Problem 100 - Chemistry: An Atoms-Focused Approach

Question

Sodium-Sulfur Batteries The low cost of sodium and sulfur relative to lithium makes voltaic cells based on sodium attractive to electric vehicle manufacturers, provided the technological hurdles of managing a battery that operates at \(300^{\circ} \mathrm{C}\) can be overcome. The overall cell reaction is $$ 2 \mathrm{Na}(s)+3 \mathrm{S}(\ell) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{3}(s) \quad E_{\mathrm{cell}}^{\circ}=2.076 \mathrm{V} $$ a. Which element is oxidized and which is reduced? b. How many electrons are transferred in the overall cell reaction? c. What is the value of \(\Delta G^{\circ}\) for the reaction? d. If a battery containing \(5.25 \mathrm{kg} \mathrm{Na}\) is \(50 \%\) discharged when it is connected to a charger with an output of 200 A, how long does it take to recharge the battery? e. Draw a Lewis structure for the \(\mathrm{S}_{3}^{2-}\) anion.

Step-by-Step Solution

Answer

a. Sodium (Na) undergoes oxidation, and sulfur (S) undergoes reduction. b. 2 electrons are transferred in the overall cell reaction. c. The standard Gibbs free energy change is -400062.12 J/mol. d. It takes 3075.26 minutes to recharge the partially discharged battery. e. The Lewis structure for \(\mathrm{S}_{3}^{2-}\) has a chain of three sulfur atoms, with a double bond between one pair of sulfurs, and single bonds between the others. Each sulfur atom has two lone pairs of electrons.

Step 1: a. Oxidation and Reduction Identifications

To determine which element is oxidized and which is reduced, we will look for a change in oxidation states between the reactants and products. Sodium (Na) starts as a neutral element in its zero oxidation state and becomes part of the compound \(\mathrm{Na}_{2}\mathrm{S}_{3}\), where it has a +1 oxidation state. Since the oxidation state increased, sodium is oxidized. Sulfur starts as a neutral element in its zero oxidation state and becomes part of the compound \(\mathrm{Na}_{2}\mathrm{S}_{3}\), where it has a -2 oxidation state. Since the oxidation state decreased, sulfur is reduced.

Step 2: b. Electron Transfer

To determine the number of electrons transferred in the overall cell reaction, we need to balance the oxidation and reduction half-reactions. For sodium: $$ \mathrm{Na} \rightarrow \mathrm{Na}^+ + e^- $$ For sulfur: $$ 2 e^- + 3 \mathrm{S} \rightarrow {\mathrm{S}}_3^{2-} $$ Balancing the two half-reactions requires multiplying the sodium half-reaction by 2 so that the total number of electrons involved is the same for both half-reactions: $$ 2 \ \mathrm{Na} \rightarrow 2 \ \mathrm{Na}^+ + 2 \ e^- $$ Now we can see that 2 electrons are transferred in the overall cell reaction.

Step 3: c. Delta G calculation

To find the value of \(\Delta G^\circ\), we can use the following equation relating standard cell potential (\(E_{cell}^\circ\)) and standard Gibbs free energy change (\(\Delta G^\circ\)): $$ \Delta G^\circ = -nFE_{cell}^\circ, $$ where \(n\) is the number of electrons transferred and \(F\) is Faraday's constant (\(96485 \ \mathrm{C/mol}\)). We found that 2 electrons are transferred in the overall cell reaction, so: $$ \Delta G^\circ = -2 \times 96485 \ \mathrm{C/mol} \times 2.076 \ \mathrm{V} = -400062.12 \ \mathrm{J/mol}. $$

Step 4: d. Recharge Time

We are given that a battery containing \(5.25 \ \mathrm{kg}\) of sodium is \(50\%\) discharged when connected to a charger with an output of \(200 \ \mathrm{A}\). To find the time needed to recharge the battery, we first need to determine the number of moles of sodium involved in the half-discharged state. The molar mass of sodium is \(22.99 \ \mathrm{g/mol}\), so: $$ \text{moles of Na} = \frac{5.25 \ \mathrm{kg} \times 1000 \ \mathrm{g/kg}}{22.99 \ \mathrm{g/mol}} = 228.36 \ \mathrm{mol}. $$ Since the battery is \(50\%\) discharged, we will only consider half of the moles of sodium for charging: $$ \text{moles of Na to recharge} = \frac{228.36 \ \mathrm{mol}}{2} = 114.18 \ \mathrm{mol}. $$ We know that 2 electrons are transferred per mole of sodium in the reaction, so the total charge needed to recharge the battery is: $$ \text{charge to recharge} = 114.18 \ \mathrm{mol} \times 2 \frac{\mathrm{e}^-}{\mathrm{mol}} \times 96485 \ \mathrm{C/e^-} = 22061856 \ \mathrm{C}. $$ Now, we can find the time required by dividing the total charge needed by the charger's current output: $$ \text{time to recharge} = \frac{22061856 \ \mathrm{C}}{200 \ \mathrm{A}} = 110309.28 \ \mathrm{s} = 3075.26 \ \mathrm{minutes}. $$

