A lead storage cell is discharged which causes the electrolyte to change from a concentration of 40% — 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 electrode reactions 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 the two half-reactions gives the overall balanced discharging 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 ($n = 2$) through the external circuit. Step 2 - Determine the Initial Mass and Composition of the Electrolyte We are given: * Initial volume of electrolyte solution ($V_{\text{initial}}$) = $1\text{ L} = 1000\text{ mL}$ * Initial density of solution ($d_{\text{initial}}$) = $1.260\text{ g/mL}$ * Initial concentration of $\ce{H2SO4}$ = $40\%$ by weight The total initial mass of the electrolyte solution ($m_{\text{initial, sol}}$) is calculated using the formula: $$m_{\text{initial, sol}} = V_{\text{initial}} \times d_{\text{initial}}$$ $$m_{\text{initial, sol}} = 1000\text{ mL} \times 1.260\text{ g/mL} = 1260\text{ g}$$ Using the percentage composition, we find the initial mass of dissolved $\ce{H2SO4}$ ($m_{\text{initial, }\ce{H2SO4}}$): $$m_{\text{initial, }\ce{H2SO4}} = 40\% \times 1260\text{ g}$$ $$m_{\text{initial, }\ce{H2SO4}} = 0.40 \times 1260\text{ g} = 504\text{ g}$$ Step 3 - Set Up Mass Balance Expressions for the Discharging Process Let $x$ represent the number of moles of solid lead ($\ce{Pb}$) oxidized at the anode. * **Change in mass of \ce{H2SO4}:** Since $2x\text{ moles}$ of $\ce{H2SO4}$ are consumed, and the molar mass of $\ce{H2SO4}$ is $98\text{ g/mol}$: $$\text{Mass of }\ce{H2SO4}\text{ consumed} = 2x\text{ mol} \times 98\text{ g/mol} = 196x\text{ g}$$ Thus, the final mass of sulfuric acid in the solution is: $$m_{\text{final, }\ce{H2SO4}} = 504\text{ g} - 196x\text{ g}$$ * **Change in mass of the solution:** During discharging, solid $\ce{Pb}$ and solid $\ce{PbO2}$ react to form solid $\ce{PbSO4}$ which deposits on the plates. The only materials entering or leaving the liquid phase are the dissolved $\ce{H2SO4}$ consumed and the liquid $\ce{H2O}$ produced. Since $2x\text{ moles}$ of $\ce{H2O}$ are produced, and the molar mass of $\ce{H2O}$ is $18\text{ g/mol}$: $$\text{Mass of }\ce{H2O}\text{ produced} = 2x\text{ mol} \times 18\text{ g/mol} = 36x\text{ g}$$ The net change in the total mass of the solution ($\Delta m_{\text{sol}}$) is: $$\Delta m_{\text{sol}} = \text{Mass of }\ce{H2O}\text{ produced} - \text{Mass of }\ce{H2SO4}\text{ consumed}$$ $$\Delta m_{\text{sol}} = 36x\text{ g} - 196x\text{ g} = -160x\text{ g}$$ Thus, the final total mass of the electrolyte solution is: $$m_{\text{final, sol}} = 1260\text{ g} - 160x\text{ g}$$ Step 4 - Calculate the Moles of Reactants Using Concentration Change The final concentration of the sulfuric acid is $28\%$ by weight. We set up the ratio: $$\text{Weight fraction of }\ce{H2SO4} = \frac{m_{\text{final, }\ce{H2SO4}}}{m_{\text{final, sol}}}$$ $$0.28 = \frac{504 - 196x}{1260 - 160x}$$ Multiplying both sides by the denominator: $$0.28 \times (1260 - 160x) = 504 - 196x$$ $$352.8 - 44.8x = 504 - 196x$$ $$196x - 44.8x = 504 - 352.8$$ $$151.2x = 151.2$$ $$x = 1.0\text{ mol}$$ This calculation establishes that exactly $1.0\text{ mole}$ of $\ce{Pb}$ reacts during this discharge process. Step 5 - Evaluate and Explain Each Option * **Option (A) is correct:** As shown in Step 1, the overall cell discharging reaction is indeed: $$\ce{Pb(s) + PbO2(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)}$$ * **Option (B) is correct:** The number of moles of $\ce{H2SO4}$ reacted is: $$\text{Moles of }\ce{H2SO4}\text{ reacted} = 2x = 2 \times 1.0\text{ mol} = 2.0\text{ moles}$$ * **Option (C) is correct:** Since the oxidation of $1\text{ mole}$ of $\ce{Pb}$ transfers $2\text{ moles}$ of electrons ($n_e = 2x = 2.0\text{ mol}$), the total 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 mass of the electrolyte solution decreases during discharge. We calculate the final mass: $$m_{\text{final, sol}} = 1260\text{ g} - 160(1.0)\text{ g} = 1100\text{ g}$$ Since $1100\text{ g} < 1260\text{ g}$, the mass of the solution decreased by exactly $160\text{ g}$. All statements are thermodynamically and stoichiometrically correct. $$\text{Correct Options: } \boxed{A,B,C,D}$$