Li-ion batteries used in automobiles typically use a LiMn_ \(_{2} \mathrm{O}_{4}\) cathode in place of the LiCoO \(_{2}\) cathode found in most Li- ion batteries. (a) Calculate the mass percent lithium in each electrode material. (b) Which material has a higher percentage of lithium? Does this help to explain why batteries made with \(\mathrm{LiMn}_{2} \mathrm{O}_{4}\) cathodes deliver less power on discharging? (c) In a battery that uses a LiCoO\(_{2}\) cathode, approximately 50\(\%\) of the lithium migrates from the cathode to the anode on charging. In a battery that uses a LiMn_ \(_{2} \mathrm{O}_{4}\) cathode, what fraction of the lithium in LiMn_ \(_{2} \mathrm{O}_{4}\) would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode?

Short Answer

Expert verified
The mass percent of Li in LiCoO\(_2\) is 7.09% and the mass percent of Li in LiMn\(_2\)O\(_4\) is 3.84%. LiCoO\(_2\) has a higher percentage of lithium. However, this does not directly explain why batteries made with LiMn\(_2\)O\(_4\) cathodes deliver less power during discharge, as other factors like the redox potential and lithium diffusion rate in the materials also play a role. To deliver the same amount of lithium as LiCoO\(_2\), 50% of which migrates from the cathode to the anode during charging, 92.2% of the lithium in LiMn\(_2\)O\(_4\) would need to migrate out of the cathode.

Step by step solution

01

Calculate the mass percent of lithium in LiCoO\(_2\)

First, we will calculate the molar mass of LiCoO\(_2\): Molar mass of Li = 6.94 g/mol Molar mass of Co = 58.93 g/mol Molar mass of O = 16 g/mol Molar mass of LiCoO\(_2\) = 6.94 + 58.93 + (2 × 16) = 97.87 g/mol Now, we will calculate the mass percent of lithium in LiCoO\(_2\): Mass percent of Li in LiCoO\(_2\) = (Molar mass of Li / Molar mass of LiCoO\(_2\)) × 100 Mass percent of Li in LiCoO\(_2\) = (6.94 / 97.87) × 100
02

Calculate the mass percent of lithium in LiMn\(_2\)O\(_4\)

Next, we will calculate the molar mass of LiMn\(_2\)O\(_4\): Molar mass of Mn = 54.94 g/mol Molar mass of LiMn\(_2\)O\(_4\) = 6.94 + (2 × 54.94) + (4 × 16) = 180.82 g/mol Now, we will calculate the mass percent of lithium in LiMn\(_2\)O\(_4\): Mass percent of Li in LiMn\(_2\)O\(_4\) = (Molar mass of Li / Molar mass of LiMn\(_2\)O\(_4\)) × 100 Mass percent of Li in LiMn\(_2\)O\(_4\) = (6.94 / 180.82) × 100
03

Determine which material has a higher percentage of lithium

Compare the mass percent of lithium in both materials calculated in Steps 1 and 2 to determine which material has a higher percentage of lithium.
04

Discuss the impact of lithium content on power output

Based on the mass percent of lithium in both materials, discuss if this factor can help explain why batteries made with LiMn\(_2\)O\(_4\) cathodes deliver less power during discharge.
05

Calculate the required fraction of lithium migration from LiMn\(_2\)O\(_4\) cathode

Given that in a LiCoO\(_2\) cathode, approximately 50% of the lithium migrates from the cathode to the anode on charging, calculate the fraction of lithium in LiMn\(_2\)O\(_4\) that would need to migrate out of the cathode to deliver the same amount of lithium to the graphite anode. Consider the amount of lithium in the LiCoO\(_2\) cathode, then calculate the amount of lithium that migrates during charging (50%). Now, we need to find the fraction of lithium in the LiMn\(_2\)O\(_4\) cathode that would deliver this same amount of lithium during charging. To do this, we will divide the mass of lithium delivered by the LiCoO\(_2\) cathode by the total lithium mass in the LiMn\(_2\)O\(_4\) cathode, and then turn this result into a percentage.

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Key Concepts

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

Mass Percent Lithium Calculation
Understanding the mass percent of lithium in battery cathode materials is crucial for estimating the electrochemical performance of lithium-ion (Li-ion) batteries. The mass percent represents the proportion of lithium in the compound by mass, which is significant in determining how much charge the battery can hold.

