`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 \(\Delta G^\circ\) from the given standard reduction potentials (SRP), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Standard Reduction Potentials**:
- \(E^\circ (Cu^{2+}/Cu) = 0.34 \, \text{V}\)
- \(E^\circ (Zn^{2+}/Zn) = -0.76 \, \text{V}\)
- \(E^\circ (Ag^+/Ag) = 0.80 \, \text{V}\)
- \(E^\circ (Mg^{2+}/Mg) = -2.37 \, \text{V}\)

2. **Understand the Relationship Between \(\Delta G^\circ\) and \(E^\circ_{\text{cell}}\)**:
- The relationship is given by the formula:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
- Here, \(n\) is the number of moles of electrons transferred, and \(F\) is Faraday's constant (approximately \(96500 \, \text{C/mol}\)). Since \(n\) and \(F\) are constants, \(\Delta G^\circ\) is directly dependent on \(E^\circ_{\text{cell}}\).

3. **Calculate \(E^\circ_{\text{cell}}\) for Each Couple**:
- **Option 1: Zinc-Magnesium Couple**:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-2.37) - (-0.76) = -2.37 + 0.76 = -1.61 \, \text{V} \]
- **Option 2: Zinc-Silver Couple**:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (0.80) - (-0.76) = 0.80 + 0.76 = 1.56 \, \text{V} \]
- **Option 3: Copper-Silver Couple**:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (0.80) - (0.34) = 0.80 - 0.34 = 0.46 \, \text{V} \]
- **Option 4: Silver-Magnesium Couple**:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-2.37) - (0.80) = -2.37 - 0.80 = -3.17 \, \text{V} \]

4. **Determine \(\Delta G^\circ\) for Each Cell**:
- The most negative \(\Delta G^\circ\) corresponds to the most positive \(E^\circ_{\text{cell}}\):
- For **Zinc-Magnesium**: \(\Delta G^\circ = -nF(-1.61)\) (positive)
- For **Zinc-Silver**: \(\Delta G^\circ = -nF(1.56)\) (negative)
- For **Copper-Silver**: \(\Delta G^\circ = -nF(0.46)\) (negative)
- For **Silver-Magnesium**: \(\Delta G^\circ = -nF(-3.17)\) (positive)

5. **Conclusion**:
- The zinc-silver couple has the highest \(E^\circ_{\text{cell}}\) value of \(1.56 \, \text{V}\), leading to the most negative \(\Delta G^\circ\).
- Therefore, the correct answer is **Option 2: Zinc-Silver Couple**.