What is the maximum theoretical discharge capacity (Ah) that can be obtained whe

Question

What is the maximum theoretical discharge capacity (Ah) that can be obtained when a cell contains 100g lead dioxide and 100g lead? Which electrode would be the limiting component of the cell assuming that excess add is available and that all the material is involved in the reaction? Assume that ail the active material of the limiting electrode was used, how many grams of the un-reacted active material of the other electrode was not involved in the discharge reaction? Atomic Mass for Pb = 207.2 g/mol; O = 16 g/mol; S = 32.1 g/mol

Explanation / Answer

The reaction in the battery will be

Pb(s) + PbO2(s) + 2H2SO4(aq) 2PbSO4(s) + 2H2O(l)

So here the limiting reagent will be lead oxide as its molecular weight is high and the moles will be less for same amount of substance (100g)

as per the balanced equation

one mole of Pb will react with one mole of lead oxide

Moles of lead present = Mass / atomic weight =100/207.2 = 0.483 moles

Moles of PbO2 present = Mass / mol weight = 100/ (207.2 + 2 X 16) = 0.418 moles

so moles of Pb left after reaction = 0.483 - 0.418 = 0.065 moles

mass of Pb left = Moles X atomic weight = 0.065 X 207.2 = 13.47 grams

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.