A lead storage battery operates for a certain time, delivering a steady current. The overall dischar — Electrochemistry Chemistry Question
Question
A lead storage battery operates for a certain time, delivering a steady current. The overall discharge reaction is $Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$. If exactly $9.8\text{ g}$ of $H_2SO_4$ (Molar mass = $98\text{ g mol}^{-1}$) is consumed during this discharge process, how much total charge in Coulombs was delivered by the battery? (Given $1\text{ F} = 96500\text{ C}$)
💡 Solution & Explanation
Moles of $H_2SO_4$ consumed = $9.8\text{ g} / 98\text{ g mol}^{-1} = 0.1\text{ mol}$. From the balanced overall cell reaction, 2 moles of $H_2SO_4$ are consumed when 2 moles of electrons ($2\text{ F}$) are transferred through the external circuit (since $Pb$ goes to $Pb^{2+}$, releasing $2e^-$). Thus, the ratio of moles of electrons to moles of $H_2SO_4$ is $1:1$. For $0.1\text{ mol}$ of $H_2SO_4$, exactly $0.1\text{ mol}$ of electrons are transferred. Total charge $Q = 0.1 \times 96500\text{ C} = 9650\text{ C}$.