Problem 48
Question
A cell in a lead-acid battery delivers exactly \(2.00 \mathrm{V}\) of cell potential in accordance with the following cell reaction: \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) What is the value of \(\Delta G_{\text {cell }} ?\)
Step-by-Step Solution
Verified Answer
Question: Calculate the Gibbs free energy change for the following cell reaction: $\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$. The cell potential is given as 2.00 V.
Answer: The Gibbs free energy change for the given cell reaction is approximately -385,940 J/mol.
1Step 1: Identify the species being oxidized and reduced, and the electrons transferred
Analyze the equation given:
$\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)
\rightarrow 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$
Pb is oxidized as it becomes part of PbSO_4.
PbO_2 is reduced as it becomes part of PbSO_4.
The balanced half-reactions are:
\(\begin{aligned} \mathrm{Reduction} &: \mathrm{~PbO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \\ \mathrm{Oxidation} &: \mathrm{~Pb}(s)+ \mathrm{SO}^{2-}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)+2 \mathrm{e}^{-} \end{aligned}\)
The number of moles of electrons transferred in this reaction is \(2\) moles.
2Step 2: Calculate Gibbs free energy change
Now, use the formula to calculate the Gibbs free energy change associated with the cell reaction:
\(\Delta G_{\text{cell}} = -nFE_{\text{cell}}\)
where:
\(n = 2\) (number of moles of electrons)
\(F = 96,485 \,\text{C/mol}\) (Faraday constant)
\(E_{\text{cell}} = 2.00 \,\text{V}\) (given cell potential)
Calculate \(\Delta G_{\text{cell}}\):
\(\Delta G_{\text{cell}} = -(2)(96,485 \,\text{C/mol})(2.00 \,\text{V})\)
\(\Delta G_{\text{cell}} = -385,940 \,\text{J/mol}\)
The value of \(\Delta G_{\text {cell }}\) is approximately \(-385,940 \,\text{J/mol}\).
Key Concepts
ElectrochemistryLead-Acid BatteryRedox Reactions
Electrochemistry
Electrochemistry is a branch of chemistry that deals with the relationship between electrical energy and chemical changes. It focuses on the processes that occur in solutions and at interfaces between different phases, which involve electric charges. At the heart of electrochemistry are redox reactions, where one substance is oxidized (loses electrons) and another is reduced (gains electrons).
Understanding electrochemistry is vital for comprehending how batteries work, including the lead-acid batteries used in cars. The electron transfer that occurs during a redox reaction can be harnessed to create an electric current. This is the principle behind galvanic cells or voltaic cells, which are the foundation of batteries. When connected in a circuit, these cells can produce a steady flow of electrons from the negative to the positive terminal, powering electrical devices.
In the context of the given exercise, a lead-acid battery cell generates 2.00 V based on a specific redox reaction. Calculating the Gibbs free energy change for this reaction involves applying principles from electrochemistry, which includes understanding the flow of electrons and the potential difference (voltage) across the cell.
Understanding electrochemistry is vital for comprehending how batteries work, including the lead-acid batteries used in cars. The electron transfer that occurs during a redox reaction can be harnessed to create an electric current. This is the principle behind galvanic cells or voltaic cells, which are the foundation of batteries. When connected in a circuit, these cells can produce a steady flow of electrons from the negative to the positive terminal, powering electrical devices.
In the context of the given exercise, a lead-acid battery cell generates 2.00 V based on a specific redox reaction. Calculating the Gibbs free energy change for this reaction involves applying principles from electrochemistry, which includes understanding the flow of electrons and the potential difference (voltage) across the cell.
Lead-Acid Battery
A lead-acid battery is a type of rechargeable battery that has been used for over a century, typically in vehicles and backup power systems. It is composed of lead dioxide (PbO2) as the positive plate, sponge lead (Pb) as the negative plate, and a sulfuric acid (H2SO4) solution as the electrolyte.
The operation of a lead-acid battery is based on redox reactions occurring at both plates. When the battery is discharging, the lead and lead dioxide are converted into lead sulfate (PbSO4), while sulfuric acid is consumed and water (H2O) is produced. Conversely, when the battery is recharged, the chemical actions are reversed.
The exercise presented shows the discharging reaction of a lead-acid battery, where both lead and lead dioxide transform into lead sulfate. The understanding of these redox reactions is crucial to determine the cell's potential and the energy changes associated with its operation, symbolized as the Gibbs free energy change.
The operation of a lead-acid battery is based on redox reactions occurring at both plates. When the battery is discharging, the lead and lead dioxide are converted into lead sulfate (PbSO4), while sulfuric acid is consumed and water (H2O) is produced. Conversely, when the battery is recharged, the chemical actions are reversed.
The exercise presented shows the discharging reaction of a lead-acid battery, where both lead and lead dioxide transform into lead sulfate. The understanding of these redox reactions is crucial to determine the cell's potential and the energy changes associated with its operation, symbolized as the Gibbs free energy change.
Redox Reactions
Redox reactions are chemical reactions involving the transfer of electrons between two species. The term 'redox' is a contraction of 'reduction-oxidation'. In such reactions, one reactant is oxidized, losing electrons, while the other is reduced, gaining electrons. These reactions are pivotal to numerous processes in chemistry and biology, including metabolism and photosynthesis, in addition to being fundamental to electrochemistry and battery technology.
In the illustration provided, the exercise explores the cell reaction in a lead-acid battery, which is a complex redox reaction. By identifying the species that get oxidized and reduced, we can understand the electron flow that drives the current. In this particular case, metallic lead is oxidized to lead sulfate, while lead dioxide is reduced to lead sulfate.
Oxidation and reduction reactions always occur together, as the electrons lost by the oxidized species are gained by the reduced one. Understanding these concepts not only allows for the calculation of the Gibbs free energy change in the exercise but also provides insight into how energy is stored and released in batteries.
In the illustration provided, the exercise explores the cell reaction in a lead-acid battery, which is a complex redox reaction. By identifying the species that get oxidized and reduced, we can understand the electron flow that drives the current. In this particular case, metallic lead is oxidized to lead sulfate, while lead dioxide is reduced to lead sulfate.
Oxidation and reduction reactions always occur together, as the electrons lost by the oxidized species are gained by the reduced one. Understanding these concepts not only allows for the calculation of the Gibbs free energy change in the exercise but also provides insight into how energy is stored and released in batteries.
Other exercises in this chapter
Problem 43
The value of \(E_{\text {cell for the reaction below is } 0.500 \mathrm{V} . \text { What is }}\) the value of \(\Delta G_{\text {cell }}^{\text {? }}\) $$\math
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What is the value of \(\Delta G_{\text {cell }}^{\circ}\) for an electrochemical cell based on a cell reaction described by the following net ionic equation? $$
View solution Problem 49
What is the function of platinum in the standard hydrogen electrode?
View solution Problem 50
Platinum is very expensive, so why is it used in standard hydrogen standard electrodes where the half-reaction is \(2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \righ
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