Question

In a Li-ion battery the composition of the cathode is \(\mathrm{LiCoO}_{2}\) when completely discharged. On charging, approximately \(50 \%\) of the \(\mathrm{Li}^{+}\) ions can be extracted from the cathode and transported to the graphite anode where they intercalate between the layers. (a) What is the composition of the cathode when the battery is fully charged? (b) If the \(\mathrm{LiCoO}_{2}\) cathode has a mass of \(10 \mathrm{~g}\) (when fully discharged), how many coulombs of electricity can be delivered on completely discharging a fully charged battery?

Short answer

The composition of the cathode when the battery is fully charged is \(\mathrm{Li_{0.5}CoO_{2}}\). A fully charged battery with a 10 g cathode can deliver approximately 4945 coulombs of electricity upon complete discharge.

Step-by-step solution

(a) Finding the composition of the cathode when fully charged

To find the composition of the cathode when the battery is fully charged, we need to find out the formula of the cathode material after 50% of the Li+ ions have been extracted. Since the cathode is represented by the formula \(\mathrm{LiCoO}_{2}\), and upon charging, approximately 50% of the Li+ ions are extracted. Thus, the number of Li+ ions in the cathode composition when fully charged is 50% less than the initial number of ions, or 0.5 moles of Li+ ions per mole of \(\mathrm{LiCoO}_{2}\). So, the composition of the cathode when the battery is fully charged is \(\mathrm{Li_{0.5}CoO_{2}}\).

(b) Calculating the charge produced by 50% of Li+ ions

First, determine the moles of \(\mathrm{LiCoO}_{2}\) in the 10 g mass of the cathode. To do this, we need to find the molar mass of \(\mathrm{LiCoO}_{2}\). Molar mass of \(\mathrm{LiCoO}_{2} = \mathrm{Li} + \mathrm{Co} + 2\mathrm{O}\) Molar mass of \(\mathrm{LiCoO}_{2} = 6.94 \mathrm{~g/mol} + 58.93 \mathrm{~g/mol} + 2 \times 16.00 \mathrm{~g/mol} = 97.87 \mathrm{~g/mol}\) Next, find the moles of \(\mathrm{LiCoO}_{2}\): \(moles\ of\ \mathrm{LiCoO}_{2} = \dfrac{10\ \mathrm{g}}{97.87\ \mathrm{g/mol}} = 0.1022\ \mathrm{mol}\) Since each \(\mathrm{LiCoO}_{2}\) contains one Li+ ion, when the battery is fully charged and 50% of Li+ ions are extracted, there will be 0.5 times the original number of Li+ ions remaining in the cathode. Therefore, the number of Li+ ions extracted is equal to: \(0.5 \times 0.1022\ \mathrm{mol} = 0.0511\ \mathrm{mol}\) The charge produced by each Li+ ion is equal to its charge (1+), multiplied by the elementary charge (e): Charge produced by one Li+ ion = \(1 \times 1.602 \times 10^{-19}\ \mathrm{C}\) Total charge (coulombs) produced by 0.0511 mol of Li+ ions is: Total charge = \((0.0511\ \mathrm{mol})(6.022 \times 10^{23}\ \mathrm{ions/mol})(1 \times 1.602 \times 10^{-19}\ \mathrm{C}) = 4945\ \mathrm{C}\) So, a fully charged battery with a 10 g cathode can deliver approximately 4945 coulombs of electricity upon complete discharge.

Key concepts

Cathode Composition

In a Li-ion battery, the cathode undergoes a transformation when the battery is charged. Originally, the cathode is composed of \( \mathrm{LiCoO}_{2} \) when fully discharged. Upon charging, half of the lithium ions (\( \mathrm{Li}^{+} \)) can be removed, altering the material's chemical structure. This means that when the battery is fully charged, the cathode's formula changes.
Here's what's happening:

• The initial formula is \( \mathrm{LiCoO}_{2} \), representing lithium, cobalt, and oxygen.
• As 50% of the \( \mathrm{Li}^{+} \) ions are extracted, the lithium amount is reduced to half.
• This results in a new cathode composition of \( \mathrm{Li_{0.5}CoO_{2}} \), reflecting the decreased lithium content.

This change is crucial for the battery's function, as it facilitates the flow of lithium ions to the anode. When the lithium returns during discharge, the battery releases stored energy to power devices.

Charge Calculation

Electric charge in a battery is crucial to understanding how long and powerful its energy output can be. When 50% of \( \mathrm{Li}^{+} \) ions are extracted from the cathode in a lithium-ion battery, it releases a measurable amount of charge. This is dissected as follows:
First, it's necessary to find the number of moles of \( \mathrm{Li}^{+} \) ions involved:

• The molar mass of \( \mathrm{LiCoO}_{2} \) is determined to be 97.87 g/mol.
• In 10 g of the cathode, calculate the moles: \( 0.1022 \) moles.
• With 50% of lithium ions moving, 0.0511 moles of \( \mathrm{Li}^{+} \) are actually participating in the charge transfer.

Each lithium ion carries a charge quantified as \( 1 \times 1.602 \times 10^{-19} \) coulombs.
When considering all extracted lithium ions:

• Total charge = \(0.0511 \) mol \( \times 6.022 \times 10^{23} \text{ ions/mol} \times 1.602 \times 10^{-19} \text{ C/ion} = 4945 \text{ C}\).

This results in approximately 4945 coulombs, which tells us how much electrical energy can be delivered when the battery fully discharges.

LiCoO2 Molar Mass

Understanding the molar mass of \( \mathrm{LiCoO}_{2} \) is key to solving energy-related problems in Li-ion batteries. It serves as a foundation for many calculations:
The molar mass is the sum of all atomic masses in the formula:

• Lithium (Li) contributes 6.94 g/mol.
• Cobalt (Co) adds 58.93 g/mol.
• Oxygen (each O) offers 16.00 g/mol.
• As \( \mathrm{LiCoO}_{2} \) includes two oxygen atoms, their combined contribution is 32.00 g/mol.

When you sum these atomic masses:

• \( 6.94 + 58.93 + 32.00 = 97.87 \text{ g/mol} \)

This value of 97.87 g/mol is essential to convert grams to moles, a vital step in further calculations around electric charge and energy capacity. With the correct molar mass, habitual calculations become smoother when dealing with the electrochemistry of lithium-ion batteries.