Problem 89 - Chemistry: An Atoms-Focused Approach

Question

Lithium-Sulfur Dioxide Batteries The U.S. military uses batteries based on the reduction of liquid sulfur dioxide at a carbon cathode in certain communications equipment. Lithium metal is used as the anode. The overall cell reaction is $$ 2 \mathrm{Li}(s)+2 \mathrm{SO}_{2}(\ell) \rightarrow \mathrm{Li}_{2} \mathrm{S}_{2} \mathrm{O}_{4}(s) $$ a. Write half-reactions for the anode and cathode reactions. b. How many electrons are transferred in the cell reaction? c. Draw a Lewis structure for the \(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\) anion.

Step-by-Step Solution

Answer

Based on the given lithium-sulfur dioxide battery's overall cell reaction, the oxidation half-reaction is Li(s) → Li^(+)(aq) + e^-, and the reduction half-reaction is SO₂(ℓ) + 2e^- → (1/2)S₂O₄^(2-)(aq). In this reaction, a total of 2 electrons are transferred. The Lewis structure for the S₂O₄^(2-) ion can be represented as: $$ \begin{array}{cc} \mathrm{O} = \mathrm{S} - \mathrm{S} = \mathrm{O}\\ \mathrm{||}\:\:\mathrm{H}\:\:\mathrm{||}\\ \mathrm{OH}\:\:\:\:\:\_\mathrm{OH} \end{array} $$

Step 1: Identify the oxidation and reduction half-reactions

First, we need to break the overall cell reaction into oxidation and reduction half-reactions. Lithium is the anode, and it will undergo oxidation. Sulfur dioxide is the cathode and will undergo reduction.

Step 2: Write the oxidation half-reaction at the anode

Oxidation half-reaction: In oxidation, an element loses electrons. The lithium (Li) metal acting as the anode will lose its electrons since it is being oxidized. To balance the half-reaction, we need to add the correct number of electrons to the product side of the half-reaction. The oxidation half-reaction is: $$ \mathrm{Li}(s) \rightarrow \mathrm{Li}^{+}(aq) + e^- $$

Step 3: Write the reduction half-reaction at the cathode

Reduction half-reaction: At the cathode, sulfur dioxide (SO₂) will gain electrons since it undergoes reduction. We need to determine how many electrons will be gained. For this, we can look at the overall cell reaction and balance the charges between the reactants and products. The reduction half-reaction is: $$ \mathrm{SO}_{2}(\ell) + 2e^- \rightarrow \frac{1}{2}\mathrm{S}_{2}\mathrm{O}_{4}^{2-}(aq) $$

Step 4: Determine the number of electrons transferred

Since each 2 Lithium (Li) loses 1 electron in the oxidation half-reaction, there are a total of 2 electrons transferred in the cell reaction.

Step 5: Draw the Lewis structure of S2O42-

To draw the Lewis structure for the S2O4^(2-) ion we will follow these steps: 1. Determine the total number of valence electrons in the ion. - Each Sulfur (S) has 6 valence electrons and we have 2 Sulfur atoms: \(\:6 \times 2 = 12\) electrons. - Each Oxygen (O) has 6 valence electrons and we have 4 Oxygen atoms: \(\:6 \times 4 = 24\) electrons. - As the ion has a 2- charge, we must add 2 electrons. - Total valence electrons: \(24 + 12 + 2 = 38\) electrons. 2. Arrange the atoms to form the skeleton structure of the ion. - Sulfur atoms form a central pair, with double bonds to the outer 2 oxygen atoms, and single bonds to the remaining 2 oxygen atoms. 3. Distribute the electrons among the atoms. - We have 38 electrons to distribute. - Starting with the double bonds, we assign 8 electrons to the O=S double bonds (4 electrons per double bond). - We assign 4 electrons to the S-O single bonds (2 electrons per bond, a total of 4 electrons). - At this point, we have used 8+4=12 electrons. - We need to distribute the remaining 26 electrons (38-12) among the outer atoms (Oxygen) to satisfy the octet rule. We assign 6 electrons to each of the double-bonded Oxygens, and 8 electrons to each of the single-bonded Oxygens. - All valence electron requirements should be met for each atom in the ion. Thus, the Lewis structure for \(\mathrm{S}_{2}\mathrm{O}_{4}^{2-}\) is: $$ \begin{array}{cc} \mathrm{O} = \mathrm{S} - \mathrm{S} = \mathrm{O}\\ \mathrm{||}\:\:\mathrm{H}\:\:\mathrm{||}\\ \mathrm{OH}\:\:\:\:\:\_\mathrm{OH} \end{array} $$

