Chapter 21.2, Problem 19LC

Interpretation Introduction

Interpretation: The reason for a lead-acid battery to produce energy needs to be explained. The standard cell potential of one voltaic cell in a lead-acid battery needs to be calculated.

Concept Introduction: Lead-acid battery is a rechargeable battery. The energy density of a lead-acid battery is low as compared to other batteries. When the battery is charged, the chemical energy is stored as the potential difference between pure lead on the negative side and lead oxide on the positive side.

Explanation of Solution

In a lead-acid battery, the reactions at the cathode and anode are as follows:

Anode-Oxidation:

  $\text{Pb}\left(s\right)+{\text{HSO}}_4^{-}\left(aq\right)\to {\text{PbSO}}_{\text{4}}\left(s\right)+{\text{H}}^{+}\left(aq\right)+2e^{-}-\text{0}\text{.36 V}$

Cathode-Reduction:

  ${\text{PbO}}_{\text{2}}\left(s\right)+{\text{HSO}}_4^{-}\left(aq\right)+3{\text{H}}^{+}\left(aq\right)+2e^{-}\to {\text{PbSO}}_4\left(s\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)+1.69\text{ V}$

Here, the overall reaction is as follows:

  $\text{Pb}\left(s\right)+{\text{PbO}}_{\text{2}}\left(s\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to 2{\text{PbSO}}_{\text{4}}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

A lead-acid battery produces energy at the time of discharge. At the time of discharging, both positive and negative plates become lead (II) sulfate and the electrolyte loses sulfuric acid (in dissolved form) to form water.

The value of the standard cell potential of one voltaic cell in a lead-acid battery is calculated as follows:

  $E^{\text{o}}=E^{\text{o}}{}_{\text{cathode}}-E^{\text{o}}{}_{\text{anode}}$

Substitute the values,

  $\begin{array}{c}{E}^{\text{o}}=1.69\text{ V}-\left(-0.36\text{ V}\right)\\ =\left(1.69+0.36\right)\text{ V}\\ =\text{2}\text{.05 V}\end{array}$

Thus, the standard cell potential is $\text{2}\text{.05 V}$ .