Step 5: e. Lewis Structure for \(\mathrm{S}_{3}^{2-}\)

To draw the Lewis structure for \(\mathrm{S}_{3}^{2-}\), we first count the total number of valence electrons. Sulfur has 6 valence electrons, and since there are 3 sulfur atoms, we have 18 valence electrons from sulfur. The 2 extra electrons due to the 2- charge bring the total to 20 valence electrons. We can arrange the three sulfur atoms in a S-S-S chain, starting with single bonds between the sulfurs. This structure uses 4 valence electrons for the two single bonds. The remaining 16 valence electrons can be distributed as lone pairs, with each sulfur having 2 lone pairs (8 electrons). Now, one of the sulfur atoms will have a total of 7 valence electrons, and to achieve a stable octet, we can form a double bond between one pair of sulfurs by converting one set of lone pairs to a bond pair. This now gives us the Lewis structure for \(\mathrm{S}_{3}^{2-}\) where one sulfur has a double bond and the other two have single bonds, each sulfur with two lone pairs.

Key Concepts

Oxidation and Reduction

In the world of chemistry, oxidation and reduction are processes that involve changes in oxidation states. These changes determine which elements lose or gain electrons.
In sodium-sulfur batteries, sodium ( \(\text{Na}\) ) starts with an oxidation state of zero in its elemental form. It transforms into \(\text{Na}^+\) in the compound \(\text{Na}_2\text{S}_3\) , indicating it is oxidized.
On the other hand, sulfur ( \(\text{S}\) ) starts with an oxidation state of zero and ends up in \(\text{Na}_2\text{S}_3\) as \(\text{S}^{2-}\) . This decrease in its oxidation state shows sulfur is reduced.

• Oxidation: Loss of electrons, increase in oxidation state (Sodium)
• Reduction: Gain of electrons, decrease in oxidation state (Sulfur)

Understanding these changes is key to analyzing electrochemical reactions like those in sodium-sulfur batteries.

Electron Transfer

Electron transfer is a fundamental concept in redox reactions and it plays a vital role in how batteries work. When sodium is oxidized, it loses electrons which are then accepted by sulfur as it gets reduced.
The balanced half-reactions for the overall cell in a sodium-sulfur battery are as follows:

• \(\text{Na} \rightarrow \text{Na}^+ + e^-\) (Sodium loses electrons)
• \(2e^- + 3\text{S} \rightarrow \text{S}_3^{2-}\) (Sulfur gains electrons)

This means that 2 electrons are transferred per reaction, allowing the battery to generate energy. This transfer sets up an electric current in the battery, crucial for its operation.

Gibbs Free Energy

Gibbs free energy ( \(\Delta G\) ) tells us about the spontaneity of a reaction. In batteries, it helps determine how much energy can be extracted as electrical work.
The relation between Gibbs free energy and cell potential ( \(E_{cell}\) ) is given by: \[ \Delta G^\circ = -nF E_{cell}^\circ \] where \(n\) is the number of electrons transferred, and \(F\) is Faraday's constant ( 96485 \text{C/mol} ).
For our sodium-sulfur battery, the calculation is: \[ \Delta G^\circ = -2 \times 96485 \times 2.076 = -400062.12 \text{ J/mol} \]This negative \(\Delta G\) confirms the reaction releases energy, making it spontaneous. This is why these batteries can efficiently power devices.

Lewis Structures

Lewis structures provide a visual representation of the distribution of electrons in molecules. For anions like \(\text{S}_3^{2-}\) , these diagrams help in understanding how the atoms connect and share electrons.
The anion \(\text{S}_3^{2-}\) involves three sulfur atoms, each contributing 6 valence electrons, plus 2 extra from the 2- charge, totalling 20 valence electrons.

• Arrange sulfur in a linear \(\text{S-S-S}\) chain.
• Form single bonds first, using 4 electrons.
• Distribute remaining valence electrons as lone pairs

Create a double bond between two sulfur atoms to give stability through octets for all atoms.