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Chapter 20: Problem 73

During a period of discharge of a lead-acid battery, \(402 \mathrm{~g}\) of \(\mathrm{Pb}\) from the anode is converted into \(\mathrm{PbSO}_{4}(s)\). (a) What mass of \(\mathrm{PbO}_{2}(s)\) is reduced at the cathode during this same period? (b) How many coulombs of electrical charge are transferred from \(\mathrm{Pb}\) to \(\mathrm{PbO}_{2}\) ?

Short Answer

Expert verified
During the discharge process, 463.86 g of PbO鈧� is reduced at the cathode, and 373960.6 coulombs of electrical charge are transferred from Pb to PbO鈧�.

Step by step solution

01

(1) Write the balanced half-reactions

The overall balanced reaction of a lead-acid battery during discharge is given by: Anode Reaction: \(\mathrm{Pb(s)} + \mathrm{SO}_{4}^{2-} \rightarrow \mathrm{PbSO}_{4}(s) + 2e^{-}\) Cathode Reaction: \(\mathrm{PbO}_{2}(s) + \mathrm{SO}_{4}^{2-} + 4H^{+} + 2e^{-} \rightarrow \mathrm{PbSO}_{4}(s) + 2H_{2}O\)
02

(2) Determine the moles of \(\mathrm{Pb}\) consumed

Given the mass of \(\mathrm{Pb}\) consumed is 402 g, we can calculate the moles of \(\mathrm{Pb}\) using its molar mass: \(n_{Pb} = \frac{mass_{Pb}}{molar\_mass_{Pb}} = \frac{402\mathrm{~g}}{207.2\mathrm{~g/mol}} = 1.94\mathrm{~moles}\)
03

(3) Determine the moles of \(\mathrm{PbO}_{2}\) produced

From the balanced half-reactions, you can see that 1 mole of \(\mathrm{Pb}\) produces 1 mole of \(\mathrm{PbO}_{2}\). Therefore, the moles of \(\mathrm{PbO}_{2}\) produced are equal to the moles of \(\mathrm{Pb}\) consumed: \(n_{PbO_2} = n_{Pb} = 1.94\mathrm{~moles}\)
04

(4) Calculate the mass of \(\mathrm{PbO}_{2}\) produced

Now that we know the moles of \(\mathrm{PbO}_{2}\), we can calculate the mass using its molar mass: \(mass_{PbO_2} = n_{PbO_2} \times molar\_mass_{PbO_2} = (1.94\mathrm{~moles}) \times (239.0\mathrm{~g/mol}) = 463.86\mathrm{~g}\) Therefore, the mass of \(\mathrm{PbO}_{2}\) reduced at the cathode during the same period is 463.86 g.
05

(5) Calculate the amount of charge transferred

From the anode reaction, we know that 1 mole of \(\mathrm{Pb}\) releases 2 moles of electrons (2e\(^-\)). We can use Faraday's constant (F = 96485 C/mol) to convert the moles of electrons transferred to the amount of electric charge in coulombs: \(charge\_transferred = n_{e^{-}} \times F = (2 \times n_{Pb}) \times F = (2 \times 1.94\mathrm{~moles})\times (96485\mathrm{~C/mol}) = 373960.6\mathrm{~C}\) Hence, 373960.6 coulombs of electrical charge are transferred from \(\mathrm{Pb}\) to \(\mathrm{PbO}_{2}\).

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Key Concepts

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

Lead-acid battery
The lead-acid battery is a type of rechargeable battery that is commonly used in automobiles and backup power systems. It's known for its reliability and ability to deliver a large amount of current. The battery consists of several cells, each containing an anode made of lead (Pb) and a cathode made of lead dioxide (PbO鈧�).

The battery operates through the conversion of chemical energy into electrical energy via electrochemical reactions. During discharge, the lead at the anode reacts with sulfuric acid ( \( \text{H}_2\text{SO}_4 \)) electrolyte to form lead sulfate (PbSO鈧�) and electrons. These electrons travel through an external circuit to the cathode, where they help convert lead dioxide into lead sulfate.
  • The balanced overall reaction can be represented as:
    At the anode: Pb + SO鈧劼测伝 鈫� PbSO鈧� + 2e鈦�
    At the cathode: PbO鈧� + SO鈧劼测伝 + 4H鈦� + 2e鈦� 鈫� PbSO鈧� + 2H鈧侽

  • The battery's performance depends heavily on the state of discharge and the concentration of sulfuric acid, making understanding of its chemistry vital for optimal usage.
Reduction-oxidation (redox) reactions
Reduction-oxidation (redox) reactions are fundamental to electrochemistry, involving the transfer of electrons between chemical species. In these reactions, reduction refers to the gain of electrons, whereas oxidation is the loss of electrons.

