Chapter 19, Problem 19.146SP

Interpretation Introduction

(a)

Interpretation:

The reaction of converting ${\text{PbSO}}_{\text{4}}(s)$ into ${\text{PbO}}_{\text{2}}(s)$ and $\text{Pb}(s)$ in a lead storage battery needs to be identified as an electrolysis cell or galvanic cell.

Concept introduction:

Lead storage batteries can store a huge amount of charge and can discharge a huge amount of current for short period of time. Discharging of the stored energy of the battery relies on both negative and positive plates becoming ${\text{PbSO}}_4(s)$ and the electrolyte losing much of its dissolved sulfuric acid. Charging of the battery produce $\text{Pb}(s)$ at the negative plate and ${\text{PbO}}_{\text{2}}(s)$ will produce at the positive plate.

Interpretation Introduction

(b)

Interpretation:

Electrical charge needed for the charging reaction of the lead storage battery should be calculated in coulombs.

Concept introduction:

The amount of substance produced at an electrode by electrolysis depends on the quantity of charge passed through the cell.

$\text{Charge}(C)=\text{ Current}(A)\times \text{Time}(s)$

Interpretation Introduction

(c)

Interpretation:

Time required to charge the lead storage battery should be identified.

Concept introduction:

The amount of substance produced at an electrode by electrolysis depends on the quantity of charge passed through the cell.

$\text{Charge}(C)=\text{ Current}(A)\times \text{Time}(s)$