In the lead storage battery, the anode reaction is Pb(s) + HSO4^- + -> PbSO4(s) + H3O^+ + 2e^-. How — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the Anode Half-Reaction and Determine the Stoichiometric Relationship In a lead storage battery, oxidation of lead occurs at the anode during discharge according to the balanced half-reaction: $$\ce{Pb(s) + HSO4^-(aq) + H2O(l) -> PbSO4(s) + H3O^+(aq) + 2e^-}$$ From the stoichiometry of this half-reaction, the oxidation of $1\text{ mole}$ of solid lead ($\ce{Pb}$) is accompanied by the release of $2\text{ moles}$ of electrons. This gives a stoichiometric ratio of: $$\frac{\text{Moles of }\ce{Pb}\text{ oxidized}}{\text{Moles of electrons released}} = \frac{1}{2}$$ Step 2 - Calculate the Total Quantity of Electricity ($Q$) Delivered The relationship between electric current ($I$), time ($t$), and the total electric charge ($Q$) passed is given by the formula: $$Q = I \times t$$ Given values: * Current ($I$) = $1\text{ A}$ * Time ($t$) = $100\text{ h} = 100 \times 3,600\text{ s} = 360,000\text{ s}$ Substitute these values into the formula: $$Q = 1\text{ A} \times 360,000\text{ s} = 360,000\text{ C}$$ Step 3 - Calculate the Moles of Electrons Transferred ($n_{e^-}$) Using Faraday's constant ($F \approx 96,500\text{ C/mol of electrons}$), we calculate the number of moles of electrons transferred: $$n_{e^-} = \frac{Q}{F}$$ Substitute the calculated charge: $$n_{e^-} = \frac{360,000\text{ C}}{96,500\text{ C/mol}} \approx 3.73\text{ mol of electrons}$$ Step 4 - Calculate the Mass of Lead ($\ce{Pb}$) Consumed Using the stoichiometric ratio from Step 1, the moles of lead consumed ($n_{\ce{Pb}}$) is: $$n_{\ce{Pb}} = \frac{n_{e^-}}{2}$$ $$n_{\ce{Pb}} = \frac{3.73\text{ mol}}{2} \approx 1.865\text{ mol of }\ce{Pb}$$ Now, calculate the mass of lead consumed ($W_{\ce{Pb}}$) using the given atomic mass of lead ($M_{\ce{Pb}} = 208\text{ g/mol}$): $$W_{\ce{Pb}} = n_{\ce{Pb}} \times M_{\ce{Pb}}$$ $$W_{\ce{Pb}} = 1.865\text{ mol} \times 208\text{ g/mol} = \mathbf{388\text{ g}}$$ Thus, the mass of lead used up is exactly $388\text{ g}$. Step 5 - Evaluate the Options * **Option (A) is incorrect:** This value ($776\text{ g}$) is obtained by failing to divide the moles of electrons by the stoichiometric factor of $2$, i.e., $3.73\text{ mol} \times 208\text{ g/mol} \approx 776\text{ g}$. * **Option (B) is correct:** As mathematically demonstrated, the correct mass of lead used up under these conditions is $388\text{ g}$. * **Option (C) is incorrect:** This value ($194\text{ g}$) represents half the correct mass, resulting from dividing the moles of electrons by $4$ instead of $2$. * **Option (D) is incorrect:** This is an extremely small value representing a calculation or unit conversion error. $$\text{Correct Option: } \boxed{\text{B}}$$