Final answer:
By applying Faraday's law, we calculate that a lead storage battery with 200 g of lead and 200 g of PbO₂ could theoretically deliver a current of 10.0 A for approximately 18557.7 seconds before needing recharging.
Step-by-step explanation:
The question asks how long a lead storage battery could deliver a current of 10.0 A using 200 g of lead and 200 g of PbO₂. To solve this, we need to consider the electrochemical reaction taking place in the battery and apply Faraday's law of electrolysis. The relevant chemical reaction at the electrodes during discharge is:
Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l).
Lead has a molar mass of 208 g/mol, so we can convert the mass of lead and PbO₂ to moles:
- 200 g of Pb ≈ 200 g / 208 g/mol ≈ 0.96 moles.
- Since Pb and PbO₂ react in a 1:1 molar ratio, Pb limits the reaction amount to 0.96 moles.
To find the total charge Q moved, we use:
Q = n * F, where n = number of moles of electrons, F = Faraday's constant (96,485 C/mol).
For every mole of Pb used, two moles of electrons are involved; hence n = 0.96 moles * 2.
Q = 0.96 moles * 2 * 96,485 C/mol ≈ 185,530 C.
Current (I) is the charge moved per unit time (t). I = Q/t or t = Q/I.
To find the time t:
t = 185,530 C / 10.0 A ≈ 18,553 s, which is option D, 18557.7 seconds.