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Chapter 17: Problem 71

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

Short Answer

Expert verified
The cell potential of the lead storage battery at 25°C with a concentration of 4.5 M for \(\mathrm{H}_{2}\mathrm{SO}_{4}\), \(\mathrm{H}^{+\), and \(\mathrm{HSO}_{4}^{-}\) is approximately \(2.06\,\text{V}\).

Step by step solution

01

Write the Nernst equation

The Nernst equation relates the cell potential of a half-cell at non-standard conditions to the standard cell potential, temperature, and concentrations: \[ \mathscr{E} = E^{\circ} - \frac{RT}{nF} \ln Q \] where \(\mathscr{E}\) is the cell potential at non-standard conditions, \(E^{\circ}\) is the standard cell potential, R is the gas constant, T is the temperature in kelvin, n is the number of moles of electrons transferred in the redox reaction, F is the Faraday constant, and Q is the reaction quotient.
02

Convert the temperature to Kelvin

The temperature is given as 25°C. Convert this temperature to Kelvin by adding 273.15: \[ T = 25 + 273.15 = 298.15\,\text{K} \]
03

Calculate the reaction quotient (Q)

Now, we can determine the reaction quotient (Q) for the reaction using the given concentrations of the species: \[ Q = \frac{[\mathrm{PbSO}_{4}]^{2}[\mathrm{H}_{2}\mathrm{O}]^{2}}{[\mathrm{Pb}][\mathrm{PbO}_{2}][\mathrm{H}^{+}]^{2}[\mathrm{HSO}_{4}^{-}]^{2}} \] Since the concentrations of all solid species are constant, we can eliminate them and obtain the expression for Q as follows: \[ Q = \frac{1}{[\mathrm{H}^{+}]^{2}[\mathrm{HSO}_{4}^{-}]^{2}} \]
04

Plug in the given concentrations and calculate Q

The concentrations of \(\mathrm{H}^{+}\) and \(\mathrm{HSO}_{4}^{-}\) are both 4.5 M. Substitute these values into the equation for Q: \[ Q = \frac{1}{(4.5)^{2}(4.5)^{2}} = \frac{1}{410.0625} \]
05

Plug values into the Nernst equation and solve for \(\mathscr{E}\)

Now, plug the values of \(E^{\circ}\), R, T, n, F, and Q from the previous steps into the Nernst equation: \[ \mathscr{E} = 2.04\,\text{V} - \frac{8.314\,\text{J}\,\text{mol}^{-1}\text{K}^{-1} \times 298.15\,\text{K}}{2 \times 96485\,\text{C}\,\text{mol}^{-1}} \ln \frac{1}{410.0625} \] Solve for \(\mathscr{E}\): \[ \mathscr{E} \approx 2.06\,\text{V} \] The cell potential of the lead storage battery at 25°C when the concentrations of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), \(\mathrm{H}^{+}\), and \(\mathrm{HSO}_{4}^{-}\) are 4.5 M is approximately 2.06 V.

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Key Concepts

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

Cell Potential
Understanding the concept of cell potential is crucial when studying electrochemistry. Cell potential, symbolized by \( \mathscr{E} \), refers to the measure of the electrical potential difference between two electrodes in an electrochemical cell. In simple terms, it's the voltage that an electrochemical cell can produce. Cell potential is determined by the specific materials (electrodes) used in the cell and by the concentrations of the ionic substances within the cell's electrolyte.

When the concentrations are not at their standard conditions (1 M for solutions, 1 atm for gases, pure solids or liquids for other substances), the cell potential can be calculated using the Nernst equation. This equation adjusts the standard cell potential by accounting for temperature and the reaction quotients, which represent the ratios of product activities to reactant activities, quantifying the departure from standard conditions.
Lead Storage Battery
A lead storage battery, also known as the lead-acid battery, is a rechargeable battery that is widely used in vehicles and for energy storage in backup power supplies. The battery consists of lead (\( \mathrm{Pb} \) ) and lead dioxide (\( \mathrm{PbO}_{2} \) ) as electrodes, and sulfuric acid (\( \mathrm{H}_{2} \mathrm{SO}_{4} \) ) as the electrolyte.

The chemical reactions that take place during discharging and charging are crucial for the battery's operation. For the lead storage battery, the overall cell reaction describes how lead, lead dioxide, and sulfuric acid react to form lead sulfate and water, releasing electric energy in the process. In practice, knowing the cell potential of this battery under different concentrations of sulfuric acid is important for predicting its performance and state-of-charge.
Reaction Quotient
The reaction quotient (\( Q \)) is a key term in chemical thermodynamics representing the relative quantities of reactants and products in a reaction at a given instant. It is used in the Nernst equation to calculate the cell potential for non-standard conditions.

It is essentially the ratio of the concentrations (or partial pressures for gases) of the reaction products raised to the power of their stoichiometric coefficients to the concentrations of the reactants raised to the power of their stoichiometric coefficients. The reaction quotient is dimensionless and changes as the reaction proceeds until it reaches the equilibrium constant (\( K \)) value when the reaction is at equilibrium. For the cell reaction in the lead storage battery, since solids don't participate in the quotient, only the concentrations of the ions in the electrolyte, \( \mathrm{H}^{+} \) and \( \mathrm{HSO}_{4}^{-} \) in this case, are considered.
Standard Cell Potential
The standard cell potential (\( E^{\circ} \)) is the cell potential under standard conditions, which typically involve all solutes at 1 M concentration, all gases at 1 atm pressure, and pure solids or liquids for other phases, at 25°C. It is a measure of the driving force behind an electrochemical cell, also known as electromotive force (EMF).

The standard cell potential is determined by the intrinsic properties of the materials involved and is specific to the particular redox reaction. It can also be considered as a quantitative value that signifies how greatly a reaction wants to occur. Positive values indicate that a reaction is spontaneous under standard conditions. In the context of the lead storage battery, \( E^{\circ} \) represents the maximal potential the battery can achieve when fully charged at standard conditions.

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Most popular questions from this chapter

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{C}_{\text {cell. }}\) 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 \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} \mathrm{M}\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at \(298 \mathrm{K}\) is $$ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83} $$ a. Calculate \(\mathscr{E}^{\circ}\) 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.

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned} &3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\ & \mathscr{E}^{\circ}=0.957 \mathrm{V} \end{aligned}$$ $$\begin{aligned} &\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) &\mathscr{E}^{\circ}=0.775 \mathrm{V} \end{aligned}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If \(31.05 \mathrm{mL}\) of \(0.0600 M\) potassium dichromate solution is required to titrate \(30.0 \mathrm{g}\) blood plasma, determine the mass percent of alcohol in the blood.

It takes \(15 \mathrm{kWh}\) (kilowatt-hours) of electrical energy to produce \(1.0 \mathrm{kg}\) aluminum metal from aluminum oxide by the Hall-Heroult process. Compare this to the amount of energy necessary to melt \(1.0 \mathrm{kg}\) aluminum metal. Why is it economically feasible to recycle aluminum cans? [The enthalpy of fusion for aluminum metal is \(10.7 \mathrm{kJ} / \mathrm{mol} \text { ( } 1 \text { watt }=1 \mathrm{J} / \mathrm{s} \text { ). }]\)
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