Problem 73 - Chemistry: The Central Science

Question

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 \(\mathrm{PbO}_{2}\) ?

Step-by-Step Solution

Answer

During the discharge process, 463.86 g of PbO₂ is reduced at the cathode, and 373960.6 coulombs of electrical charge are transferred from Pb to PbO₂.

Step 1: (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: \(\mathrm{PbO}_{2}(s) + \mathrm{SO}_{4}^{2-} + 4H^{+} + 2e^{-} \rightarrow \mathrm{PbSO}_{4}(s) + 2H_{2}O\)

Step 2: (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: \(n_{Pb} = \frac{mass_{Pb}}{molar\_mass_{Pb}} = \frac{402\mathrm{~g}}{207.2\mathrm{~g/mol}} = 1.94\mathrm{~moles}\)

Step 3: (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: \(n_{PbO_2} = n_{Pb} = 1.94\mathrm{~moles}\)

Step 4: (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.

Step 5: (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 \(\mathrm{PbO}_{2}\).

Key Concepts

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 (PbO₂).

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 (PbSO₄) 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 + SO₄²⁻ → PbSO₄ + 2e⁻
  At the cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O


• 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 (PbSO₄), a heavier compound.


• Simultaneously, at the cathode, lead dioxide (PbO₂) 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.