The cell notation indicates that the left side (Ag|Ag+) is the anode (oxidation occurs), and the right side (Cu²⁺|Cu) is the cathode (reduction occurs). In a galvanic (voltaic) cell, electrons flow from anode to cathode.

The standard cell potential (E° cell) is the difference in potential between the cathode and anode. It's calculated as :

E°_cell = E°_cathode - E°_anode

At LHS (oxidation) :

$$ 2 \times\left(\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+\mathrm{e}^{-}\right); \quad E_{\text {oxi }}^{\circ}=-x $$

At RHS (reduction) :

$$ \mathrm{Cu}^{2+}+2 e^{-} \longrightarrow \mathrm{Cu}; \quad \mathrm{E}_{\text {red }}^{\circ}=+y $$

$$ 2 \mathrm{Ag}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Cu}+2 \mathrm{Ag}^{+}, E_{\text {cell }}^{\circ}=(y-x) $$
_____________________________________________________

$$ \therefore $$ E°_cell = y - x

Therefore, the correct answer is :

Option C : y - x.