Calculate `Delta G^(Theta)` and `E_(cell)` for the cell `Al |Al^(3+) (0.01 M) || Fe^(2+) (0.02 M) |Fe`
Given that `E_((Al^(3+)|Al))^(Theta) = -1.66 V` and `E_((Fe^(2+) |Fe))^(Theta) = -0.44 V`

Explanation

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) and the cell potential (E_cell) for the given electrochemical cell:

**Cell Notation:**
Al | Al³⁺ (0.01 M) || Fe²⁺ (0.02 M) | Fe

**Given Data:**
- Standard reduction potential for Al³⁺/Al: E°(Al³⁺/Al) = -1.66 V
- Standard reduction potential for Fe²⁺/Fe: E°(Fe²⁺/Fe) = -0.44 V

### Step 1: Identify the half-reactions

1. **Anode (oxidation)**:
- Aluminum is oxidized:
\[ \text{Al} \rightarrow \text{Al}^{3+} + 3e^- \]
The standard oxidation potential (E°) for this reaction is the negative of the reduction potential:
\[ E°_{\text{oxidation}} = +1.66 \, V \]

2. **Cathode (reduction)**:
- Iron is reduced:
\[ \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \]
The standard reduction potential is given:
\[ E°_{\text{reduction}} = -0.44 \, V \]

### Step 2: Balance the half-reactions

To balance the electrons transferred, we need to multiply the half-reactions so that the number of electrons is the same.

- Multiply the aluminum oxidation half-reaction by 2:
\[ 2 \text{Al} \rightarrow 2 \text{Al}^{3+} + 6e^- \]

- Multiply the iron reduction half-reaction by 3:
\[ 3 \text{Fe}^{2+} + 6e^- \rightarrow 3 \text{Fe} \]

### Step 3: Write the overall balanced reaction

Combining the balanced half-reactions gives:
\[ 2 \text{Al} + 3 \text{Fe}^{2+} \rightarrow 2 \text{Al}^{3+} + 3 \text{Fe} \]

### Step 4: Calculate the standard cell potential (E°_cell)

Using the formula:
\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} \]
Substituting the values:
\[ E°_{\text{cell}} = (-0.44 \, V) - (-1.66 \, V) = 1.22 \, V \]

### Step 5: Calculate the cell potential under non-standard conditions using the Nernst equation

The Nernst equation is given by:
\[ E_{\text{cell}} = E°_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
Where:
- n = number of moles of electrons transferred = 6 (from the balanced equation)
- Concentrations: [Al³⁺] = 0.01 M, [Fe²⁺] = 0.02 M

Substituting into the Nernst equation:
\[ E_{\text{cell}} = 1.22 - \frac{0.0591}{6} \log \left( \frac{(0.01)^2}{(0.02)^3} \right) \]

Calculating the logarithm:
\[ \log \left( \frac{(0.01)^2}{(0.02)^3} \right) = \log \left( \frac{0.0001}{0.000008} \right) = \log(12.5) \approx 1.096 \]

Now substituting back:
\[ E_{\text{cell}} = 1.22 - \frac{0.0591}{6} \times 1.096 \]
\[ E_{\text{cell}} = 1.22 - 0.0091 \approx 1.209 \, V \]

### Step 6: Calculate ΔG°

Using the formula:
\[ \Delta G° = -nFE°_{\text{cell}} \]
Where:
- F (Faraday's constant) = 96485 C/mol

Substituting the values:
\[ \Delta G° = -6 \times 96485 \times 1.22 \]
\[ \Delta G° = -706380 \, J \approx -706.38 \, kJ \]

### Final Answers:
- \( E_{\text{cell}} \approx 1.209 \, V \)
- \( \Delta G° \approx -706.38 \, kJ \)