What is the value fo `DeltaG^(@0` for the reactions?
`2Al(s) + 3Cu^(2+)(aq) rightarrow + 3Cu(s) E^(@) = 2.02V`

Options

A. `1170kJ`
B. `-585kJ`
C. `-390kJ`
D. `-195kJ`

Answer

To find the value of \(\Delta G^\circ\) for the given reaction:

**Given Reaction:**
\[ 2 \text{Al}(s) + 3 \text{Cu}^{2+}(aq) \rightarrow 3 \text{Cu}(s) + 2 \text{Al}^{3+}(aq) \]

**Given:**
- Standard cell potential, \(E^\circ = 2.02 \, \text{V}\)
- Faraday's constant, \(F = 96500 \, \text{C/mol}\)

### Step 1: Identify the number of electrons transferred (n)

In the given reaction:
- Each aluminum atom loses 3 electrons to become \(\text{Al}^{3+}\).
- Since there are 2 aluminum atoms, the total number of electrons lost is:
\[ n = 2 \times 3 = 6 \, \text{electrons} \]

### Step 2: Use the formula for \(\Delta G^\circ\)

The relationship between Gibbs free energy change (\(\Delta G^\circ\)) and cell potential (\(E^\circ\)) is given by the equation:
\[ \Delta G^\circ = -nFE^\circ \]

### Step 3: Substitute the values into the equation

Now, substituting the values we found:
- \(n = 6\)
- \(F = 96500 \, \text{C/mol}\)
- \(E^\circ = 2.02 \, \text{V}\)

So,
\[ \Delta G^\circ = -6 \times 96500 \, \text{C/mol} \times 2.02 \, \text{V} \]

### Step 4: Calculate \(\Delta G^\circ\)

Calculating the above expression:
\[ \Delta G^\circ = -6 \times 96500 \times 2.02 \]
\[ \Delta G^\circ = -6 \times 96500 \times 2.02 = -1167690 \, \text{J/mol} \]
Converting to kilojoules:
\[ \Delta G^\circ = -1167.69 \, \text{kJ/mol} \approx -1170 \, \text{kJ/mol} \]

### Final Answer

Thus, the value of \(\Delta G^\circ\) for the reaction is:
\[ \Delta G^\circ \approx -1170 \, \text{kJ/mol} \]

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