To determine the overall voltage of the galvanic cell, we use the Nernst equation and the standard reduction potentials from a table of standard electrode potentials.

Step 1: Write the half-reactions

1. Oxidation half-reaction (Aluminum being oxidized): $\text{Al(s)} \to \text{Al}^{3+}(aq) + 3e^-$ Standard reduction potential for $\text{Al}^{3+}/\text{Al}$: $E^\circ = -1.66 \, \text{V}$.

2. Reduction half-reaction (Bromine being reduced in HBr): $\text{Br}_2(l) + 2e^- \to 2\text{Br}^-(aq)$ Standard reduction potential for $\text{Br}_2/2\text{Br}^-$: $E^\circ = +1.07 \, \text{V}$.

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Step 2: Calculate the standard cell potential

$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ Here:

• Cathode: Reduction of bromine ($+1.07 \, \text{V}$),
• Anode: Oxidation of aluminum ($-1.66 \, \text{V}$).

$E^\circ_{\text{cell}} = 1.07 - (-1.66) = 2.73 \, \text{V}.$

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Step 3: Account for non-standard conditions (Nernst Equation)

The Nernst equation is: $E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q$ Where:

• $E_{\text{cell}}$: Non-standard cell potential,
• $E^\circ_{\text{cell}}$: Standard cell potential,
• $R = 8.314 \, \text{J/mol·K}$,
• $T = 298 \, \text{K}$ (room temperature),
• $n$: Number of electrons transferred,
• $F = 96485 \, \text{C/mol}$,
• $Q$: Reaction quotient.

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Step 4: Determine the reaction quotient ($Q$)

The balanced overall reaction is: $2\text{Al}(s) + 3\text{Br}_2(l) \to 2\text{Al}^{3+}(aq) + 6\text{Br}^-(aq)$

The reaction quotient ($Q$) is: [ Q = \frac{[\text{Al}^{3+}]^2 \cdot [\text{Br}^-]^6}{[\text{Br}_2]^3} ]

• $[\text{Al}^{3+}] = 0.01 \, \text{M}$,
• $[\text{Br}^-] = 0.1 \, \text{M}$,
• $[\text{Br}_2]$: Bromine is a liquid, so its activity is considered $1$.

Thus: $Q = \frac{(0.01)^2 \cdot (0.1)^6}{1^3} = (10^{-2})^2 \cdot (10^{-1})^6 = 10^{-4} \cdot 10^{-6} = 10^{-10}.$

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Step 5: Apply the Nernst equation

Substitute values:

• $E^\circ_{\text{cell}} = 2.73 \, \text{V}$,
• $n = 6$ (since each Al loses 3 electrons, and 2 Al atoms are oxidized),
• $Q = 10^{-10}$.

The term $\frac{RT}{nF}$ at $298 \, \text{K}$ simplifies to: $\frac{RT}{nF} = \frac{8.314 \cdot 298}{6 \cdot 96485} \approx 0.00426 \, \text{V}.$

The Nernst equation becomes: $E_{\text{cell}} = 2.73 - (0.00426) \ln(10^{-10}).$

Simplify $\ln(10^{-10}) = -10 \ln(10) = -10 (2.3026) = -23.026$: $E_{\text{cell}} = 2.73 - (0.00426)(-23.026).$

Calculate: $E_{\text{cell}} = 2.73 + 0.0981 = 2.8281 \, \text{V}.$

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Final Answer:

The overall voltage of the galvanic cell is approximately: $\boxed{2.83 \, \text{V}}.$

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Do you have questions, or want details on any step?

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Related Questions:

1. What is the role of the Nernst equation in electrochemistry?
2. Why does the activity of pure liquids and solids not appear in the reaction quotient?
3. How do we determine the number of electrons transferred ($n$) in a reaction?
4. What is the significance of standard electrode potentials in predicting reaction spontaneity?
5. How would temperature changes affect the cell potential?

Tip:

Always check that your units are consistent when using the Nernst equation to avoid calculation errors.