Problem 34 - Chemistry The Molecular Science

Question

In principle, a battery could be made from aluminum metal and chlorine gas. (a) Write a balanced equation for the reaction that would occur in a battery using \(\mathrm{Al}^{3+}(\mathrm{aq}) \mid \mathrm{Al}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{~g}) \mid \mathrm{Cl}^{-}(\) aq \()\) half-cells. (b) Identify the half-reaction at the anode and at the cathode. Do electrons flow from the \(\mathrm{Al}\) electrode when the cell does work? Explain. (c) Calculate the standard potential, \(E_{\text {cell }}^{\circ}\), for the battery.

Step-by-Step Solution

Answer

(a) 2Al(s) + 3Cl₂(g) → 2Al³⁺(aq) + 6Cl⁻(aq). (b) Anode: Al(s), Cathode: Cl₂(g). Yes, electrons flow from Al. (c) E°cell = 3.02 V.

Step 1: Write Half-Reactions

First, we need to identify the half-reactions for the aluminum and chlorine half-cells. Aluminum will be oxidized, and chlorine will be reduced. The half-reactions are: - Anode (oxidation): \( \mathrm{Al}(\mathrm{s}) \rightarrow \mathrm{Al}^{3+}(\mathrm{aq}) + 3\mathrm{e}^{-} \)- Cathode (reduction): \( \mathrm{Cl}_{2}(\mathrm{~g}) + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Cl}^{-}(\mathrm{aq}) \)

Step 2: Balance the Electron Transfer

Next, we must balance the electron transfer between the oxidation and reduction half-reactions. Multiply the aluminum reaction by 2 and the chlorine reaction by 3, so that both transfer a total of 6 electrons:- Adjusted anode: \( 2\mathrm{Al}(\mathrm{s}) \rightarrow 2\mathrm{Al}^{3+}(\mathrm{aq}) + 6\mathrm{e}^{-} \)- Adjusted cathode: \( 3\mathrm{Cl}_{2}(\mathrm{~g}) + 6\mathrm{e}^{-} \rightarrow 6\mathrm{Cl}^{-}(\mathrm{aq}) \)

Step 3: Write the Balanced Equation

Combine the adjusted half-reactions to form the balanced overall reaction:\[ 2\mathrm{Al}(\mathrm{s}) + 3\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2\mathrm{Al}^{3+}(\mathrm{aq}) + 6\mathrm{Cl}^{-}(\mathrm{aq}) \]

Step 4: Identify Anode and Cathode Reactions

The anode, where oxidation occurs, is the aluminum reaction: \( \mathrm{Al}(\mathrm{s}) \rightarrow \mathrm{Al}^{3+}(\mathrm{aq}) + 3\mathrm{e}^{-} \). The cathode, where reduction occurs, is the chlorine reaction: \( \mathrm{Cl}_{2}(\mathrm{~g}) + 2\mathrm{e}^{-} \rightarrow 2\mathrm{Cl}^{-}(\mathrm{aq}) \). In a battery, electrons flow from the anode to the cathode, so electrons flow from the aluminum electrode.

Step 5: Calculate Standard Cell Potential

Standard electrode potentials are: \( E^\circ(\mathrm{Al}^{3+}/\mathrm{Al}) = -1.66 \, \text{V} \) and \( E^\circ(\mathrm{Cl}_{2}/\mathrm{Cl}^{-}) = +1.36 \, \text{V} \). The calculation for the standard cell potential is:\[ E_{\text{cell}}^{\circ} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.36 \, \text{V} - (-1.66 \, \text{V}) = 3.02 \, \text{V} \]

Key Concepts

Half-Reaction

In electrochemistry, the concept of half-reactions is essential to understanding how batteries and other electrochemical cells operate. A half-reaction is a part of the overall reaction that occurs in an electrochemical cell, particularly in the context of electrons being transferred.
In this problem, aluminum and chlorine are used to form a battery, and identifying their respective half-reactions is the first step to solving this exercise.

