Question

A lead-acid battery is designed so that the lead anode is the limiting component; all other components (lead oxide, sulfuric acid, water) are in excess. The design specifies that the battery contains 60 grams of lead. If the battery was completely discharged, how much charge (in coulombs) could the battery provide?

If current is drawn at a constant rate of 1.0 A, how long will the battery last?

Determine the minimum mass (grams) required for the cathode.

Answer 1

Discharge of Lead- Acid Battery:

During discharge, the lead dioxide (positive plate- Cathode) and lead (negative plate - Anode) react with the electrolyte sulfuric acid to produce lead sulfate, water and energy.

Reaction at the negative plate (Anode) :

Pb(s) + HSO₄^- (aq) ? PbSO₄(s) + H⁺⁽aq) + 2e^? (Oxidation)

Reaction at the positive plate (Cathode) :

PbO₂(s) + HSO₄^- (aq) + 3H⁺(aq) + 2e^? ? PbSO₄(s) + 2H₂O(l) (Reduction)

On adding the two equations we get the total cell reaction as

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

CALCULATION OF CHARGE PROVIDED BY THE BATTERY

According to the cell reaction 1 mol of Pb can produce 2 Faradays of electrical charge (because 2 molesof electrons are involved in the cell reactions) - Faradays second law of electrolysis

1 mol of Pb = atomic mass of lead expressed in gram = 207.2 g

1 Faraday = 96500 Coulombs

2Faradays = 2 x 96500 Coulombs = 193,000 Coulombs

The amount of charge 207.2 g of Pb can provide = 193,000 Coulombs

The amount of charge 60g of Pb can provide = 193,000 Coulombs x 60 g / 207.2g

                                                                        = 55,888 Coulombs

CALCULATION OF LIFE OF BATTERY

The quantity of electricity is defined as

Q = I t

Q = quantity of electricity in Coulombs = 55,888 Coulombs

I = Current in Ampere = 1 A

t   = time in second

t =55,8888/1 = 55,888 sec = 55,888sec

To convert the time in second to time hour divide by 60 x 60

Time the battery will last = 55,888/60x60 = 15.52 hours

CALCULATION OF MINIMUM MASS REQUIRED FOR CATHODE

The cathode is PbO2(s).

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

According to the cell reaction 1 mole of Pb (207.2g of Pb) requires 1mole of PbO2

1 mole of PbO₂ = 207.2 + 2 x 16 =239.2 g

207.2 g of Pb require 239.2 g of PbO₂

Therfore 60 g of Pb will require 239.2 g x 60 g / 207.2 g of PbO₂ = 69.2 g of PbO₂

                                                         

The minimum mass (grams) required for the cathode PbO₂ = 69.2 g