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In a Li-ion battery the composition of the cathode is LiCoO \(_{2}\) when completely discharged. On charging, approximately 50\(\%\) of the Lit 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 LiCo \(_{2}\) 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

Expert verified
The composition of the cathode when the battery is fully charged is Li0.5CoO2. The fully charged Li-ion battery with a LiCoO2 cathode of 10 g can deliver approximately 4925 coulombs of electricity when it is fully discharged.

Step by step solution

01

Determine the composition of the cathode when the battery is fully charged

Since approximately 50% of the Li+ ions are extracted from the cathode upon charging, the composition of the cathode must change to represent that. Let's denote the charged cathode's composition as LiXCoO2, where X signifies the proportion of Li ions remaining in the cathode after charging. Given that 50% of Li ions remain: X = (1 - 0.5) = 0.5 Thus, the composition of the cathode when the battery is fully charged is Li0.5CoO2.
02

Calculate the number of moles of LiCoO2 in 10g of the cathode

To find the number of moles of LiCoO2 in the cathode (10 g), we first need to calculate the molar mass of LiCoO2. For that, we can use the periodic table to find the masses of individual elements: Molar mass of Li = 6.939 g/mol Molar mass of Co = 58.933 g/mol Molar mass of O = 15.999 g/mol Molar mass of LiCoO2 = 6.939 + 58.933 + 2 * 15.999 = 97.87 g/mol Now, we can calculate the number of moles (n) of LiCoO2 using the formula: n = mass / molar mass n = 10 g / 97.87 g/mol = 0.1021 mol
03

Calculate the number of coulombs of electricity using Faraday's Law

Faraday's Law states that the electric charge (Q) required for the redox reaction is related to the number of moles of electrons (n) being exchanged by the Faraday constant (F) as follows: Q = nfF Where: - n = number of moles of reacting species (in this case, Li ions) - f = number of electrons involved in the redox reaction per ion (in this case, 1, because each Li+ ion carries one positive charge) - F = Faraday constant (approximately 96485 C/mol) Since 50% of the Li ions are extracted from the cathode during charging, there are 0.1021 mol * 0.5 = 0.05105 mol of Li ions extracted. Now, we can calculate the number of coulombs of electricity (Q) when discharging: Q = 0.05105 mol * 1 * 96485 C/mol ≈ 4925 C The fully charged Li-ion battery with a LiCoO2 cathode of 10 g can deliver approximately 4925 coulombs of electricity when it is fully discharged.

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Key Concepts

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

Cathode Composition in Li-ion Batteries
In the context of Li-ion batteries, understanding the cathode composition is crucial as it determines many of the battery's key characteristics, such as capacity, voltage, and safety. The cathode is the positive electrode in the battery and is typically made from a lithium metal oxide. LiCoO2 is a common cathode material due to its stable chemistry and reasonable capacity.

During charging, Li ions are extracted from the cathode and move towards the anode. When the battery is fully charged, the cathode composition changes as it loses some of its lithium content. In our example, the cathode shifts from being fully lithiated, LiCoO2, to a partially delithiated state, Li0.5CoO2, as 50% of the Li ions have been extracted and moved to the anode. This extraction modifies the charge and capacity of the battery, and it's this dynamism that allows the battery to store and release energy.
Faraday's Law and Electric Charge in Batteries
Faraday's Law is a fundamental principle that relates the amount of electric charge needed to liberate a substance in a redox reaction with the substance's quantity, expressed in moles, and the number of electrons involved in the reaction. In the context of a Li-ion battery, this principle allows us to calculate the amount of electrical energy involved in charging and discharging processes.

In the example given, we use Faraday's Law to determine how many coulombs, the unit of electric charge, can be delivered when the Li-ion battery is discharged from its fully charged state. Since charging the battery involves extracting lithium ions from the cathode, the discharge process involves reintegrating them, leading to an electric charge transfer that can be quantified by Faraday's constant and the number of moles of lithium involved.
Molar Mass Calculation for Battery Materials
Calculating the molar mass of a substance is essential in many chemistry applications and is particularly relevant when analyzing battery materials. In the case of a LiCoO2 cathode, knowing its molar mass allows us to convert between mass and moles, granting insight into the amount of material available for electrochemical reactions.

The process involves adding the molar masses of each element in the compound based on their proportional representation. Here, the atomic weights of lithium (Li), cobalt (Co), and oxygen (O) from the periodic table are summed, considering the number of atoms of each in the compound. For instance, LiCoO2 has one atom of Li, one of Co, and two of O. The calculated molar mass informs us about the quantity of LiCoO2 in a given mass, serving as a stepping stone to further calculations related to the battery's electrical properties and energy capacity.
Understanding Electric Charge in Redox Reactions
Electric charge plays a vital role in redox reactions, which are at the heart of how batteries operate. In a redox reaction within a battery, one material loses electrons (oxidation) while another gains electrons (reduction).

The ability to quantify electric charge in these reactions enables the prediction and measurement of a battery's energy capacity. By using Faraday's Law, we can calculate the total charge that flows during the battery's charge and discharge cycles based on the moles of electrons exchanged. This is fundamental to understanding battery performance and is vital for designing batteries with higher capacities and longer life cycles. Each lithium ion contributes a known charge as it moves between electrodes—this is what generates the electric current we harness from Li-ion batteries.

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Most popular questions from this chapter

In a galvanic cell the cathode is an \(\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}(s)\) half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing 0.10\(M\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and 0.050 \(\mathrm{M}\) sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\right) .\) The measured cell voltage is 1.030 \(\mathrm{V}\) . What is the \(\mathrm{p} K_{\mathrm{a}}\) of benzoic acid?

In some applications nickel-cadmium batteries have been replaced by nickel- zinc batteries. The overall cell reaction for this relatively new battery is: $$ \begin{aligned} 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{Zn}(s) & \\ & \longrightarrow 2 \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Zn}(\mathrm{OH})_{2}(s) \end{aligned} $$ (a)What is the cathode half-reaction? (b)What is the anode half-reaction? (c) A single nickel-cadmium cell has a voltage of 1.30 \(\mathrm{V}\) . Based on the difference in the standard reduction potentials of \(\mathrm{Cd}^{2+}\) and \(\mathrm{Zn}^{2+},\) what voltage would you estimate a nickel-zinc battery will produce? (d) Would you expect the specific energy density of a nickel-zinc battery to be higher or lower than that of a nickel-cadmium battery?

A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode half-cell consists of a silver strip placed in a solution of \(\mathrm{AgNO}_{3},\) and the other has an iron strip placed in a solution of \(\mathrm{FeCl}_{2}\) . The overall cell reaction is $$ \mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\) . (a) Why is an aqueous solution of MgCl_ not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are 96\(\%\) efficient in producing the desired products in electrolysis, what mass of Mg is formed by passing a current of \(97,000\) A for a period of 24 \(\mathrm{h} ?\)

Is each of the following substances likely to serve as an oxidant or a reductant: (a) \(\mathrm{Ce}^{3+}(a q),\)(b) \(\mathrm{Ca}(s),\) (c) \(\mathrm{ClO}_{3}(a q)\) (d) \(\mathrm{N}_{2} \mathrm{O}_{5}(g) ?\)

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