Consider the following ${\mathrm{E}}^{0}\mathrm{Value}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{half}\mathrm{cell}\ {{\mathrm{E}}^{0}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}=0.8\mathrm{V},{{\mathrm{E}}^{0}}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}=-0.76\mathrm{V}\ {{\mathrm{E}}^{0}}_{{\mathrm{Cu}}^{2+}/\mathrm{Cu}}=0.34\mathrm{V}{{\mathrm{E}}^{0}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}=-2.35\mathrm{V}\ \mathrm{Then}\mathrm{which}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{will}\mathrm{have}\mathrm{the}\mathrm{most}\mathrm{negative}\mathrm{value}\mathrm{of}∆{\mathrm{G}}^{0}?$ (Memory Based JEE Mains 23/01/2025 ,Shift -2)

Question

YESPYQ · Accepted Answer

Correct answer: b) $Mg|M{g}^{2+}∥A{g}^{+}|Ag$. The cell which has most positive value of will have most negative value of. ${{\mathrm{E}}_{\mathrm{cell}}^{\mathrm{o}}={\mathrm{E}}^{0}}_{{\mathrm{Ag}}^{+}/\mathrm{Ag}}-{{\mathrm{E}}^{0}}_{{\mathrm{Mg}}^{2+}/\mathrm{Mg}}\ =0.8\mathrm{V}-(-2.35\mathrm{V})=3.15\mathrm{V}\ ∆{\mathrm{G}}^{0}=-{\mathrm{nFE}}_{\mathrm{cell}}^{\mathrm{o}}$

Consider the following \({\mathrm{E}}^{0}\mathrm{Value}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{half}\mathrm{cell}\…

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