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Chapter 20: Problem 71

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

Short Answer

Expert verified
During the discharge of the lead-acid battery, the mass of \(PbO_{2}(s)\) reduced at the cathode is approximately \(464 \mathrm{~g}\).

Step by step solution

01

Write the balanced chemical equation for the lead-acid battery discharging process.

The overall balanced chemical equation for a lead-acid battery's discharging process is: \[Pb(s) + PbO_{2}(s) + 2H_{2}SO_{4}(aq) \rightarrow 2PbSO_{4}(s) + 2H_{2}O(l)\] This equation tells us that 1 mole of \(\mathrm{Pb}\) reacts with 1 mole of \(\mathrm{PbO}_{2}\) and 2 moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) to produce 2 moles of \(\mathrm{PbSO}_{4}\) and 2 moles of \(\mathrm{H}_{2}\mathrm{O}\). We will use this information to determine the mass of \(\mathrm{PbO}_{2}\) reduced at the cathode.
02

Convert the mass of \(\mathrm{Pb}\) to moles.

Given the mass of \(\mathrm{Pb}\) converted into \(\mathrm{PbSO}_{4}\), we can convert it to moles using the molar mass of \(\mathrm{Pb}\). The molar mass of \(\mathrm{Pb}\) is \(207.2 \mathrm{~g/mol}\). To convert the mass of \(\mathrm{Pb}\) (\(402 \mathrm{~g}\)) to moles: \[\text{moles of Pb} = \frac{402 \mathrm{~g}}{207.2 \mathrm{~g/mol}} = 1.94 \mathrm{~moles}\]
03

Use the stoichiometry to find moles of \(\mathrm{PbO}_{2}\) reduced.

According to the balanced chemical equation, moles of \(\mathrm{Pb}\) and \(\mathrm{PbO}_{2}\) have a 1:1 stoichiometric ratio. Therefore, the moles of \(\mathrm{PbO}_{2}\) reduced at the cathode are equal to the moles of \(\mathrm{Pb}\) converted at the anode: \[\text{moles of PbO}_{2} = 1.94 \mathrm{~moles}\]
04

Convert moles of \(\mathrm{PbO}_{2}\) to mass.

Now that we have the moles of \(\mathrm{PbO}_{2}\) reduced, we can convert it to mass using the molar mass of \(\mathrm{PbO}_{2}\). The molar mass of \(\mathrm{PbO}_{2}\) is \(239.2 \mathrm{~g/mol}\). To convert the moles of \(\mathrm{PbO}_{2}\) to mass: \[\text{mass of PbO}_{2} = 1.94 \mathrm{~moles} \times 239.2 \mathrm{~g/mol} = 464 \mathrm{~g}\]
05

State the final mass of \(\mathrm{PbO}_{2}\) reduced.

During the discharge of the lead-acid battery, the mass of \(\mathrm{PbO}_{2}(s)\) reduced at the cathode is approximately \(464 \mathrm{~g}\).

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Key Concepts

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

Chemical Equation Balancing
When dealing with reactions, balancing the chemical equation is crucial. It ensures that the same number of each type of atom appears on both sides of the equation. In the context of a lead-acid battery, the discharging process involves chemical transformations. The balanced equation for this reaction is: \[Pb(s) + PbO_{2}(s) + 2H_{2}SO_{4}(aq) \rightarrow 2PbSO_{4}(s) + 2H_{2}O(l)\]This equation depicts how reactants (left side) convert into products (right side).
Each compound's coefficient represents the number of moles involved. Here, 1 mole of lead and 1 mole of lead(IV) oxide react with 2 moles of sulfuric acid, resulting in the formation of 2 moles of lead sulfate and water. Balancing is vital because it reflects conservation of mass, aligning with the first law of thermodynamics.
Stoichiometry
Stoichiometry is the method we use to relate quantities of reactants to products in a chemical reaction. It's a way to convert between moles, mass, and number of particles, based on the balanced chemical equation. In the lead-acid battery discharge example, stoichiometry helps in determining how much of the various substances are required or produced.
  • The balanced equation tells us that the stoichiometric ratio between lead and lead(IV) oxide is 1:1.
  • This means for every mole of lead undergoing reaction, one mole of lead(IV) oxide is needed.
Using stoichiometry, we can further calculate the conversion of various materials in processes, ensuring we have the right amounts to react and produce expected products.
Molar Mass Conversion
Molar mass is a critical concept in chemistry as it allows for the conversion of mass in grams to moles. In exercises like the one given, you often have to start by converting the mass of a substance to moles, using its molar mass.For instance, in the conversion:
Lead (\(Pb\)) has a molar mass of 207.2 g/mol, and Lead(IV) oxide (\(PbO_{2}\)) has a molar mass of 239.2 g/mol.
To find the moles: \[\text{moles of Pb} = \frac{\text{mass of Pb}}{\text{molar mass of Pb}} = \frac{402 \text{ g}}{207.2 \text{ g/mol}} = 1.94 \text{ moles}\]The same approach applies when converting moles back to mass. This calculation is crucial because it bridges the gap between the lab scale (grams) and the reaction scale (moles).
Electrochemical Reactions
In the realm of chemistry, electrochemical reactions describe processes where electrons transfer between species. Lead-acid batteries utilize these reactions during charging and discharging.
  • During discharge, lead at the anode converts to lead sulfate, releasing electrons.
  • Simultaneously, lead(IV) oxide at the cathode is reduced (gains electrons) and also forms lead sulfate.
These reactions are vital as they create the flow of electrons, which provides electrical energy to devices. Understanding the oxidation (loss of electrons) and reduction (gain of electrons) processes helps us comprehend how batteries store and release energy. It's the backbone of how most rechargeable batteries function.

