QJEEM02Chf15f: JEE Main 2002 Chemistry - Class 12 JEEM02Chf15f Solution: For a cell given below Ag + + e − ⟶ Ag ; E ∘ = x Cu 2 + + 2 e − …

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) :

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

At RHS (reduction) :

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

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$$2\mathrm{Ag}+{\mathrm{Cu}}^{2+}⟶\mathrm{Cu}+2{\mathrm{Ag}}^{+},E_{\text{cell }}^{\circ }=(y-x)$$
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$$\therefore$$ E°_cell = y - x

Therefore, the correct answer is :

Option C : y - x.