Standard electrode potentials for a few half cells are mentioned below :
$$E_{Cu^{2+}/Cu}^{\circ}=0.34 V,E_{Zn^{2+}/Zn}^{\circ}=-0.76 V\\E_{Ag^{+}/Ag}^{\circ}=0.80 V,E_{Mg^{2+}/Mg}^{\circ}=-2.37 V$$ Which one of the following cells gives the most negative value of $$\Delta G^{\circ}$$ ?

We need to find which cell gives the most negative $$\Delta G°$$.

Key formula: $$\Delta G° = -nFE°_{cell}$$. Most negative $$\Delta G°$$ requires largest $$nE°_{cell}$$.

Option 1: Zn|Zn²⁺||Ag⁺|Ag

$$E°_{cell} = E°_{cathode} - E°_{anode} = 0.80 - (-0.76) = 1.56$$ V

n = 2 (Zn → Zn²⁺ + 2e⁻, 2Ag⁺ + 2e⁻ → 2Ag)

$$nE° = 2 \times 1.56 = 3.12$$

Option 2: Zn|Zn²⁺||Mg²⁺|Mg

$$E°_{cell} = -2.37 - (-0.76) = -1.61$$ V (negative, not spontaneous)

Option 3: Ag|Ag⁺||Mg²⁺|Mg

$$E°_{cell} = -2.37 - 0.80 = -3.17$$ V (negative, not spontaneous)

Option 4: Cu|Cu²⁺||Ag⁺|Ag

$$E°_{cell} = 0.80 - 0.34 = 0.46$$ V

n = 2, $$nE° = 0.92$$

The largest $$nE°_{cell}$$ is for Option 1 (3.12), giving the most negative $$\Delta G°$$.

The correct answer is Option 1: Zn|Zn²⁺(1M)||Ag⁺(1M)|Ag.