`E^(c-)._(red)` of different half cell are given as `:`
`E^(c-)._(Cu^(2+)|Cu)=0.34V,E^(c-)._(Zn^(2+)|Zn)=-0.76V`.
`E^(c-)._(Ag^(o+)|Ag)=0.80 V, E^(c-)._(Mg^(2+)|Mg)=-2.37V.`
In which cell `DeltaG^(c-)` is most negative ?

Options

A. `Zn|Zn^(2+)(1M)||Mg^(2+)(1M)|Mg`
B. `Zn|Zn^(2+)(1M)||Ag^(o+)(1M)|Ag`
C. `Cu|Cu^(2+)(1M)||Ag^(o+)(1M)|Ag`
D. `Ag|Ag^(o+)(1M)||Mg^(2+)(1M)|Mg`

Explanation

To determine which cell has the most negative ΔG, we can use the relationship between ΔG and the cell potential (E) given by the equation:

\[ \Delta G = -nFE \]

Where:
- \( \Delta G \) is the change in Gibbs free energy,
- \( n \) is the number of moles of electrons transferred,
- \( F \) is Faraday's constant (approximately 96485 C/mol),
- \( E \) is the cell potential in volts.

To find the cell with the most negative ΔG, we need to calculate the cell potential (E) for each half-cell reaction and then determine which corresponds to the highest (most positive) E value.

### Step-by-Step Solution:

1. **Identify the half-reactions and their standard reduction potentials:**
- \( E^\circ (Cu^{2+}/Cu) = 0.34 \, V \)
- \( E^\circ (Zn^{2+}/Zn) = -0.76 \, V \)
- \( E^\circ (Ag^{+}/Ag) = 0.80 \, V \)
- \( E^\circ (Mg^{2+}/Mg) = -2.37 \, V \)

2. **Calculate the cell potential for each possible combination:**
- **Cell 1: Zn oxidized and Mg reduced**
- \( E = E^\circ (Mg^{2+}/Mg) - E^\circ (Zn^{2+}/Zn) \)
- \( E = (-2.37) - (-0.76) = -2.37 + 0.76 = -1.61 \, V \)

- **Cell 2: Zn oxidized and Ag reduced**
- \( E = E^\circ (Ag^{+}/Ag) - E^\circ (Zn^{2+}/Zn) \)
- \( E = 0.80 - (-0.76) = 0.80 + 0.76 = 1.56 \, V \)

- **Cell 3: Cu oxidized and Ag reduced**
- \( E = E^\circ (Ag^{+}/Ag) - E^\circ (Cu^{2+}/Cu) \)
- \( E = 0.80 - 0.34 = 0.46 \, V \)

- **Cell 4: Ag oxidized and Mg reduced**
- \( E = E^\circ (Mg^{2+}/Mg) - E^\circ (Ag^{+}/Ag) \)
- \( E = (-2.37) - 0.80 = -3.17 \, V \)

3. **Determine ΔG for each cell:**
- For Cell 1: \( \Delta G = -nF(-1.61) \) (positive)
- For Cell 2: \( \Delta G = -nF(1.56) \) (negative)
- For Cell 3: \( \Delta G = -nF(0.46) \) (negative)
- For Cell 4: \( \Delta G = -nF(-3.17) \) (positive)

4. **Identify the cell with the most negative ΔG:**
- The most negative ΔG corresponds to the cell with the highest positive E value.
- Cell 2 has the highest E value of 1.56 V, which means it has the most negative ΔG.

### Conclusion:
The cell with the most negative ΔG is the one where Zn is oxidized and Ag is reduced.