In order to charge a lead storage battery (Section 19.10) 500.0 g of PbSO₄(s) must be converted into PbO₂(s) and Pb(s). (c) If a current of 500 A is used, how long will it take?

Verified step by step guidance

1. First, write the balanced chemical equation for the conversion of PbSO4(s) to PbO2(s) and Pb(s). The overall reaction involves the reduction of PbSO4 to Pb and the oxidation of PbSO4 to PbO2.
2. Calculate the molar mass of PbSO4 using the atomic masses of lead (Pb), sulfur (S), and oxygen (O). This will help in determining the number of moles of PbSO4 present in 500.0 g.
3. Determine the number of moles of electrons required for the conversion. For each mole of PbSO4, two moles of electrons are needed: one for the reduction to Pb and one for the oxidation to PbO2.
4. Use Faraday's law of electrolysis, which states that the amount of substance transformed in an electrochemical reaction is directly proportional to the quantity of electricity (in coulombs) passed through the substance. The formula is: \( Q = n \times F \), where \( Q \) is the total charge in coulombs, \( n \) is the number of moles of electrons, and \( F \) is Faraday's constant (approximately 96485 C/mol).
5. Finally, calculate the time required using the formula \( t = \frac{Q}{I} \), where \( t \) is the time in seconds, \( Q \) is the total charge calculated in the previous step, and \( I \) is the current in amperes (500 A in this case).