Question

During a period of discharge of a lead-acid battery, \(402 \mathrm{~g}\) of \(\mathrm{Pb}\) from the anode is converted into \(\mathrm{PbSO}_{4}(s) .\) What mass of \(\mathrm{PbO}_{2}(s)\) is reduced at the cathode during this same period?

Short answer

Approximately \(464\ \mathrm{g}\) of \(\mathrm{PbO}_{2}\) is reduced at the cathode during the period of discharge of the lead-acid battery.

Step-by-step solution

Calculate the number of moles of \(\mathrm{Pb}\)

Given mass of \(\mathrm{Pb} = 402\ \mathrm{g}\) and molar mass of \(\mathrm{Pb} = 207.2\ \mathrm{g\:mol^{-1}}\). Dividing the mass by its molar mass gives us the number of moles: Moles of \(\mathrm{Pb} = \frac{402}{207.2} \approx 1.94\ \mathrm{mol}\)

Use stoichiometric ratios to determine moles of \(\mathrm{PbO}_{2}\)

From the balanced chemical equation above, we can see that 1 mole of \(\mathrm{Pb}\) reacts with 1 mole of \(\mathrm{PbO}_{2}\). Therefore, the number of moles of \(\mathrm{PbO}_{2}\) that got reduced will also be the same as the number of moles of \(\mathrm{Pb}\). Moles of \(\mathrm{PbO}_{2} = 1.94\ \mathrm{mol}\)

Determine the mass of reduced \(\mathrm{PbO}_{2}\)

Finally, we need to convert the number of moles of \(\mathrm{PbO}_{2}\) to mass. We will use the molar mass, which can be calculated as follows: Molar mass of \(\mathrm{PbO}_{2} = 207.2\ (\mathrm{Pb}) + (16\times 2)\ (\mathrm{O}) = 239.2\ \mathrm{g\:mol^{-1}}\) Now we can determine the mass using the number of moles: Mass of \(\mathrm{PbO}_{2} = 1.94 \times 239.2 \approx 464\ \mathrm{g}\) So, during the period of discharge of the lead-acid battery, approximately \(464\ \mathrm{g}\) of \(\mathrm{PbO}_{2}\) is reduced at the cathode.