Question

Use the following lead-acid battery cell to answer the related questions:

PbO2 (s) │PbSO4 (s) │H2SO4 (aq) (1 M) ‖ H2SO4 (aq) (1 M)│Pb (s) │PbSO4 (s)

A. (5 pts) Determine the Cell potential for the battery at 25°C

B. (5 pts) A high output alternator for a KIA soul can produce 220 Amp current. How long would it take to recharge/reform 4.03 kg of Pb at that rate (assuming your battery is near dead)?

Answer 1

A. Cell potential

ER Pb2+/Pb = - 0.13 V

EL Pb4+/Pb2+ = +1.67 V

Ecell = ER -EL

         = -0.13 -(+1.67)

Ecell = - 1.8 V

     E = E0 - 0.0592/2 log(1)

      E = E0 - 0.0

Ecell = E0cell

B. Time required to recharge

Mass of deposition = Z * I * t

          where, mass = 4030 g ;

                             I = 220 Amp;

                             Z=$\frac{equivalent weight of copper}{n*96500}$

i.e., 'n' is the number of electrons transfered; here n = 2

                  4030 = $\frac{63.5}{2*96500}$ * 220 * t

                          t = $\frac{4030*2*96500}{63.5*220}$

                          t = 777790000/13970

                          t = 55675.7 sec