Problem 61 - Chemistry: An Atoms-Focused Approach

Question

Which of the following voltaic cells, \(\boldsymbol{A}\) or \(\mathbf{B},\) will produce the greater quantity of electric charge per gram of anode material? Cell \(\mathbf{A}: \mathbf{C d}(s)+2 \mathrm{NiO}(\mathrm{OH})(s)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) \(2 \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{Cd}(\mathrm{OH})_{2}(\mathrm{s})\) Cell \(\mathbf{B}: 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(\ell)+4 \mathrm{OH}^{-}(a q) \rightarrow\) \(4 \mathrm{Al}(\mathrm{OH})_{4}-(a q)\)

Step-by-Step Solution

Answer

Answer: Cell B with an aluminum anode produces a greater quantity of electric charge per gram of anode material than cell A with a cadmium anode.

Step 1: Determine the half-reactions and identify the anode in each cell

For cell A, the half reactions are: Cadmium: Cd(s) -> Cd(OH)2(s) + 2e^- Nickel: 2NiO(OH)(s) + 2H2O(l) + 2e^- -> 2Ni(OH)2(s) For cell B, the half reactions are: Aluminum: 4Al(s) -> 4Al(OH)4^(-) + 12e^- Oxygen: 3O2(g) + 6H2O(l) + 12e^- -> 8OH^(-) Cadmium in cell A and Aluminum in cell B are the anode materials that get oxidized.

Step 2: Calculate moles of electrons produced per mole of anode material

For cell A: 1 mole of Cd gives 2 moles of electrons For cell B: 1 mole of Al gives 3 moles of electrons

Step 3: Calculate electric charge produced per mole of anode material

For cell A: Charge produced per mole of Cd = 2 moles of electrons * 96485 C/mol = 192970 C/mol For cell B: Charge produced per mole of Al = 3 moles of electrons * 96485 C/mol = 289455 C/mol

Step 4: Calculate electric charge produced per gram of anode material

For cell A: Molar mass of Cd = 112.42 g/mol Charge produced per gram of Cd = 192970 C/mol / 112.42 g/mol = 1717 C/g For cell B: Molar mass of Al = 26.98 g/mol Charge produced per gram of Al = 289455 C/mol / 26.98 g/mol = 10732 C/g

Step 5: Comparing both cells

Comparing the charge produced per gram of anode material: Cell A (Cd): 1717 C/g Cell B (Al): 10732 C/g Since cell B (Al) produces a greater quantity of electric charge per gram of anode material (10732 C/g) than cell A (Cd) at 1717 C/g, cell B will produce a greater quantity of electric charge per gram of anode material.

Key Concepts

Anode Material

In a voltaic cell, the anode material is crucial because it undergoes the oxidation reaction. It loses electrons, which then flow through an external circuit to the cathode. In cell A, cadmium (Cd) is used as the anode material, while in cell B, aluminum (Al) serves this role.
These materials are chosen based on their ability to provide electrons effectively. The choice of anode material affects the cell's overall efficiency and output of electric charge. Because these materials participate in the redox process, they determine how many electrons each mole can release, impacting the electric charge produced per gram.

• Cadmium (Cell A): Loses 2 electrons per mole.
• Aluminum (Cell B): Loses 3 electrons per mole.

Electric Charge

Electric charge in a voltaic cell is generated through electron transfer from the anode. Electrons flow from the anode to the cathode, creating an electric current that can be harnessed for energy.
The amount of charge produced depends on the number of electrons released by the anode material per mole. For our cells, the charge is calculated using Faraday's constant (\(96485 \text{ C/mol} \), the charge of a mole of electrons). The total electric charge produced per mole of anode material is:

• Cell A (Cadmium): \(2 \times 96485 = 192970 \text{ C/mol} \)
• Cell B (Aluminum): \(3 \times 96485 = 289455 \text{ C/mol} \)

To find the electric charge per gram, divide by the molar mass:

• Cell A: 1717 C/g
• Cell B: 10732 C/g

Half-Reactions

Half-reactions are essential parts of understanding how voltaic cells work. They break down the overall reaction into oxidation and reduction parts.
For voltaic cells, the anode is where oxidation occurs, and the cathode is where reduction happens. Each cell has specific half-reactions:

• Cell A's Anode (Cd): \(\text{Cd(s)} \rightarrow \text{Cd(OH)}_2(s) + 2\text{e}^-\)
• Cell B's Anode (Al): \(4\text{Al(s)} \rightarrow 4\text{Al(OH)}_4^-(\text{aq}) + 12\text{e}^-\)

Breaking down these reactions helps identify the number of electrons involved, which is crucial for calculating electric charge.

Oxidation Reactions

Oxidation reactions in voltaic cells involve the loss of electrons from the anode material. This process releases electrons, which create a current as they move to the cathode.
In the context of our cells:

• Cell A's Oxidation (Cd): The cadmium anode loses electrons, turning into cadmium hydroxide.
• Cell B's Oxidation (Al): The aluminum anode loses more electrons than cadmium, transforming it to aluminum hydroxide complexes.

The efficiency of these reactions influences the amount of charge a cell can produce per gram of its anode material. Cell B produces more charge due to aluminum's ability to release more electrons.