`E^(@) (SRP)` of different half cell given
`{:(E_(Cu^(2+)//Cu)^(@) =0.34"volt",,,E_(Zn^(2+)//Zn)^(@) =- 0.76"volt"),(E_(Ag^(+)//Ag)^(@) = 0.8"volt",,,E_(Mg^(2+)//Mg)^(@) =- 2.37 "volt"):}`
In which cell `Delta^(@)` is most negative:-

Options

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

Explanation

To determine which cell has the most negative ΔG° (Gibbs free energy change), we can use the relationship between ΔG° and the cell potential (E°cell):

\[ \Delta G° = -nFE°_{cell} \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction (we'll assume it to be 1 for simplicity unless specified otherwise)
- \( F \) = Faraday's constant (approximately 96485 C/mol)
- \( E°_{cell} \) = standard cell potential

### Step-by-Step Solution:

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

2. **Calculate E°cell for each possible cell configuration**:
- **Cell 1**: \( Zn \) oxidation and \( Mg^{2+} \) reduction
\[ E°_{cell1} = E_{Mg^{2+}/Mg} - E_{Zn^{2+}/Zn} = (-2.37) - (-0.76) = -2.37 + 0.76 = -1.61 \, \text{V} \]

- **Cell 2**: \( Zn \) oxidation and \( Ag^{+} \) reduction
\[ E°_{cell2} = E_{Ag^{+}/Ag} - E_{Zn^{2+}/Zn} = 0.80 - (-0.76) = 0.80 + 0.76 = 1.56 \, \text{V} \]

- **Cell 3**: \( Cu \) oxidation and \( Ag^{+} \) reduction
\[ E°_{cell3} = E_{Ag^{+}/Ag} - E_{Cu^{2+}/Cu} = 0.80 - 0.34 = 0.46 \, \text{V} \]

- **Cell 4**: \( Ag \) oxidation and \( Mg^{2+} \) reduction
\[ E°_{cell4} = E_{Mg^{2+}/Mg} - E_{Ag^{+}/Ag} = (-2.37) - 0.80 = -3.17 \, \text{V} \]

3. **Calculate ΔG° for each cell**:
- For all calculations, we will assume \( n = 1 \) and use \( F = 96485 \, \text{C/mol} \):
- **Cell 1**:
\[ \Delta G°_{cell1} = -1 \times 96485 \times (-1.61) = 155,000.85 \, \text{J/mol} \]
- **Cell 2**:
\[ \Delta G°_{cell2} = -1 \times 96485 \times (1.56) = -150,000.6 \, \text{J/mol} \]
- **Cell 3**:
\[ \Delta G°_{cell3} = -1 \times 96485 \times (0.46) = -44,400.1 \, \text{J/mol} \]
- **Cell 4**:
\[ \Delta G°_{cell4} = -1 \times 96485 \times (-3.17) = 305,000.45 \, \text{J/mol} \]

4. **Determine which ΔG° is most negative**:
- Comparing the ΔG° values:
- Cell 1: \( 155,000.85 \, \text{J/mol} \)
- Cell 2: \( -150,000.6 \, \text{J/mol} \)
- Cell 3: \( -44,400.1 \, \text{J/mol} \)
- Cell 4: \( 305,000.45 \, \text{J/mol} \)

The most negative ΔG° is for Cell 2, which is \( -150,000.6 \, \text{J/mol} \).

### Final Answer:
The cell with the most negative ΔG° is **Cell 2** (Zn oxidation and Ag+ reduction).