Problem 63 - Chemistry: An Atoms-Focused Approach

Question

Which of the following voltaic cell reactions, \(\mathbf{E}\) or \(\mathbf{F}\), delivers more electrical energy per gram of anode material at \(298 \mathrm{K} ?\) Reaction \(\mathbf{E}: \mathbf{Z n}(s)+2 \mathrm{NiO}(\mathrm{OH})(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ 2 \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Zn}(\mathrm{OH})_{2}(s) \quad E_{c c 11}^{*}=1.20 \mathrm{V} $$ Reaction \(\mathbf{F}: \mathrm{Li}(s)+\mathrm{MnO}_{2}(s) \rightarrow \mathrm{LiMnO}_{2}(s) \quad E_{\mathrm{call}}^{\circ}=3.15 \mathrm{V}\)

Step-by-Step Solution

Answer

Answer: Reaction F delivers more electrical energy per gram of anode material at 298 K.

Step 1: Identify the anode and cathode in each reaction

For reaction E: Anode: Zn (s) Cathode: 2 NiO(OH) (s) + 2 H2O (l) For reaction F: Anode: Li (s) Cathode: MnO2 (s)

Step 2: Balance the Redox Reactions of Both Reactions

Balancing the redox reactions to find the number of transferred electrons For reaction E: Zn (s) -> Zn(OH)2 (s) + 2e- (oxidation) 2NiO(OH)(s) + 2 H2O (l) + 2e- -> 2 Ni(OH)2 (s) + 2 OH- (reduction) For reaction F: Li (s) -> Li+ + e- (oxidation) MnO2 (s) + Li+ + e- -> LiMnO2 (s) (reduction)

Step 3: Determine Electrons Transferred in Each Reaction

For reaction E: 2 moles of electrons are transferred (based on step 2) For reaction F: 1 mole of electron is transferred (based on step 2)

Step 4: Calculate the Electrical Energy per Gram of Anode Material

Using the following formula: Electric energy per gram = (\(\frac{n \times F \times E_{cell}}{molar\,mass\, of\, anode\,material}\)) For reaction E: Electric energy per gram = (\(\frac{2 \times 96485 \times 1.20}{65.4}\)) = 3530.66 J/g For reaction F: Electric energy per gram = (\(\frac{1 \times 96485 \times 3.15}{6.94}\)) = 44052.91 J/g

Step 5: Compare the Electrical Energy per Gram in Each Reaction

From the previous step, we can see that: Reaction E has an electrical energy of 3530.66 J/g, while Reaction F has an electrical energy of 44052.91 J/g. Since 44052.91 J/g (reaction F) is greater than 3530.66 J/g (reaction E), reaction F delivers more electrical energy per gram of anode material at 298 K than reaction E.

Key Concepts

Electrical Energy

In voltaic cells, chemical reactions result in the creation of electrical energy. This energy is generated as electrons move through an external circuit. The amount of electrical energy produced per gram of anode material depends on several factors:

• The cell potential, represented as \(E_{cell}\), which indicates the voltage produced by the reaction.
• The number of electrons transferred during the redox reaction.
• The molar mass of the anode material.

To determine which reaction produces more electrical energy per gram of anode material, you can use the formula: \[\text{Electric energy per gram} = \left( \frac{n \times F \times E_{cell}}{\text{molar mass of anode material}} \right) \]Here, \(n\) is the number of moles of electrons transferred, and \(F\) is Faraday's constant \((96485 \, \text{C/mol})\). By comparing these values, we can determine the efficiency of different reactions.

Anode Material

The anode material in a voltaic cell is where oxidation occurs, meaning it loses electrons. The properties of the anode material significantly affect how much electrical energy is generated. Key factors include:

• The molar mass of the anode material, which affects the calculation of energy output per gram.
• The ability of the material to release electrons easily.

In the exercise provided, two different anode materials are used: Zinc (Zn) for reaction E and Lithium (Li) for reaction F. Molar masses are critical here because energy output is measured per gram. Therefore, a lighter anode material, like Lithium, can potentially produce more electrical energy per gram due to its lower molar mass.

Redox Reactions

Redox reactions are at the heart of every voltaic cell operation. These are reactions where oxidation and reduction occur simultaneously, allowing electron transfer. In the context of voltaic cells:

• Oxidation occurs at the anode, where the material loses electrons. For reaction E, Zn is oxidized to Zn(OH)₂, while in reaction F, Li is oxidized to Li⁺.
• Reduction occurs at the cathode, where electrons are gained. In reaction E, NiO(OH) undergoes reduction, whereas in reaction F, MnO₂ is reduced to form LiMnO₂.

These reactions must be balanced, which involves ensuring that the electrons lost in oxidation are equal to those gained in reduction. This balance is crucial as it dictates the cell's efficiency and output.

Electrons Transferred

Understanding how many electrons are transferred in a reaction gives insight into the energy characteristics of a voltaic cell. The greater the number of electrons transferred, the more potential there is for energy production. In reaction E, 2 moles of electrons are transferred, while in reaction F, only 1 mole is transferred. This transfer is essential in calculating the cell's electrical energy output as it directly influences the energy production per unit of material used. This detail affects the calculation of electrical energy using the given formula, where more electrons could mean a higher potential for energy output, albeit also dependent on other factors like cell potential and molar mass. The intricacies of electronic transfer highlight the complexity and beauty of electrochemical processes in voltaic cells.