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Most popular questions from this chapter

Given the following reduction half-reactions: \(\mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q)\) \(E_{\mathrm{red}}^{\circ}=+0.77 \mathrm{~V}\) \(\mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\) \(E_{\mathrm{red}}^{\circ}=+0.60 \mathrm{~V}\) \(\mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) \(E_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V}\) \(\mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) \(E_{\mathrm{red}}^{\circ}=+1.00 \mathrm{~V}\) (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}{ }^{2-}(a q)\), by \(\mathrm{N}_{2} \mathrm{O}(a q)\), and \(\mathrm{by} \mathrm{VO}_{2}{ }^{+}(a q)\). (b) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

This oxidation-reduction reaction in acidic solution is spontaneous: $$ \begin{array}{r} 5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)-\rightarrow \\ 5 \mathrm{Fe}^{3+}(a q)+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ A solution containing \(\mathrm{KMnO}_{4}\) and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is poured into one beaker, and a solution of \(\mathrm{FeSO}_{4}\) is poured into another. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and a wire that passes through a voltmeter connects the two solutions. (a) Sketch the cell, indicating the anode and the cathode, the direction of electron movement through the external circuit, and the direction of ion migrations through the solutions. (b) Sketch the process that occurs at the atomic level at the surface of the anode. (c) Calculate the emf of the cell under standard conditions. (d) Calculate the emf of the cell at \(298 \mathrm{~K}\) when the concentrations are the following: \(\mathrm{pH}=0.0, \quad\left[\mathrm{Fe}^{2+}\right]=0.10 \mathrm{M}, \quad\left[\mathrm{MnO}_{4}^{-}\right]=1.50 \mathrm{M}\) \(\left[\mathrm{Fe}^{3+}\right]=2.5 \times 10^{-4} \mathrm{M},\left[\mathrm{Mn}^{2+}\right]=0.001 \mathrm{M}\)

(a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). (b) Write the balanced half-reactions involved in the air oxidation of \(\mathrm{Fe}^{2+}(a q)\) to \(\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}\)

(a) \(\mathrm{A} \mathrm{Cr}^{3+}(a q)\) solution is electrolyzed, using a current of \(7.60 \mathrm{~A}\). What mass of \(\mathrm{Cr}(s)\) is plated out after \(2.00\) days? (b) What amperage is required to plate out \(0.250 \mathrm{~mol} \mathrm{Cr}\) from a \(\mathrm{Cr}^{3+}\) solution in a period of \(8.00 \mathrm{~h}\) ?

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & E_{\mathrm{red}}^{\circ}=-0.858 \mathrm{~V} \\ & E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{~V} \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(\mathrm{aq})+2 \mathrm{OH}^{-}(a q) \\ \mathrm{En}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}(s) & E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \\ E_{\mathrm{red}}^{\circ}=-0.14 \mathrm{~V} \end{aligned} $$ (a) Write the cell reaction for the combination of these half-cell reactions that leads to the largest positive cell emf, and calculate the value. (b) Write the cell reaction for the combination of half-cell reactions that leads to the smallest positive cell emf, and calculate that value.
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