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Chapter 18: Problem 67

The overall reaction in the lead storage battery is \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow\) Calculate \(\mathscr{8}\) at \(25^{\circ} \mathrm{C}\) for this battery when \(\left[\mathrm{H}_{2} \mathrm{SO}_{4}\right]=4.5 M\), that is, \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{HSO}_{4}^{-}\right]=4.5 M .\) At \(25^{\circ} \mathrm{C}, \mathscr{b}^{\circ}=2.04 \mathrm{~V}\) for the lead storage battery.

Short Answer

Expert verified
The cell potential of the lead-acid battery at the given conditions, with [\(\text{H}^{+}\)] = [HSO鈧勨伝] = 4.5 M and T = 25掳C (298 K), is approximately 2.0073 V.

Step by step solution

01

Write down the Nernst equation

The Nernst Equation relates the cell potential (E), the standard cell potential (E掳), the number of electrons transferred (n), the gas constant (R), the temperature (T), and the concentrations of species involved in the reaction. The Nernst equation is given as follows: \[E = E掳 - \frac{RT}{nF} \ln Q\] where E掳 is the standard cell potential, R is the gas constant, T is the temperature in Kelvin, n is the number of electrons transferred, F is Faraday's constant, and Q is the reaction quotient.
02

Identify the given values and constants

From the question, we have E掳 = 2.04 V, concentration of H鈦� and HSO鈧勨伝 both equal to 4.5 M, and temperature T = 25掳C = 298 K. The gas constant (R) is 8.314 J/(mol路K) and Faraday's constant (F) is 96485 C/mol. For the lead-acid battery, two electrons are transferred in the reaction, so n = 2.
03

Calculate the reaction quotient (Q)

For the given reaction, \[Pb(s) + PbO_2(s) + 2 H鈦�(aq) + 2 HSO鈧勨伝(aq) \longrightarrow\] the reaction quotient Q is defined as follows: \[Q = \frac{[H鈦篯^2[H_{2}SO_4]^2}{1}\] Since [H鈦篯 = [HSO鈧勨伝] = 4.5 M, Q simplifies to \[Q = (4.5)^2(4.5)^2 = 410.0625\]
04

Calculate the cell potential (E) using the Nernst Equation

By substituting the given values and constants into the Nernst Equation, we can solve for the cell potential E: \[E = E掳 - \frac{RT}{nF} \ln Q\] \[E = 2.04 - \frac{8.314脳298}{2脳96485} \ln 410.0625\] \[E \approx 2.04 - 0.0211 \ln 410.0625 \approx 2.0073\, V\] So, the cell potential of this battery at the given conditions is approximately 2.0073 V.

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Key Concepts

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

Lead Storage Battery
Lead storage batteries are a type of rechargeable battery commonly used in cars. They consist of lead dioxide (\(\text{PbO}_2\)), lead (\(\text{Pb}\)), and a sulfuric acid (\(\text{H}_2\text{SO}_4\)) electrolyte. These batteries can produce a high electric current quickly, making them ideal for engines requiring a lot of power to start.

  • Components: Each cell contains plates of lead and lead dioxide submerged in sulfuric acid.
  • Functionality: During discharge, lead and lead dioxide react with sulfuric acid to create lead sulfate and water. This reaction releases electrons, producing electricity.
  • Rechargeability: The process can be reversed by applying an external electrical charge, restoring the original materials.
Understanding these basics helps explain why the lead storage battery has a high starting power and how it utilizes chemical reactions to generate energy.
Cell Potential Calculation
Cell potential is crucial for determining how much voltage a battery can produce. For the lead storage battery, this involves using the Nernst Equation to find the cell potential under non-standard conditions.

The Nernst Equation is:\[E = E^\circ - \frac{RT}{nF} \ln Q\]

  • \(E^\circ\) is the standard cell potential, in this case, 2.04 V.
  • Variables: \(R\) is the gas constant (8.314 J/(mol路K)), \(T\) is the temperature in Kelvin (298 K for 25掳C), \(n\) is the number of electrons transferred (2 for this reaction), and \(F\) is Faraday's constant (96485 C/mol).
  • Reaction Quotient: \(Q\) is calculated from the concentrations of the reacting species.
Plugging in these values, you can calculate the actual cell potential, which gives insight into the battery's efficiency under these specific conditions.
Reaction Quotient
The reaction quotient, \(Q\), helps determine how far a reaction is from equilibrium. It's a ratio of the products' concentrations to the reactants' concentrations, each raised to the power of their coefficients in the balanced equation.

For the lead storage battery reaction:\[\text{Pb}(s) + \text{PbO}_2(s) + 2 \text{H}^+(aq) + 2 \text{HSO}_4^-(aq) \longrightarrow \]\The reaction quotient is:\[Q = \frac{{[\text{H}^+]^2 [\text{HSO}_4^-]^2}}{1}\]
  • Here, both \([\text{H}^+]\) and \([\text{HSO}_4^-]\) are given as 4.5 M.
  • Calculation: Substitute these into the equation to find \(Q = (4.5)^2(4.5)^2 = 410.0625\).
By understanding \(Q\), you can see how the concentrations affect the potential and how close the reaction is to reaching equilibrium. This is vital for adjusting and optimizing battery performance.

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Most popular questions from this chapter

What volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) at STP are produced from the electrolysis of water by a current of \(2.50 \mathrm{~A}\) in \(15.0 \mathrm{~min} ?\)

Assign oxidation numbers to all the atoms in each of the following. a. \(\mathrm{HNO}_{3}\) e. \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) i. \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) b. \(\mathrm{CuCl}_{2}\) f. Ag j. \(\mathrm{CO}_{2}\) c. \(\mathrm{O}_{2}\) g. \(\mathrm{PbSO}_{4}\) k. \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}\) d. \(\mathrm{H}_{2} \mathrm{O}_{2}\) h. \(\mathrm{PbO}_{2}\) 1\. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\)

Consider the galvanic cell based on the following half-reactions: $$\begin{array}{ll}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{~V} \end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\mathrm{ccll}}^{\circ}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{E}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} M\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

Calculate \(K_{\mathrm{sp}}\) for iron(II) sulfide given the following data: $$\begin{aligned}\mathrm{FeS}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s)+\mathrm{S}^{2-}(a q) & & \mathscr{b}^{\circ} &=-1.01 \mathrm{~V} \\\\\mathrm{Fe}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & & \mathscr{b}^{\circ} &=-0.44 \mathrm{~V} \end{aligned}$$

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{~K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(I) \quad K=1.28 \times 10^{83}$$ a. Calculate \(\mathscr{C}^{\circ}\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.
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