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Chapter 17: Problem 55

Write the chemical equation for the overall cell reaction that occurs in a lead storage automobile battery. What species is oxidized in such a battery? What species is reduced? Why can such a battery be "recharged"?

Short Answer

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The overall cell reaction that occurs in a lead storage automobile battery is: Pb(s) + PbO2(s) + 2HSO4鈭�(aq) + 2H+(aq) 鈫� 2PbSO4(s) + 2H2O(l) In this reaction, lead (Pb) is the species being oxidized, and lead dioxide (PbO2) is the species being reduced. Such a battery can be recharged because the chemical reactions during discharging are reversible. When an external electric current is applied, the lead sulfate (PbSO4) formed during discharging reacts back to form lead (Pb) and lead dioxide (PbO2), regenerating the sulfuric acid electrolyte and restoring the original composition of the battery electrodes.

Step by step solution

01

Identify the components involved in the lead storage battery reaction.

A lead storage battery is also known as a lead-acid battery. It consists of two electrodes: the positive electrode made of lead dioxide (PbO2) and the negative electrode made of lead (Pb). These electrodes are immersed in an electrolyte solution of sulfuric acid (H2SO4).
02

Write the half-cell reactions for the oxidation and reduction processes.

In order to determine the overall cell reaction, we need to first write the half-cell reactions for oxidation and reduction. The negative electrode (lead) is oxidized to form lead sulfate (PbSO4), while the positive electrode (lead dioxide) is reduced, forming lead sulfate as well. The half-cell reactions are as follows: Oxidation (anode) half-cell reaction: Pb(s) + HSO4鈭�(aq) 鈫� PbSO4(s) + H+(aq) + 2e鈭� Reduction (cathode) half-cell reaction: PbO2(s) + HSO4鈭�(aq) + 3H+(aq) + 2e鈭� 鈫� PbSO4(s) + 2H2O(l)
03

Combine the half-cell reactions to form the overall cell reaction.

Now, we can add the half-cell reactions together to get the overall cell reaction. Since the number of electrons involved in both reactions is the same (2e鈭�), we do not need to balance the number of electrons. The overall cell reaction is as follows: Pb(s) + PbO2(s) + 2HSO4鈭�(aq) + 2H+(aq) 鈫� 2PbSO4(s) + 2H2O(l)
04

Identify the oxidized and reduced species.

In the lead storage battery reaction, lead (Pb) is oxidized, losing two electrons to form lead sulfate (PbSO4). In the same reaction, lead dioxide (PbO2) is reduced, gaining two electrons to form lead sulfate as well. Thus, the oxidized species is lead (Pb), and the reduced species is lead dioxide (PbO2).
05

Explain why the battery can be recharged.

A lead storage battery can be recharged because the chemical reactions involved in the discharge (draining) process are reversible. When an external electric current is applied to the battery, the reverse reactions occur. The lead sulfate (PbSO4) formed during discharging reacts back to form lead (Pb) and lead dioxide (PbO2), while the sulfuric acid electrolyte is regenerated. This restores the original composition of the battery electrodes and allows it to be used again.

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Most popular questions from this chapter

What is the oxidation state of the atoms in an uncombined element? Does it depend on whether the element occurs as a diatomic molecule \(\left(\mathrm{O}_{2}, \mathrm{N}_{2}\right)\) or as a larger molecule \(\left(\mathrm{P}_{4}, \mathrm{S}_{8}\right) ?\)

Why must the sum of all the oxidation states of the atoms in a neutral molecule be zero?

For each of the following oxidation-reduction reactions of metals with nonmetals, identify which element is oxidized and which is reduced. a. \(6 \mathrm{Na}(s)+\mathrm{N}_{2}(g) \rightarrow 2 \mathrm{Na}_{3} \mathrm{N}(s)\) b. \(\mathrm{Mg}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{MgCl}_{2}(s)\) c. \(2 \mathrm{Al}(s)+3 \mathrm{Br}_{2}(l) \rightarrow 2 \mathrm{AlBr}_{3}(s)\) d. \(4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\)

In each of the following reactions, identify which element is oxidized and which is reduced by assigning oxidation numbers. a. \(\mathrm{Zn}(s)+2 \mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{H}_{2}(g)+\operatorname{CuSO}_{4}(a q) \rightarrow \operatorname{Cu}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)\) c. \(\mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(l) \rightarrow 2 \mathrm{NBr}_{3}(g)\) d. \(2 \mathrm{KBr}(a q)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{KCl}(a q)+\mathrm{Br}_{2}(l)\)

In assigning oxidation states for a covalently bonded molecule, we assume that the more ___________ element controls both electrons of the covalent bond.
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