Question

During a period of discharge of a lead-acid battery, \(300 \mathrm{~g}\) of \(\mathrm{PbO}_{2}(s)\) from the cathode is converted into \(\mathrm{PbSO}_{4}(s)\). (a) What mass of \(\mathrm{Pb}(s)\) is oxidized at the anode during this same period? (b) How many coulombs of electrical charge are transferred from \(\mathrm{Pb}\) to \(\mathrm{PbO}_{2}\) ?

Short answer

(a) The mass of oxidized \(Pb\) during the discharge period is approximately 259.5 g. (b) The total charge transferred from \(Pb\) to \(PbO_2\) during this period is approximately 2.42 x 10^5 C.

Step-by-step solution

(Step 1: Write the balanced chemical equation for the lead-acid battery reaction)

The overall reaction in a lead-acid battery during discharge can be written as: \(Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)\) Here, we can see that 1 mole of \(Pb(s)\) is converted to 1 mole of \(PbSO_4(s)\) during the reaction.

(Step 2: Calculate the moles of converted \(PbO_{2}\))

We are given that 300g of \(PbO_2\) is converted to \(PbSO_4\). To find the moles of \(PbO_2\), we will use its molar mass: Molar mass of \(PbO_2 = 207.2 + (16.0 \times 2) = 239.2 \mathrm{~g/mol}\) Moles of \(PbO_2 = \dfrac{300 \mathrm{~g}}{239.2 \mathrm{~g/mol}} \approx 1.255 \mathrm{~mol}\)

(Step 3: Calculate the moles and mass of oxidized \(Pb\))

Since 1 mole of \(Pb(s)\) is converted to 1 mole of \(PbSO_4(s)\), the moles of \(Pb\) oxidized are equal to the moles of \(PbO_2\) converted: Moles of oxidized \(Pb\) = 1.255 mol To find the mass of oxidized \(Pb\), we will multiply the moles by its molar mass: Molar mass of \(Pb = 207.2 \mathrm{~g/mol}\) Mass of oxidized \(Pb\) = \(1.255 \mathrm{~mol} \times 207.2 \mathrm{~g/mol} \approx 259.5 \mathrm{~g}\)

(Step 4: Calculate the charge transferred from \(Pb\) to \(PbO_{2}\))

In the balanced chemical equation, the oxidation state of \(Pb\) in \(Pb(s)\) is 0 and in \(PbO_2\) is +4. Therefore, 2 electrons are transferred from \(Pb\) to \(PbO_2\) per atom. Knowing the moles of \(Pb\) that react, we can find the total charge transferred: Total \(Pb\) atoms = \(1.255 \mathrm{~mol} \times 6.022 \times 10^{23} \mathrm{~atoms/mol} \approx 7.553 \times 10^{23} \mathrm{~atoms}\) Charge transferred per atom = 2 electrons = \(2 \times 1.602 \times 10^{-19} \mathrm{~C}\) Total charge transferred = \((7.553 \times 10^{23} \mathrm{~atoms}) \times (2 \times 1.602 \times 10^{-19} \mathrm{~C/atom}) = 2.42 \times 10^5 \mathrm{~C}\) (a) The mass of \(Pb(s)\) oxidized at the anode during this period is approximately 259.5 g. (b) The total charge transferred from \(Pb\) to \(PbO_2\) during this period is approximately 2.42 x 10^5 C.

Key concepts

Chemical Reaction Balancing

Balancing chemical equations is a fundamental skill in chemistry. It ensures that the mass and charge are conserved during a chemical reaction. In the context of lead-acid batteries, the balanced equation depicts the chemical process occurring during battery discharge. The equation given is:\[Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)\]This equation tells us that one mole of lead \( (Pb) \) reacts with one mole of lead dioxide \( (PbO_2) \) and two moles of sulfuric acid \( (H_2SO_4) \) to produce two moles of lead sulfate \( (PbSO_4) \) and two moles of water \( (H_2O) \). By balancing the atoms on each side, we confirm no atoms are lost or gained, adhering to the law of conservation of mass.

Oxidation-Reduction

Oxidation-reduction, or redox reactions, involve the transfer of electrons between substances. In a lead-acid battery, redox reactions are crucial as they facilitate the discharge and recharge cycle. During discharge, lead \( (Pb) \) is oxidized and loses electrons, transforming into lead sulfate \( (PbSO_4) \). Conversely, lead dioxide \( (PbO_2) \) is reduced, gaining electrons to also form lead sulfate.

• Oxidation: Loss of electrons (e.g., \( Pb \to Pb^{2+} + 2e^- \))
• Reduction: Gain of electrons (e.g., \( PbO_2 + 4H^+ + 2e^- \to PbSO_4 + 2H_2O \))

The electron transfer is what generates electrical energy in the battery, which can then be used to power devices.

Coulombs Law

Coulomb's Law is essential for understanding the charge transfer in redox reactions within batteries. This law relates the force between two electric charges to the product of their charges and inversely to the square of their distance apart. In simpler terms, it helps calculate the amount of electrical charge transferred during a reaction, which in this exercise involves moving electrons from lead to lead dioxide.With the calculation of transferred electrons, each having a charge of \( 1.602 \times 10^{-19} \) coulombs, we can compute the total charge transferred by multiplying by the total number of reacting atoms. In the given solution, the transfer of charge calculated is \( 2.42 \times 10^5 \) coulombs, illustrating the major role of electrons in the discharge process of a battery.

Molar Mass Calculation

Molar mass is a measure of the mass of one mole of a substance, typically expressed in grams per mole \( (g/mol) \). Calculating molar mass is a crucial step in stoichiometry, allowing chemists to convert between grams and moles.For instance, to determine how much \( Pb \) is oxidized in the lead-acid battery, we first calculate the molar mass of \( PbO_2 \), which is \( 239.2 \ g/mol \). Given 300 grams of \( PbO_2 \) convert to lead sulfate, we calculate moles by dividing mass by molar mass:\[\frac{300 \ ext{g}}{239.2 \ ext{g/mol}} \approx 1.255 \ ext{mol}\]This mole calculation then helps find the mass of lead oxidized by using the molar mass of lead \( (207.2 \ g/mol) \), equating to approximately 259.5 grams oxidized.