/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 142 The overall reaction in the lead... [FREE SOLUTION] | 91影视

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Chapter 18: Problem 142

The overall reaction in the lead storage battery is \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow\) $$\quad\quad\quad\quad\quad\quad\quad 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ a. For the cell reaction \(\Delta H^{\circ}=-315.9 \mathrm{kJ}\) and \(\Delta S^{\circ}=\) 263.5 \(\mathrm{J} / \mathrm{K}\) . Calculate \(\mathscr{E}^{\circ} \mathrm{at}-20 .^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. b. Calculate \(\mathscr{E}\) at \(-20 .^{\circ} \mathrm{C}\) when \(\left[\mathrm{HSO}_{4}^{-}\right]=\left[\mathrm{H}^{+}\right]=4.5 \mathrm{M}\) . c. Consider your answer to Exercise 69. Why does it seem that batteries fail more often on cold days than on warm days?

Short Answer

Expert verified
The standard cell potential E掳 at -20掳C is 0.0905 V, and the cell potential E at -20掳C with given ion concentrations is -0.00582 V. Batteries are more likely to fail on cold days because the cell potential, which represents the driving force of the chemical reaction, decreases significantly at lower temperatures. This decrease adversely affects the electrical output, making it difficult for the battery to provide enough power, leading to weaker battery performance and a higher likelihood of failure on colder days compared to warmer ones.

Step by step solution

01

a. Calculating the standard cell potential E掳 at -20掳C

We will start by using the relation between the Gibbs free energy change 鈭咷掳, the standard enthalpy change 鈭咹掳, and the standard entropy change 鈭哠掳: 鈭咷掳 = 鈭咹掳 - T鈭哠掳 Next, we need to relate 鈭咷掳 to the standard electromotive force (EMF) E掳 of the cell through the equation: 鈭咷掳 = -nFE掳 Where n is the number of moles of electrons transferred in the overall reaction, and F is Faraday's constant (96485 C/mol). Step 1: Determine the number of moles of electrons transferred (n) The balanced overall reaction is: Pb(s) + PbO鈧�(s) + 2 H鈦�(aq) + 2 HSO鈧勨伝(aq) 鈫� 2 PbSO鈧�(s) + 2 H鈧侽(l) It can be seen that the redox reaction involves the transfer of 2 moles of electrons. Step 2: Calculate the standard Gibbs free energy change 鈭咷掳 Using the relationship 鈭咷掳 = 鈭咹掳 - T鈭哠掳, we need to first convert -20掳C to Kelvin: T = -20 + 273.15 = 253.15 K Now, plug in the given values of 鈭咹掳 (-315.9 kJ) and 鈭哠掳 (263.5 J/K) and calculate 鈭咷掳: 鈭咷掳 = -(-315900 J) - (253.15 K)(263.5 J/K) = 17495.475 J Step 3: Calculate the standard cell potential E掳 Finally, we can calculate the standard cell potential E掳 using the equation: 鈭咷掳 = -nFE掳 E掳 = -鈭咷掳 / (nF) = -17495.475 J / (2 mol * 96485 C/mol) = 0.0905 V So, the standard cell potential E掳 at -20掳C is 0.0905 V.
02

b. Calculating the cell potential E at -20掳C with given ion concentrations

Now, we have to find the cell potential E at -20掳C with the given concentrations of HSO鈧勨伝 and H鈦� ions. To do this, we'll use the Nernst equation: E = E掳 - (RT/nF) ln(Q) Where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin, n is the number of moles of electrons transferred, F is Faraday's constant, and Q is the reaction quotient. Step 1: Calculate the reaction quotient Q For the balanced overall reaction, the reaction quotient Q can be written as: Q = ([PbSO鈧刔虏[H鈧侽]虏) / ([Pb][PbO鈧俔[H鈦篯虏[HSO鈧勨伝]虏) Since the concentrations of solid and liquid species remain constant, we'll only include the concentrations of H鈦� and HSO鈧勨伝 ions: Q = ([H鈦篯虏[HSO鈧勨伝]虏) = (4.5 M)虏(4.5 M)虏 = 410.0625 Step 2: Calculate the cell potential E Plugging the known values into the Nernst equation: E = 0.0905 V - (8.314 J/mol K)(253.15 K) / (2 mol * 96485 C/mol) ln(410.0625) = -0.00582 V So, the cell potential E at -20掳C with given ion concentrations is -0.00582 V.
03

c. Explaining the influence of cold temperatures on battery performance

Based on the results obtained in this exercise, it can be observed that the cell potential E decreases significantly at lower temperatures (-20掳C). The cell potential E, which represents the driving force of the chemical reaction, directly affects the electrical output of the battery. The decrease in cell potential at low temperatures results in less electrical output, leading to a weaker battery performance. Batteries are more likely to fail on cold days because the decreased cell potential adversely affects the electrical output. Lower electrical output makes it difficult for the battery to provide enough power to start a vehicle or run electrical systems adequately. So, batteries tend to fail more often on colder days than on warmer ones.