During the discharge of a lead-acid battery, both the oxidation and reduction reactions take place:
  • At the anode, lead (Pb) is oxidized. This means it loses electrons to form lead sulfate (PbSO鈧�), a heavier compound.

  • Simultaneously, at the cathode, lead dioxide (PbO鈧�) undergoes reduction. Here, it gains electrons and reacts with hydrogen ions and sulfate ions to also form lead sulfate.

  • The flow of electrons from the anode to the cathode through the external circuit provides electrical energy.
The half-reactions help in understanding how materials change and energy is transferred within the cell.

This detailed exchange is key to the battery's ability to recharge, as the process can be reversed, allowing the battery to return to its original state through applied external electrical energy.
Faraday's constant
Faraday's constant is a pivotal concept in electrochemistry that relates the amount of electric charge carried by one mole of electrons. Its value is approximately 96485 coulombs per mole (C/mol).

This constant enables us to calculate the total charge transfer in an electrochemical reaction, like those occurring in a lead-acid battery. By knowing the number of moles of electrons involved, we can use Faraday's constant to determine how much electric charge is involved in these reactions.
  • Within the lead-acid battery, the reaction at the anode produces electrons: for each mole of lead oxidized, two moles of electrons are released.

  • Faraday's constant allows us to calculate the total charge as:
    \[\text{Charge} = \text{number of moles of electrons} \times \text{Faraday's constant}\]

  • In this exercise, 1.94 moles of lead contribute to the release of electrons, leading to a calculated charge of 373960.6 coulombs transferred during discharge.
This measure of electric charge is essential in designing and evaluating battery capacity and efficiency, illustrating how electrochemical processes convert chemical into electrical energy.

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Most popular questions from this chapter

The hydrogen-oxygen fuel cell has a standard emf of \(1.23 \mathrm{~V}\). What advantages and disadvantages are there to using this device as a source of power compared to a \(1.55\) - \(\mathrm{V}\) alkaline battery?

Hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) and dinitrogen tetroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) form a self-igniting mixture that has been used as a rocket propellant. The reaction products are \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). (a) Write a balanced chemical equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance serves as the reducing agent and which as the oxidizing agent?

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. (a) \(\mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s)\) (acidic solution) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q)\) (acidic solution) (c) \(\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(\mathrm{g})\) (acidic solution) (d) \(\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (acidic solution) (e) \(\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (basic solution) (f) \(\mathrm{Mn}^{2+}(\) aq \() \longrightarrow \mathrm{MnO}_{2}(s)\) (basic solution) (g) \(\mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}\) (aq) (basic solution)

Gold exists in two common positive oxidation states, \(+1\) and \(+3\). The standard reduction potentials for these oxidation states are $$ \begin{aligned} A u^{*}(a q)+\mathrm{e}^{-} & \longrightarrow A u(s) \quad E_{\text {red }}^{o}=+1.69 \mathrm{~V} \\ A u^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow A u(s) \quad E_{\text {red }}^{o}=+1.50 \mathrm{~V} \end{aligned} $$ (a) Can you use these data to explain why gold does not tarnish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold-containing ores in an aqucous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction $$ \begin{gathered} 4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g) \\ \longrightarrow 4 \mathrm{Na}\left[\mathrm{Au}(\mathrm{CN})_{2}\right](a q)+4 \mathrm{NaOH}(a q) \end{gathered} $$ What is being oxidized and what is being reduced in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with \(\mathrm{Zn}\) dust to get gold metal. Write a balanced redox reaction for this process. What is being exidized, and what is being reduced?

The cell in Figure \(20.9\) could be used to provide a measure of the \(\mathrm{pH}\) in the cathode half-cell. Calculate the \(\mathrm{pH}\) of the cathode half-cell solution if the cell emf at \(298 \mathrm{~K}\) is measured to be \(+0.684 \mathrm{~V}\) when \(\left[\mathrm{Zn}^{2+}\right]=0.30 \mathrm{M}\) and \(P_{\mathrm{H}_{2}}=0.90 \mathrm{~atm}\).
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