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Chapter 20: Problem 73

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

Expert verified
(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

01

(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.
02

(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 \(PbO_2 = \dfrac{300 \mathrm{~g}}{239.2 \mathrm{~g/mol}} \approx 1.255 \mathrm{~mol}\)
03

(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 \(Pb\) = \(1.255 \mathrm{~mol} \times 207.2 \mathrm{~g/mol} \approx 259.5 \mathrm{~g}\)
04

(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.

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Key Concepts

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

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.

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Most popular questions from this chapter

A voltaic cell utilizes the following reaction and operates at 298 K: $$ 3 \mathrm{Ce}^{4+}(a q)+\mathrm{Cr}(s) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+\mathrm{Cr}^{3+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when \(\left[\mathrm{Ce}^{4+}\right]=3.0 \mathrm{M},\) \(\left[\mathrm{Ce}^{3+}\right]=0.10 \mathrm{M},\) and \(\left[\mathrm{Cr}^{3+}\right]=0.010 \mathrm{M} ?(\mathbf{c})\) What is the emf of the cell when \(\left[\mathrm{Ce}^{4^{+}}\right]=0.010 \mathrm{M},\left[\mathrm{Ce}^{3+}\right]=2.0 \mathrm{M}\) and \(\left[\mathrm{Cr}^{3+}\right]=1.5 \mathrm{M} ?\)

Cytochrome, a complicated molecule that we will represent as \(\mathrm{CyFe}^{2+}\), reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions (Section 19.7). At \(\mathrm{pH} 7.0\) the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}\) $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & & E_{\mathrm{red}}^{\circ}=+0.82 \mathrm{~V} \\\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} & \longrightarrow \mathrm{CyFe}^{2+}(a q) & E_{\mathrm{red}}^{\circ} &=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air? \((\mathbf{b})\) If the synthesis of \(1.00 \mathrm{~mol}\) of ATP from adenosine diphosphate (ADP) requires a \(\Delta G\) of \(37.7 \mathrm{~kJ},\) how many moles of ATP are synthesized per mole of \(\mathrm{O}_{2} ?\)

Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about \(7 \times 10^{8} \mathrm{~mol}\) of \(\mathrm{H}_{2}\) to provide the buoyancy to lift the ship (J. Chem. Educ., 1973, Vol. 50, 61). (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) if the pressure on the gases at the depth of the wreckage \((3 \mathrm{~km})\) is \(30 \mathrm{MPa} ?(\mathbf{c})\) What is the minimum electrical energy required to raise the Titanic by electrolysis? (d) What is the minimum cost of the electrical energy required to generate the necessary \(\mathrm{H}_{2}\) if the electricity costs 85 cents per kilowatt-hour to generate at the site?

A voltaic cell is constructed that is based on the following reaction: $$ \mathrm{Sn}^{2+}(a q)+\mathrm{Pb}(s) \longrightarrow \mathrm{Sn}(s)+\mathrm{Pb}^{2+}(a q) $$ (a) If the concentration of \(\mathrm{Sn}^{2+}\) in the cathode half-cell is \(1.00 M\) and the cell generates an emf of \(+0.22 \mathrm{~V},\) what is the concentration of \(\mathrm{Pb}^{2+}\) in the anode half-cell? \((\mathbf{b})\) If the anode half-cell contains \(\left[\mathrm{SO}_{4}^{2-}\right]=1.00 M\) in equilibrium with \(\mathrm{PbSO}_{4}(s),\) what is the \(K_{s p}\) of \(\mathrm{PbSO}_{4} ?\)

Indicate whether each of the following statements is true or false: (a) If something is oxidized, it is formally losing electrons. (b) For the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\) \(\mathrm{Co}^{3+}(a q), \mathrm{Fe}^{3+}(a q)\) is the reducing agent and \(\mathrm{Co}^{2+}(a q)\) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction.
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