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Key Concepts

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

Standard Cell Potential
The standard cell potential, denoted as \(E^\circ\), is a measure of the voltage produced by a cell under standard conditions. These conditions typically include solutions at a concentration of 1 M, a pressure of 1 atm, and a temperature of 25掳C. In the context of a lead storage battery, the standard cell potential is an important parameter because it reflects the maximum potential difference between electrodes when the cell is not delivering current.
Calculating \(E^\circ\) involves finding the difference in standard reduction potentials of the cathode and anode. This helps us understand the efficiency and energy output of the battery. When the conditions deviate, such as changes in concentration or temperature, the actual cell potential can differ from \(E^\circ\). Understanding this helps us anticipate how the battery will perform in various environmental conditions.
Gibbs Free Energy
Gibbs free energy (\(\Delta G^\circ\)) is a thermodynamic quantity that provides information about the spontaneity of a chemical process. If \(\Delta G^\circ\) is negative, the reaction occurs spontaneously. For the lead storage battery, the relationship between \(\Delta G^\circ\) and the standard cell potential is given by \(\Delta G^\circ = -nFE^\circ\), where \(n\) is the number of moles of electrons transferred, and \(F\) is Faraday's constant.
This equation helps us determine how much energy can be obtained from the reaction. The value of \(\Delta G^\circ\) influences the amount of work the battery can perform as it discharges. Understanding Gibbs free energy is crucial for predicting the life and efficiency of the battery.
Nernst Equation
The Nernst equation is a fundamental tool in electrochemistry that lets us calculate the actual cell potential under non-standard conditions. It adjusts the standard cell potential, \(E^\circ\), based on the temperature and the concentration of reactants and products. The equation is expressed as: \[ E = E^\circ - \frac{RT}{nF} \ln Q \] Where \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred in the cell reaction, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.
The Nernst equation is invaluable in real-world applications as it helps predict how variations in concentration and temperature affect the battery output. For example, in a lead storage battery, deviating from standard concentrations and temperatures will alter \(E\), highlighting the practical limits of battery performance.
Battery Performance
Battery performance is primarily determined by how effectively it can maintain its voltage while delivering current. It depends on the internal reaction kinetics, which are largely influenced by temperature and concentration changes. For lead-acid batteries, performance is optimal at moderate temperatures and standard conditions.
Factors that affect battery performance include:
  • Internal Resistance: Resistance slows down electron flow, reducing current and potential.
  • Concentration of Electrolytes: Optimal concentration is crucial for maintaining voltage and efficiency.
  • Temperature: Affects reaction rates, internal resistance, and electrolyte viscosity.
Understanding battery performance is key to predicting when a battery will need replacing and for optimizing its use across different temperature ranges and conditions.
Temperature Effect on Batteries
Temperature has a critical impact on battery operation. Low temperatures can slow down the chemical reactions within a battery, reducing its efficiency and capacity. In the case of the lead storage battery, colder environments result in decreased cell potential and higher internal resistance.
These effects occur because:
  • Chemical Reaction Rates: Temperature drop slows reaction kinetics.
  • Electrolyte Viscosity: Lower temperatures increase the viscosity of electrolytes, hindering ion movement.
  • Cell Potential: Lower temperatures reduce the cell potential as predicted by the Nernst equation.
The practical implication is that batteries might deliver insufficient power at low temperatures, making them less effective. Understanding these effects helps in the design and usage of batteries in varied environmental conditions.

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Most popular questions from this chapter

Define oxidation and reduction in terms of both change in oxidation number and electron loss or gain.

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M(\text { right side })\) and \(1.0 \times 10^{-4} M(\text { left side })\) a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & \\\ & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M} .\)

Consider a galvanic cell based on the following half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\ \hline {\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}} & {1.50} \\\ {\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}} & {-2.37}\end{array}$$ a. What is the standard potential for this cell? b. A nonstandard cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Mg}^{2+}\right]=\) \(1.00 \times 10^{-5} \mathrm{M}\) . The cell potential is observed to be 4.01 \(\mathrm{V}\) . Calculate \(\left[\mathrm{Au}^{3}+\right]\) in this cell.

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? a. molten \(\mathrm{KF} \quad\) b. molten \(\mathrm{CuCl}_{2} \quad\) c. molten \(\mathrm{MgI}_{2}\)

A galvanic cell is based on the following half-reactions at \(25^{\circ} \mathrm{C} :\) $$\begin{array}{c}{\mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}} \\\ {\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}}\end{array}$$ Predict whether \(\mathscr{E}_{\text{cell}}\) is larger or smaller than \(\mathscr{E}^{\circ}_{\text{cell}}\) for the following cases. a. [Ag1] 5 1.0 a. \(\left[\mathrm{Ag}^{+}\right]=1.0 M,\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=2.0 M,\left[\mathrm{H}^{+}\right]=2.0 \mathrm{M}\) b. \(\left[\mathrm{Ag}^{+}\right]=2.0 \mathrm{M},\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]=1.0 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-7} \mathrm{M}\)
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