Calculate `Delta_(r)G^(@)` for the reaction :
`Mg(s)+Cu^(2+)(aq) to Mg^(2+)(aq)+Cu(s)`
[Given `E_(cell)^(@)=+2.71" V ", 1F=96500" C "`]

Explanation

To calculate the standard Gibbs free energy change (ΔG°) for the reaction:

\[ \text{Mg(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Mg}^{2+}(aq) + \text{Cu(s)} \]

we can use the relationship between Gibbs free energy change and cell potential:

\[ \Delta G° = -nFE°_{cell} \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (\( 96500 \, C/mol \))
- \( E°_{cell} \) = standard cell potential (given as \( +2.71 \, V \))

### Step-by-Step Solution:

1. **Identify the number of electrons transferred (n)**:
In the given reaction, magnesium (Mg) is oxidized to magnesium ions (Mg²⁺) and copper ions (Cu²⁺) are reduced to copper (Cu). Each magnesium atom loses 2 electrons, and each copper ion gains 2 electrons. Therefore, the number of electrons transferred, \( n \), is 2.

**Hint**: Look at the oxidation states of the reactants and products to determine how many electrons are involved.

2. **Use Faraday's constant (F)**:
The value of Faraday's constant is given as \( F = 96500 \, C/mol \).

**Hint**: Remember that Faraday's constant relates the amount of electric charge per mole of electrons.

3. **Substitute the values into the equation**:
Now, we can substitute the values into the equation:

\[ \Delta G° = -nFE°_{cell} \]

Substituting \( n = 2 \), \( F = 96500 \, C/mol \), and \( E°_{cell} = 2.71 \, V \):

\[ \Delta G° = -2 \times 96500 \, C/mol \times 2.71 \, V \]

4. **Calculate ΔG°**:
Performing the calculation:

\[ \Delta G° = -2 \times 96500 \times 2.71 = -524,830 \, J \]

Converting to kilojoules:

\[ \Delta G° = -524.83 \, kJ \]

5. **Final result**:
Rounding to three significant figures, we get:

\[ \Delta G° \approx -525 \, kJ \]

### Final Answer:
\[ \Delta G° \approx -525 \, kJ \]