• Oxidation Half-Reaction: Occurs at the anode. For aluminum in this exercise, the half-reaction is represented as: \[ \text{Al(s)} \rightarrow \text{Al}^{3+}(\text{aq}) + 3\text{e}^- \]
• Reduction Half-Reaction: Occurs at the cathode. For chlorine, the half-reaction is: \[ \text{Cl}_2(\text{g}) + 2\text{e}^- \rightarrow 2\text{Cl}^- (\text{aq})\]

Understanding these half-reactions helps clarify which substances are gaining electrons (reduction) and which are losing electrons (oxidation). Each half-reaction contributes uniquely to the flow of electricity in the electrochemical cell.

Standard Cell Potential

The standard cell potential, denoted as \(E_{\text{cell}}^\circ\), is a measure of the driving force behind the electrochemical reaction. It determines how much voltage the cell can provide under standard conditions where temperatures are 25°C, 1 atm pressure, and 1 M concentrations of all solutions involved.
The standard cell potential is calculated by subtracting the anode's standard reduction potential from that of the cathode:

• Anode Potential: For aluminum oxidation, it is given as \(E^\circ(\text{Al}^{3+}/\text{Al}) = -1.66 \, \text{V}\).
• Cathode Potential: For chlorine reduction, we use \(E^\circ(\text{Cl}_2/\text{Cl}^-) = +1.36 \, \text{V}\).

The standard cell potential calculation is:\[E_{\text{cell}}^{\circ} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.36 \, \text{V} - (-1.66 \, \text{V}) = 3.02 \, \text{V}\]This indicates that the battery can produce 3.02 V under standard conditions, demonstrating a robust potential.

Oxidation and Reduction

Oxidation and reduction are fundamental processes in chemistry, especially in electrochemical contexts. Understanding them is crucial for tackling any problem involving electrochemical cells.

• Oxidation: This process involves the loss of electrons by a substance. In an electrochemical cell, it occurs at the anode. With aluminum, the reaction is: \[ \text{Al(s)} \rightarrow \text{Al}^{3+}(\text{aq}) + 3\text{e}^- \]
• Reduction: This involves the gain of electrons. It occurs at the cathode in electrochemical cells. In our example, chlorine undergoes reduction: \[ \text{Cl}_2(\text{g}) + 2\text{e}^- \rightarrow 2\text{Cl}^- (\text{aq}) \]

The mnemonic "LEO says GER" is helpful for remembering that "Loss of Electrons is Oxidation" and "Gain of Electrons is Reduction." These terms guide us in determining how electrons move and where they result in the electrochemical reactions.

Balancing Redox Equations

Balancing redox (reduction-oxidation) equations is critical to ensuring the law of conservation of mass and charge. It ensures that both sides of the equation have the same number of atoms for each element and equivalent charges. In our example, this procedure involves several steps:

• Writing Initial Half-Reactions: Start by identifying the half-reactions for the oxidation and reduction processes.
• Balancing Electron Transfer: Adjust coefficients in the half-reactions to equalize the number of electrons lost and gained. In the problem, we balance by multiplying the aluminum oxidation by 2 and chlorine reduction by 3.

Adjusted reactions:

• Anode: \(2\text{Al}(\text{s}) \rightarrow 2\text{Al}^{3+}(\text{aq}) + 6\text{e}^-\)
• Cathode: \(3\text{Cl}_2(\text{g}) + 6\text{e}^- \rightarrow 6\text{Cl}^- (\text{aq})\)

Combine the balanced half-reactions to form the complete balanced equation:\[ 2\text{Al}(\text{s}) + 3\text{Cl}_2(\text{g}) \rightarrow 2\text{Al}^{3+}(\text{aq}) + 6\text{Cl}^-(\text{aq}) \]This step is significant because it provides a clear and precise picture of the overall redox reaction occurring within the electrochemical cell, ensuring accuracy in chemical representation.