Key Concepts

Oxidation and Reduction Reactions

Oxidation and reduction reactions are essential in understanding how batteries work. These reactions involve the transfer of electrons between chemical species. Oxidation refers to the loss of electrons, while reduction refers to the gaining of electrons. In a battery, oxidation occurs at the anode, releasing electrons into an external circuit, and reduction occurs at the cathode, absorbing those electrons to complete the circuit.
With a Lithium-Sulfur Dioxide battery, lithium metal acts as the anode and suffers oxidation. It loses electrons, turning into positively charged lithium ions (\( \mathrm{Li}^+ \)). Conversely, sulfur dioxide at the cathode undergoes reduction, gaining electrons to transform into the \(\mathrm{S}_2\mathrm{O}_4^{2-}\) ion. This process is what allows the battery to generate electricity efficiently.

Half-Reactions

In electrochemistry, overall reactions can be divided into two parts called half-reactions. Each half-reaction accurately depicts either oxidation or reduction. This separation allows for clearer analysis and balance of electron transfer.
For the lithium in the Lithium-Sulfur Dioxide battery, the half-reaction is

• **Oxidation**: \( \mathrm{Li} \rightarrow \mathrm{Li}^{+} + e^- \)

Here, lithium loses one electron per atom, showing its oxidation process.
The reduction half-reaction of sulfur dioxide is as follows:

• **Reduction**: \( \mathrm{SO}_2 + 2e^- \rightarrow \frac{1}{2} \mathrm{S}_2\mathrm{O}_4^{2-} \)

This demonstrates how sulfur dioxide gains electrons. Combining these half-reactions gives the battery's total reaction.

Lewis Structure

A Lewis structure is a visual representation of molecules showing all valence electrons. They help contextualize chemical bonds and lone pairs using dots and lines.
In the \(\mathrm{S}_2\mathrm{O}_4^{2-}\) ion, Lewis structures depict the positioning of molecules and the shared electron pairs to achieve each atom's octet.
We begin by calculating the total number of valence electrons:

• 2 Sulfur atoms contributing 12 electrons
• 4 Oxygen atoms contributing 24 electrons
• Plus 2 for the anionic charge: 38 in total

The sulfur atoms pair centrally, forming double bonds with two oxygen atoms and single bonds with the other two. Electron distribution ensures all satisfy the octet rule, and charge balance reflects the ion's -2 charge.

Electron Transfer

Electron transfer is the fundamental process whereby electrons move from one molecule to another in redox reactions. In a battery, this is the core action that facilitates electricity flow.
During electron transfer in a Lithium-Sulfur Dioxide battery, lithium atoms release electrons (oxidation). These electrons travel through an external circuit to reach sulfur dioxide molecules (reduction), forming \(\mathrm{S}_2\mathrm{O}_4^{2-}\).
This electron movement creates a potential difference, driving electric current through devices powered by the battery.

Chemical Equations

Chemical equations succinctly express chemical reactions using symbols for elements and compounds. In the case of batteries, they describe transitions within anodes and cathodes.
The complete chemical equation for a Lithium-Sulfur Dioxide battery is:\[2 \mathrm{Li}(s) + 2 \mathrm{SO}_2(\ell) \rightarrow \mathrm{Li}_2\mathrm{S}_2\mathrm{O}_4(s)\]This equation tells us what each substance starts as and what it becomes post-reaction. Balancing this equation ensures that atom and charge quantities remain constant.
Understanding chemical equations is crucial for predicting product quantities and calculating reactant needs within electrochemical processes.