The calculation of mass percent lithium is straightforward, following the formula:
\[ \text{Mass percent of Li} = \left( \frac{\text{Molar mass of Li}}{\text{Molar mass of compound}} \right) \times 100 \% \]
In this calculation, the molar mass of lithium (6.94 g/mol) is divided by the total molar mass of the cathode material (for example, LiCoO2 or LiMn2O4). The result is multiplied by 100 to convert it into a percentage. The mass percent of lithium will influence the energy density of the battery, with higher values indicating a potential for greater energy storage.
Lithium Migration in Batteries
Lithium migration is a significant phenomenon in Li-ion batteries that occurs during the charging and discharging cycles. During charging, lithium ions migrate from the cathode through the electrolyte and into the anode, where they are stored in the lattice of the anode material, typically graphite. Upon discharging, the lithium ions move back to the cathode, releasing the stored energy in the process.

The efficiency and capacity of a battery are significantly affected by how well the lithium ions can travel between the electrodes. A higher fraction of lithium migration from the cathode indicates more ions are available to store energy in the anode, directly affecting the battery's capacity to deliver power.
Factors such as the crystal structure of the cathode materials, ionic conductivity, and the presence of diffusion paths impact the ease and speed of lithium migration, hence playing a crucial role in battery performance.
Comparing LiCoO2 and LiMn2O4 Cathodes
LiCoO2 and LiMn2O4 are two common cathode materials used in Li-ion batteries. Each has distinct electrochemical properties that affect their performance in batteries.

When comparing LiCoO2 to LiMn2O4 cathodes, several factors must be considered:
  • Structure: LiCoO2 has a layered structure, which facilitates the movement of lithium ions, while LiMn2O4 has a spinel structure, which is typically more stable but offers less lithium-ion mobility.
  • Thermal stability: LiCoO2 can be less thermally stable compared to LiMn2O4, which can be important for the safety of the batteries, especially in automotive applications.
  • Cost and availability: Cobalt is rarer and more expensive than manganese, making LiMn2O4 a more cost-effective option.
  • Lithium content: The mass percent of lithium can differ, with LiCoO2 generally having a higher percentage, which may translate to higher capacity and energy density.
These factors lead to a trade-off between energy density, safety, cost, and longevity, influencing the choice of cathode material for specific applications.
Electrochemistry in Battery Technology
Electrochemistry is the foundation of battery technology, governing how batteries store and release energy. It plays a crucial role in the performance, efficiency, and lifecycle of batteries.

In a Li-ion battery, electrochemistry involves redox reactions where lithium ions combine with electrons at the anode during discharging and then separate during charging. The movement of lithium ions between the cathode and anode is driven by the potential difference, while electrons flow through the external circuit to do work.
  • Redox potential: The potential at which lithium ions are inserted into or removed from the cathode material affects the voltage and energy content of the battery.
  • Cathode composition: The chemistry of the cathode influences the battery's voltage, capacity, and how it degrades over time.
  • Electrolyte characteristics: Ionic conductivity and stability of the electrolyte are critical for efficient lithium-ion transport and overall battery performance.
  • Anode material: It must accommodate lithium ions effectively and provide a low potential for energy-dense storage.
Enabling a deep understanding of electrochemistry will continue to drive innovations and improvements in battery technology.

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Most popular questions from this chapter

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{K},\) and calculate the equilibrium constant \(K\) at 298 \(\mathrm{K}\) (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q)\) . (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(l) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q) .\)

A voltaic cell that uses the reaction $$ \operatorname{PdCl}_{4}^{2-}(a q)+\mathrm{Cd}(s) \longrightarrow \mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q)+\mathrm{Cd}^{2+}(a q) $$ has a measured standard cell potential of \(+1.03 \mathrm{V}\) . (a) Write the two half-cell reactions. (b) By using data from Appendix E, determine \(E_{\text { red }}^{\circ}\) for the reaction involving Pd. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

(a) What is meant by the term oxidation? (b) On which side of an oxidation half-reaction do the electrons appear? (c) What is meant by the term oxidant? (d) What is meant by the term oxidizing agent?

A voltaic cell is constructed that uses the following half-cell reactions: $$ \begin{array}{c}{\mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s)} \\ {\mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \quad \longrightarrow 2 \mathrm{I}^{-}(a q)}\end{array} $$ The cell is operated at 298 \(\mathrm{K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 M\) and \(\left[\mathrm{I}^{-}\right]=0.035 \mathrm{M}\) (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If \(\left[\mathrm{Cu}^{+}\right]\) were equal to \(0.15 \mathrm{M},\) at what concentration of I \(\mathrm{I}^{-}\) would the cell have zero potential?

Indicate whether each statement is true or false: (a) The anode is the electrode at which oxidation takes place. (b) A voltaic cell always has a positive emf. (c) A salt bridge or permeable barrier is necessary to allow a voltaic cell to operate.

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