Problem 79 - Chemistry: The Central Science

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?

Step-by-Step Solution

Answer

The composition of the fully charged cathode is \(Li_{0.5}CoO_2\), and the amount of electricity that can be delivered on completely discharging a fully charged battery with a 10 g LiCoO2 cathode is 4921.635 coulombs.

Step 1: (a) Finding the composition of the fully charged cathode

Initially, the composition of the cathode is LiCoO2. When 50% of the Li+ ions are extracted during charging, the number of lithium ions will be reduced to half. This means in the fully charged cathode the composition will be: \(Li_{0.5}CoO_2\) So, the composition of the fully charged cathode is \(Li_{0.5}CoO_2\).

Step 2: (b) Calculation of coulombs of electricity delivered on discharging

To find the amount of electricity delivered, we need to consider the extracted Li+ ions, which account for 50% of the initial amount. Also, the total mass of the cathode when fully discharged is 10 g. First, we need to find the number of moles of \(LiCoO_2\). We can do this by using the molar mass of Li, Co, and O: Molar mass of \(LiCoO_2 = Li + Co + 2 * O\) Molar mass of \(LiCoO_2 = 6.939 + 58.933 + 2 * 16 = 97.865 \frac{g}{mol}\) Now, we can find the moles of \(LiCoO_2\) in the 10 g cathode: Moles of \(LiCoO_2 = \frac{10}{97.865} = 0.102 \, mol\) Since 50% of the Li+ ions are extracted, the moles of extracted Li+ ions will be: Moles of Li+ extracted = 0.5 * moles of \(LiCoO_2\) Moles of Li+ extracted = 0.5 * 0.102 = 0.051 \, mol Now, we need to find the amount of electricity (in coulombs) that corresponds to this number of extracted Li+ ions. Considering the basic relationship of charge (Q) with moles and Faraday's constant (F): \(Q = n * F\) Where Q is the charge in coulombs, n is the number of moles of an ion, and F is Faraday's constant equal to about 96485 C/mol. Calculating the charge for the extracted Li+ ions: \(Q = 0.051 * 96485 = 4921.635 \, C\) Therefore, 4921.635 coulombs of electricity can be delivered on completely discharging a fully charged battery with a 10 g LiCoO2 cathode.

Key Concepts

Cathode Composition

In Li-ion batteries, the cathode composition plays a crucial role in how the battery functions. When a Li-ion battery is fully discharged, the cathode is composed of \[ \text{LiCoO}_2 \]. This means that for every cobalt and two oxygen atoms, there is one lithium ion in its structure. During the charging process, around 50% of the \( \text{Li}^+ \) ions are removed from the cathode and transported to the anode. Once fully charged, the cathode composition changes to \[ \text{Li}_{0.5}\text{CoO}_2 \]. Here, the reduction in lithium means that only half remains in the structure. This change significantly influences how the battery stores and releases electrical energy. Understanding this transformation is vital for improving battery performance and developing new technologies.

Coulombs of Electricity

The concept of coulombs of electricity relates directly to how much charge a battery can deliver. In this context, a fully charged battery with a 10 g \( \text{LiCoO}_2 \) cathode can contribute to the production of electricity upon complete discharge. To calculate the electricity delivered, you first need to determine how much \( \text{Li}^+ \) ion is extracted during charging, which is about 50%. This partial removal of ions corresponds to a specific amount of charge calculated in coulombs. A crucial part of this calculation involves understanding the number of ions being moved and their capacity to carry charge.

• 1 mole of ions carries a charge equal to Faraday's constant, approximately 96485 coulombs per mole.
• The extracted \( \text{Li}^+ \) ions determine the total charge the battery can release.

Knowing how to quantify this charge helps engineers and scientists optimize battery capacity and usage.

Moles Calculation

Calculating moles is a fundamental step in finding out the total charge a battery can deliver. This process begins with the molar mass of the compound in question, \( \text{LiCoO}_2 \) in our example. Here's how you determine it:

• Find the molar mass of each element (Li, Co, O) and add them to obtain the molar mass of the whole compound.

The formula becomes:\[ \text{Molar Mass of } \text{LiCoO}_2 = 6.939 + 58.933 + 2 \times 16 = 97.865 \, \text{g/mol} \]With this information, calculating the number of moles in a given mass is straightforward: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{10}{97.865} \approx 0.102 \, \text{mol} \]Half of these moles correspond to the extracted \( \text{Li}^+ \) ions, leading you to:\[ \text{Moles of } \text{Li}^+ \text{ extracted} = 0.5 \times 0.102 = 0.051 \, \text{mol} \]Accurate molar calculations allow for precise determinations of the battery's electrical capabilities, as each mole relates to a quantifiable amount of charge (coulombs) based on known constants, like Faraday's constant.