Calculate emf and `DeltaG` for the following reaction at 298 K
`Mg(s)|Mg^(2+)(0.01 M)||Ag^(+)(0.0001 M)|Ag(s)`
Given `E_((Mg^(2+)//Mg))^(@)=-2.37" V ", E_((Ag^(+)//Ag))^(@)=+0.80" V "`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the half-reactions and determine the anode and cathode.
The given cell reaction is:
\[ \text{Mg}(s) | \text{Mg}^{2+}(0.01 \, M) || \text{Ag}^{+}(0.0001 \, M) | \text{Ag}(s) \]

- **Oxidation (Anode)**:
\[ \text{Mg}(s) \rightarrow \text{Mg}^{2+} + 2e^- \]
Standard reduction potential, \( E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37 \, V \)

- **Reduction (Cathode)**:
\[ \text{Ag}^{+} + e^- \rightarrow \text{Ag}(s) \]
Standard reduction potential, \( E^\circ_{\text{Ag}^{+}/\text{Ag}} = +0.80 \, V \)

### Step 2: Calculate the standard cell potential \( E^\circ_{\text{cell}} \).
Using the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

Substituting the values:
\[ E^\circ_{\text{cell}} = 0.80 \, V - (-2.37 \, V) = 0.80 \, V + 2.37 \, V = 3.17 \, V \]

### Step 3: Write the balanced overall cell reaction.
The overall reaction can be written as:
\[ \text{Mg}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Mg}^{2+}(aq) + 2\text{Ag}(s) \]

### Step 4: Determine the number of electrons transferred \( N \).
From the balanced equation, we see that 2 electrons are transferred:
\[ N = 2 \]

### Step 5: Use the Nernst equation to calculate the cell potential \( E_{\text{cell}} \).
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{N} \log \left( \frac{[\text{Mg}^{2+}]}{[\text{Ag}^{+}]^2} \right) \]

Substituting the concentrations:
- \([\text{Mg}^{2+}] = 0.01 \, M\)
- \([\text{Ag}^{+}] = 0.0001 \, M\)

Thus:
\[ E_{\text{cell}} = 3.17 \, V - \frac{0.0591}{2} \log \left( \frac{0.01}{(0.0001)^2} \right) \]

Calculating the logarithm:
\[ \log \left( \frac{0.01}{(0.0001)^2} \right) = \log \left( \frac{0.01}{0.00000001} \right) = \log(10^8) = 8 \]

Now substituting back:
\[ E_{\text{cell}} = 3.17 \, V - \frac{0.0591}{2} \times 8 \]
\[ E_{\text{cell}} = 3.17 \, V - 0.2364 \]
\[ E_{\text{cell}} = 2.9336 \, V \]

### Step 6: Calculate \( \Delta G \) using the formula.
The Gibbs free energy change \( \Delta G \) is given by:
\[ \Delta G = -NFE_{\text{cell}} \]

Where:
- \( F = 96500 \, C/mol \) (Faraday's constant)

Substituting the values:
\[ \Delta G = -2 \times 96500 \times 2.9336 \]
\[ \Delta G = -2 \times 96500 \times 2.9336 \approx -566,000 \, J/mol \]

Converting to kilojoules:
\[ \Delta G \approx -566 \, kJ/mol \]

### Final Answers:
- **EMF (E_cell)**: \( 2.9336 \, V \)
- **Delta G (ΔG)**: \( -566 \, kJ/mol \)