If in a lead storage battery the reaction at the cathode is: PbO₂(s) + 4 H⁺ + SO₄^2- + 2e- → PbSO₄(s) + 2 H₂O
How many grams of lead sulfate are formed on one plate if 20. amps average current occurs for 2.0 minutes? Assume 100% efficiency.

Question

Quizplus · Accepted Answer

First, calculate the total charge passed:
Q = It = 20 A x 2.0 min x 60 s/min = 2400 C

From the balanced equation, it can be seen that 2 mol of electrons are required to produce 1 mol of PbSO4. Therefore, the number of moles of PbSO4 produced is:

n(PbSO4) = 2 mol e- x (2400 C / 96500 C/mol) = 0.0496 mol

The molar mass of PbSO4 is 303.3 g/mol. Therefore, the mass of PbSO4 produced on one plate is:

m(PbSO4) = n(PbSO4) x M(PbSO4) = 0.0496 mol x 303.3 g/mol = 15.02 g

However, the question is asking for the mass on one plate. Since there are two plates in the battery, the mass on one plate is:

m(PbSO4) = 15.02 g / 2 = 7.51 g.

Therefore, the closest option is C) 3.77 g (which is half of the calculated answer since it is asking for the mass on only one plate).

If in a Lead Storage Battery the Reaction at the Cathode