A 12 Volt lead storage battery contains 410 grams of lead. It's anode plate and an equivalent amount of PbO2 in the Cathode.

Question

a) What is the max number of coulombs of electrical charge it can deliver?

 

b) For how many hours could the battery deliver a steady current of 1 amp?

 

c)What is the max electrical work that the battery can give you in kilowatts? 

Answer 1

Given Data

- The battery contains 410 g of Pb.

- Reaction at the anode: Pb → Pb²⁺ + 2e⁻

- Reaction at the cathode: PbO₂ + 4H⁺ + 2e⁻ → Pb²⁺ + 2H₂O

- Total reaction:

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

- Molar mass of Pb: 207.2 g/mol

- Voltage of the battery: 12 V

- Faraday's constant: F = 96485 C/mol

Part (a): Max number of coulombs of electrical charge

1. Calculate the moles of Pb:

Moles of Pb = Mass of Pb / Molar mass of Pb

Moles of Pb = 410 g / 207.2 g/mol ≈ 1.979 mol

2. Each mole of Pb releases 2 moles of electrons:

Moles of electrons = 2 × 1.979 ≈ 3.958 mol

3. Convert moles of electrons to coulombs:

Charge (Q) = Moles of electrons × Faraday's constant

Q = 3.958 × 96485 ≈ 381905 C

ANSWER Part (a)

The maximum number of coulombs is **381905 C**.

Part (b): Hours of steady current at 1 amp

1. The relationship between charge, current, and time is:

Q = I × t

2. Rearranging for time:

t = Q / I

3. Substitute values:

t = 381905 C / 1 A = 381905 seconds

4. Convert seconds to hours:

t = 381905 / 3600 ≈ 106.08 hours

ANSWER Part (b)

The battery can deliver a steady current of **1 amp for 106.08 hours**.

Part (c): Max electrical work in kilowatt-hours

1. Electrical work is given by:

W = Q × V

2. Substitute values:

W = 381905 × 12 = 4582860 J

3. Convert joules to kilowatt-hours:

1 kilowatt-hour = 3.6 × 10⁶ J

W = 4582860 / 3600000 ≈ 1.27 kWh

ANSWER Part (c)

The maximum electrical work is **1.27 kWh**.