Question

The density of a sulphuric acid solution in a charged lead accumulator should be equal to ρ = 1.28 g cm^-3 which corresponds to the solution containing 36.87% of H₂SO₄ by mass. In a discharged lead accumulator it should not decrease under the value of ρ = 1.10 g cm^-3 which corresponds to the 14.35 % solution of sulphuric acid. (Faraday's constant F is equal to 26.8 Ah mol^-1.) 

1. Write the equation for a total electrochemical reaction that takes place in the lead accumulator when it is charged and discharged. 

2. Calculate the masses of H₂O and H₂SO₄ being consumed or formed according to the equation in No 1. 

3. Calculate the mass of H₂SO₄ that is required to be added to a led accumulator with a capacity of 120 Ah if the content of H₂SO₄ is to be in the range as given in the task. 

4. Calculate the difference in volumes of the sulphuric acid solutions in a charged and a discharged lead accumulator with a capacity of 120 Ah.  

Answer 1

2. n(H₂SO₄) = 2 mol n(H₂O) = 2 mol 

m(H₂SO₄) = 196 g m(H₂O) = 36 g 

Discharging: ∆m(H₂SO₄) = − 196 g 

∆m(H₂O) = + 36 g 

Charging: ∆m(H₂SO₄) = + 196 g 

∆m(H₂O) = − 36 g 

3. The mass of H₂SO₄ required: 

26.8 Ah corresponds to 98 g H₂SO₄ 

120 Ah corresponds to 438.8 g H₂SO₄ 

Analogously: 

26.8 Ah corresponds to 18 g H₂O 

120 Ah corresponds to 80.6 g H₂O 

Discharged lead accumulator: 

mass of H₂SO₄ solution − m 

mass of H₂SO₄ − m₁

mass fraction of H₂SO₄ − w₁ = 0.1435 

density of H₂SO₄ solution − ρ₁ = 1.10 g cm^-3

Charged lead accumulator: 

mass of H₂SO₄ formed − m₂ = 438.8 g 

mass of H₂O consumed − m₃ = 80.6 g

mass fraction of H₂SO₄ − w₂ = 0.3687 

density of the H₂SO₄ solution − ρ₂ = 1.28 g cm^-3

Because: 

We get a system of equations (a) and (b) which are solved for m₁ and m: 

m₁ = 195.45 g 

 m = 1362 g 

4. Volume of the electrolyte V₁ in a discharged lead accumulator: 

Volume of the electrolyte V₂ in a charged lead accumulator: 

Difference in the volumes: 

∆V = V₂ − V₁ = 1343.9 − 1238.2 = 105.7 cm³