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Chapter 9: Problem 116

When a lead storage battery is discharged (a) lead is formed (b) lead sulphate is consumed (c) \(\mathrm{SO}_{2}\) is evolved (d) sulphuric acid is consumed

Short Answer

Expert verified
(d) Sulfuric acid is consumed.

Step by step solution

01

Understanding the Lead Storage Battery

A lead storage battery, commonly found in cars, consists of lead dioxide and lead electrodes in a sulfuric acid solution. When it discharges, chemical reactions occur.
02

Identifying the Discharge Reaction

During discharge, lead ( ext{Pb}) at the negative electrode and lead dioxide ( ext{PbO}_2) at the positive electrode react with sulfuric acid ( ext{H}_2 ext{SO}_4) to form lead sulfate ( ext{PbSO}_4) and water ( ext{H}_2 ext{O}).
03

Analyzing the Chemical Changes

The overall reaction can be summarized as: \[ \text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O} \]. This means that lead sulfate forms at both electrodes, and sulfuric acid is consumed in the process.
04

Evaluating the Statements

(a) Lead is not formed; it is used instead. (b) Lead sulfate is produced, not consumed. (c) Sulfur dioxide is not evolved in this electrode reaction. (d) Sulfuric acid is consumed as it gets transformed into water and lead sulfate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrochemical Reactions
Lead storage batteries utilize electrochemical reactions to store and release energy. These reactions involve the transfer of electrons between substances. When the battery is discharging, these reactions convert chemical energy into electrical energy, powering devices such as cars. In a lead storage battery, we primarily encounter two reactants: lead and lead dioxide. As the discharge process begins, electrons move from the negative electrode (lead) through the external circuit, to the positive electrode (lead dioxide). This electron flow is what we harness as electricity.

The fascinating part of these reactions is how they are reversible when charging the battery, converting electrical energy back into chemical energy. Understanding these electrochemical processes provides insights into how energy conversion occurs efficiently within batteries and similar systems.
Sulfuric Acid Consumption
Sulfuric acid plays a crucial role in the operation of lead storage batteries. It acts as the electrolyte, facilitating the movement of ions between electrodes. During the discharge of a battery, sulfuric acid molecules are consumed, as they participate in the chemical reactions that power the battery.

As the acid decomposes, it reacts with lead and lead dioxide to produce lead sulfate and water. The chemical equation summarizing this is: \[ \text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O} \].

This consumption of sulfuric acid leads to a decrease in its concentration in the electrolyte solution. Over time, this change can be measured to gauge the battery's charge state—less sulfuric acid means the battery is more discharged.
Lead Sulfate Formation
The production of lead sulfate is a key marker of a lead battery's discharge cycle. When discharge occurs, lead and lead dioxide react with sulfuric acid, leading to the formation of lead sulfate on the surfaces of the battery plates.

Lead sulfate is essentially the result of the battery's active materials having reacted with sulfate ions from the electrolyte. This compound builds up as a white crusty layer on the plates. If the battery remains in the discharged state for too long, the lead sulfate can become more stable and harder to reconvert back during charging.

Regular charging can help to reverse the formation of lead sulfate, dissolving it back into the electrolyte as sulfate ions, thus restoring the battery’s ability to store energy effectively.
Discharge Reaction Analysis
To fully appreciate the discharge reactions in a lead storage battery, it's essential to analyze the overall chemical changes. During the discharge phase, the battery releases energy by converting the stored chemical energy into electrical energy. The overall reaction involves both electrodes—the lead anode and the lead dioxide cathode. These react with sulfuric acid to form lead sulfate and water, indicating a complete reaction.\[ \text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O} \]

By analyzing this conversion process, we can see the critical role that the different compounds play. Lead sulfate and water are the final products, illustrating the transformation that occurs at the molecular level. This discharge reaction explains why maintaining the right amount of sulfuric acid and regular charging is essential for battery longevity.

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Most popular questions from this chapter

The standard reduction potentials at \(298 \mathrm{~K}\) for the following half- reactions are given against each \(\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Zn}(\mathrm{s})-0.762\) \(\mathrm{Cr}^{3+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Cr}(\mathrm{s}) \quad-0.740\) \(2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g}) \quad 0.000\) \(\mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Fe}^{2+}\) (aq) \(0.770\) Which is the strongest reducing agent? (a) \(\mathrm{H}_{2}(\mathrm{~g})\) (b) \(\mathrm{Cr}(\mathrm{s})\) (c) \(\mathrm{Zn}(\mathrm{s})\) (d) \(\mathrm{Fe}^{2+}(\mathrm{aq})\)

Equal quantities of electricity are passed through three voltameters containing \(\mathrm{FeSO}_{4}, \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), and \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\). Consider the following statements in this regard (1) the amount of iron deposited in \(\mathrm{FeSO}_{4}\) and \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is equal (2) the amount of iron deposited in \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) is two thirds of the amount of iron deposited in \(\mathrm{FeSO}_{4}\) (3) the amount of iron deposited in \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) is equal Of these statements (a) 1 and 2 are correct (b) 2 and 3 are correct (c) 1 and 3 are correct (d) 1,2 and 3 are correct

The hydrogen electrode is dipped in a solution of \(\mathrm{pH}=\) \(3.0\) at \(25^{\circ} \mathrm{C}\). The potential of hydrogen electrode would be (a) \(-0.177 \mathrm{~V}\) (b) \(0.177 \mathrm{~V}\) (c) \(1.77 \mathrm{~V}\) (d) \(0.277 \mathrm{~V}\)

In electrolysis of dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\), what is liberated at anode? (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{SO}_{4}^{2-}\) (c) \(\mathrm{SO}_{2}\) (d) \(\mathrm{O}_{2}\)

Given that \(\mathrm{E}_{\mathrm{N}^{1+} / \mathrm{Ni}}^{0}=-0.25 \mathrm{~V} ; \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{0}=+0.34 \mathrm{~V}\) \(E_{A \mathrm{~B}^{+} / \mathrm{Ag}}^{0}=+0.80 \mathrm{~V} ; \mathrm{E}_{\mathrm{Zn}^{2+} / Z \mathrm{n}}^{0}=-0.76 \mathrm{~V}\) Which of the following redox processes will not take place in specified direction? (a) \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\) (b) \(\mathrm{Cu}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{~g})\) (c) \(\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})\) (d) \(\mathrm{Ni}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s}) \rightarrow \mathrm{Ni}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq})\)
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