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The simplest way to write the reaction for discharge in a lithium-ion battery is Li(on carbon)(s) \(+\mathrm{CoO}_{2}(\mathrm{s}) \rightarrow 6 \mathrm{C}(\mathrm{s})+\mathrm{LiCoO}_{2}(\mathrm{s})\) (a) What are the oxidation numbers for cobalt in the two substances in the battery? (b) In such a battery, what reaction occurs at the cathode? At the anode? (c) An electrolyte is needed for ion conduction within the battery. From what you know about lithium chemistry, can the electrolyte in the battery be dissolved in water?

Short Answer

Expert verified
(a) Co in \(\mathrm{CoO}_2\) is +4, and in \(\mathrm{LiCoO}_2\) is +3. (b) Cathode: CoO\(_2\) reduction; Anode: Li oxidation. (c) Electrolyte cannot be dissolved in water due to reactivity.

Step by step solution

01

Determine Oxidation Numbers

To find the oxidation numbers, we start with the compound \(\mathrm{CoO}_{2}\). Oxygen typically has an oxidation number of \(-2\). Since there are two oxygen atoms, the total oxidation charge from oxygen is \(-4\). Therefore, to ensure the compound is neutral, cobalt must have an oxidation number of \(+4\). For \(\mathrm{LiCoO}_{2}\), lithium has \(+1\) and each oxygen is \(-2\). This totals \(-4\) from oxygen. Thus, cobalt here must be \(+3\) to balance the compound.
02

Identify Cathode Reaction

In a discharge reaction, the cathode is where the reduction takes place. We see cobalt moving from an oxidation state of \(+4\) in \(\mathrm{CoO}_{2}\) to \(+3\) in \(\mathrm{LiCoO}_{2}\). This indicates a gain of electrons, meaning reduction is occurring, making \(\mathrm{CoO}_{2}\) the material reduced at the cathode.
03

Identify Anode Reaction

The anode hosts the oxidation process. From the reaction, \(\mathrm{Li}\) is oxidized on carbon as it originally is neutral (oxidation state of \(0\)) and ends up in \(\mathrm{LiCoO}_{2}\) as Li\(+1\). Thus, lithium is oxidized and is the reaction occurring at the anode.
04

Determine Suitability of Water as Electrolyte Solvent

Lithium-ion batteries contain lithium, which reacts violently with water. Lithium reacts to form lithium hydroxide and hydrogen gas, meaning using water as an electrolyte solvent would be dangerous and ineffective. Thus, non-aqueous solvents are used in such batteries.

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Key Concepts

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

Oxidation Numbers
Oxidation numbers are critical in understanding the chemical dynamics of a reaction. They help us determine the charge distribution in components of reactions, like those in lithium-ion batteries. Let's break this down with examples from the original exercise.

In the compound \( \mathrm{CoO}_{2} \), oxygen usually has an oxidation number of \(-2\), and there are two oxygen atoms. This results in a combined oxidation number of \(-4\) for the oxygen. To keep the compound neutral, cobalt must have an oxidation number of \(+4\).

For the compound \( \mathrm{LiCoO}_{2} \), lithium contributes \(+1\), while each oxygen still contributes \(-2\), totaling to a \(-4\) contribution from oxygen. Thus, the cobalt here must be \(+3\) to balance the total charge. Understanding these numbers is essential as they indicate which elements are oxidized or reduced during reactions.
Electrochemical Reactions
Electrochemical reactions are the foundation of how lithium-ion batteries operate. These reactions involve the transfer of electrons, and they occur at two main sites in the battery: the cathode and the anode.

In the discharge process, the cathode is the site of reduction. Specifically, cobalt in \( \mathrm{CoO}_{2} \) reduces from an oxidation state of \(+4\) to \(+3\) when it forms \( \mathrm{LiCoO}_{2} \). This means cobalt gains an electron and undergoes a reduction, releasing energy.

Conversely, the anode is where oxidation occurs. In this exercise, lithium, originally holding no charge (neutral), gets oxidized to \(+1\) when forming part of \( \mathrm{LiCoO}_{2} \). This electron loss by lithium signifies oxidation, completing the electrochemical cycle within the battery.
Electrolytes
Electrolytes are crucial in lithium-ion batteries as they facilitate ion conduction. They must be carefully chosen due to the inherent reactivity of lithium.

Lithium is known for its violent reaction with water. When lithium reacts with water, it forms lithium hydroxide and hydrogen gas, a process that is both dangerous and unsuitable for a stable battery environment.

Therefore, non-aqueous solvents are utilized as electrolytes in lithium-ion batteries. These solvents provide a safe medium for lithium ions to move between the anode and cathode without causing explosive chemical reactions. The choice of electrolyte is pivotal to the battery’s safety and efficiency.

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Most popular questions from this chapter

A hydrogen-oxygen fuel cell operates on the simple reaction $$ \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell) $$ If the cell is designed to produce 1.5 A of current and if the hydrogen is contained in a 1.0 -L. tank at 200 atm pressure at \(25^{\circ} \mathrm{C},\) how long can the fuel cell operate before the hydrogen runs out? (Assume there is an unlimited supply of \(\mathbf{O}_{2}\).)

Two \(\mathrm{Ag}^{+}(\mathrm{aq}) | \mathrm{Ag}(\mathrm{s})\) half-cells are constructed. The first has \(\left|\mathrm{Ag}^{+}\right|=1.0 \mathrm{M},\) the second has \(\left[\mathrm{Ag}^{+}\right]=\) \(1.0 \times 10^{-5} \mathrm{M} .\) When linked together with a salt bridge and external circuit, a cell potential is observed. (This kind of voltaic cell is referred to as a concentration cell.) (a) Draw a picture of this cell, labeling all components. Indicate the cathode and the anode, and indicate in which direction electrons flow in the external circuit. (b) Calculate the cell potential at \(298 \mathrm{K}\)

The standard potential, \(E^{\circ},\) for the reaction of \(\mathrm{Zn}(\mathrm{s})\) and \(\mathrm{C}_{2}(\mathrm{g})\) is \(+2.12 \mathrm{V}\). What is the standard free energy change, \(\Delta_{i} G^{\circ},\) for the reaction?

Balance the following redox equations. All occur in acid solution. (a) \(\mathrm{Ag}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}_{2}(\mathrm{g})+\mathrm{Ag}^{+}(\mathrm{aq})\) (b) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) \rightarrow\) \(\mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}\) (c) \(\mathrm{Zn}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{N}_{2} \mathrm{O}(\mathrm{g})\) (d) \(\mathrm{Cr}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})\)

A (a) Is it easier to reduce water in acid or base? To evaluate this, consider the half-reaction \(2 \mathrm{H}_{2} \mathrm{O}(\ell)+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g})\) \(E^{*}=-0.83 \mathrm{V}\) (b) What is the reduction potential for water for solutions at \(\mathrm{pH}=7\) (neutral) and \(\mathrm{pH}=1\) (acid)? Comment on the value of \(E^{\circ}\) at \(\mathrm{pH}=1\)

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