reactions gives the overall balanced discharging cell reaction: $$\ce{Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)}$$ According to the stoichiometry of this reaction, the consumption of $1\text{ mol}$ of $\ce{Pb(s)}$ is accompanied by: - The consumption of $1\text{ mol}$ of $\ce{PbO2(s)}$ - The consumption of $2\text{ mol}$ of $\ce{H2SO4(aq)}$ - The production of $2\text{ mol}$ of $\ce{H2O(l)}$ - The transfer of $2\text{ mol}$ of electrons ($n = 2$) from the anode to the cathode through the external circuit. This confirms that **Option (A) is correct**. Step 2 - Calculate the Initial Composition of the Electrolyte Solution We are given: * Initial volume of the electrolyte solution ($V_{\text{initial}}$) = $1\text{ L} = 1000\text{ mL}$ * Initial density of the solution ($d_{\text{initial}}$) = $1.260\text{ g/mL}$ * Initial concentration of $\ce{H2SO4}$ = $40\%$ by weight Using the density and volume, we calculate the total initial mass of the electrolyte solution ($m_{\text{initial, sol}}$): $$m_{\text{initial, sol}} = V_{\text{initial}} \times d_{\text{initial}}$$ $$m_{\text{initial, sol}} = 1000\text{ mL} \times 1.260\text{ g/mL} = 1260\text{ g}$$ Now, we calculate the mass of dissolved sulfuric acid solute ($m_{\text{initial, }\ce{H2SO4}}$): $$m_{\text{initial, }\ce{H2SO4}} = 40\% \times 1260\text{ g}$$ $$m_{\text{initial, }\ce{H2SO4}} = 0.40 \times 1260\text{ g} = 504\text{ g}$$ Step 3 - Set Up the Mass Balance and Solve for Moles of Reactant Let $x$ be the number of moles of solid lead ($\ce{Pb}$) oxidized at the anode during the discharging process. According to the stoichiometry of our balanced overall reaction: - For every $x\text{ mol}$ of $\ce{Pb}$ oxidized, $2x\text{ mol}$ of $\ce{H2SO4}$ is consumed and $2x\text{ mol}$ of $\ce{H2O}$ is produced. - The molar mass of $\ce{H2SO4}$ is $98\text{ g/mol}$. - The molar mass of $\ce{H2O}$ is $18\text{ g/mol}$. Thus: * Mass of $\ce{H2SO4}$ consumed: $$m_{\text{consumed, }\ce{H2SO4}} = 2x\text{ mol} \times 98\text{ g/mol} = 196x\text{ g}$$ * Mass of $\ce{H2O}$ produced: $$m_{\text{produced, }\ce{H2O}} = 2x\text{ mol} \times 18\text{ g/mol} = 36x\text{ g}$$ Now, we can express the final parameters of the solution: * Final mass of dissolved $\ce{H2SO4}$: $$m_{\text{final, }\ce{H2SO4}} = 504 - 196x\text{ g}$$ * Final total mass of the solution: During discharge, the solid reactants on the plates take in sulfate ions from the liquid phase and deposit as solid lead sulfate ($\ce{PbSO4}$), while liquid water is added to the solvent. The net change in the mass of the solution is: $$m_{\text{final, sol}} = m_{\text{initial, sol}} + \text{mass of }\ce{H2O}\text{ produced} - \text{mass of }\ce{H2SO4}\text{ consumed}$$ $$m_{\text{final, sol}} = 1260 + 36x - 196x = 1260 - 160x\text{ g}$$ We are given that the final weight percentage of the acid is $28\%$: $$\frac{m_{\text{final, }\ce{H2SO4}}}{m_{\text{final, sol}}} = 0.28$$ $$\frac{504 - 196x}{1260 - 160x} = 0.28$$ Now, solve for $x$: $$504 - 196x = 0.28 \times (1260 - 160x)$$ $$504 - 196x = 352.8 - 44.8x$$ $$196x - 44.8x = 504 - 352.8$$ $$151.2x = 151.2$$ $$x = 1.0\text{ mol}$$ This calculation proves that exactly $1.0\text{ mole}$ of $\ce{Pb}$ is oxidized. Step 4 - Evaluate options (B), (C), and (D) * **Option (B) is correct:** The total number of moles of $\ce{H2SO4}$ reacted is: $$\text{Moles of }\ce{H2SO4}\text{ reacted} = 2x = 2 \times 1.0\text{ mol} = 2.0\text{ mol}$$ * **Option (C) is correct:** The oxidation of $1\text{ mol}$ of $\ce{Pb(s)}$ releases $2\text{ mol}$ of electrons at the anode. Since $x = 1.0\text{ mol}$, the total moles of electrons transferred is: $$n_{e^-} = 2x = 2.0\text{ mol}$$ The total electrical charge ($Q$) released from the anode is: $$Q = n_{e^-} \times F$$ $$Q = 2.0\text{ mol} \times 96,500\text{ C/mol} = 1.93 \times 10^5\text{ C}$$ * **Option (D) is correct:** The final total mass of the electrolyte solution is: $$m_{\text{final, sol}} = 1260 - 160(1.0) = 1100\text{ g}$$ Since the mass decreased from $1260\text{ g}$ to $1100\text{ g}$ (a net loss of $160\text{ g}$), the mass of the electrolytic solution has indeed decreased. Therefore, all statements are correct. $$\text{Correct Options: } \boxed{A,B,C,D}$$