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During a period of discharge of a lead-acid battery,402g of Pb from the anode is converted into PbSO4(s). What mass of PbO2(s) is reduced at the cathode during this same period.

Answer:
Write the reactions involved.

At anode,
Pb ----> PbSO4
Pb ----> Pb(+2) + 2e-

At cathode
PbO2 -----> PbSO4
Pb(+4) + 2e(-) -----> Pb(+2)

Net reaction:
Pb(+4) + Pb(0) ----> 2Pb(+2)

Thus, for each mole of lead being oxidised at the anode, there is one mole of PbO2 being reduced at the cathode. The number of moles of Pb being oxidised = Mass of Pb/ Atomic weight of Pb = 0.402/207.2 = 1.94*10^-3 moles of Pb

So, the number of moles of PbO2 being reduced = 1.94*10^-3 moles
Mass of PbO2 involved = Number of moles * Molecular weight = 1.94*10^-3*239.2 = 0.464 grams

Thus, we have 0.464 grams of PbO2 being reduced.

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