Question

A Pb / H₂SO₄ / PbO₂ lead-acid accumulator consists of Pt sheets coated with active, active lead bonded to the negative pole and lead-dioxide bound to the positive pole, introduced into 500 cm³ of 8.2 N sulfuric acid. When supplying the accumulator 10 A hour, calculate: a) the amount of PbO₂ transformed into PbSO₄ on the positive plates, and b) the final concentration of the sulfuric acid.

Answer 1

V = 500 cm3 = 500 mL = 0.5 L

8.2 N = 16.4 M o fH2SO4

Total I = 10 A

t = 1 hour = 3600 s

Total Charge = I*t = 10*3600 = 36000 C

1 mol of e- = 96500 C

x mol of e- = 360000

x = 360000/96500 = 3.7305 mol of e-

The reaction:

Pb(s) + PbO2(s) + 2H2SO 4(aq) = 2PbSO4(s) + 2H2O(l)

then

2 mol of e- per reaction...

a)

PbO2 formed to PbSO4

2 mol of e- = 1 mol of PbO2 to PbSO4

3.7305 mol of e- -->3.7305 /2 = 1.86525 mol of PbO2 will convert to PBSO4

mass of PbO2 = mol*MW = 1.86525 *239.2 = 446.1678 g of PbO2

b)

initial H2SO4 --> MV = 16.4*0.5 = 8.2 mol of H2SO4

reacted H2SO4 -->

2 mol of e- = 2 mol of H2SO4

we have:

3.7305 mol of e- then 3.7305 mol ofH2SO4 react

change in moles = 8.2-3.7305 = 4.4695 mol

[H2SO4] = mol/V = 4.4695 /0.5 = 8.939 M

Normality = 8.939/2 = 4.4695 N

final [H2SO4]