Verified Solution • 2025–26 Session moderate

QJEEM25Ch739d: JEE Main 2025 Chemistry - Class 12 JEEM25Ch739d Solution: Standard electrode potentials for a few half cells are…

Standard electrode potentials for a few half cells are mentioned below :

amp;ECu2+/Cu∘=0.34 V,EZn2+/Zn∘=−0.76 Vamp;EAg+/Ag∘=0.80 V,EMg2+/Mg∘=−2.37 V\begin{aligned}& \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V}, \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V} \\& \mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}=0.80 \mathrm{~V}, \mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2.37 \mathrm{~V}\end{aligned}

Which one of the following cells gives the most negative value of ΔG∘\Delta \mathrm{G}^{\circ} ?

Multiple Choice Options

  • A Zn∣Zn2+(1M)∥Ag+(1M)∣Ag\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Ag}^{+}(1 \mathrm{M})\right| \mathrm{Ag}
  • B Ag∣Ag+(1M)∣∣Mg2+(1M)∣Mg\mathrm{Ag}\left|\mathrm{Ag}^{+}(1 \mathrm{M})\right|\left|\mathrm{Mg}^{2+}(1 \mathrm{M})\right| \mathrm{Mg}
  • C Zn∣Zn2+(1M)∣∣Mg2+(1M)∣Mg\mathrm{Zn}\left|\mathrm{Zn}^{2+}(1 \mathrm{M})\right|\left|\mathrm{Mg}^{2+}(1 \mathrm{M})\right| \mathrm{Mg}
  • D Cu∣Cu2+(1M)∣∣Ag+(1M)∣Ag\mathrm{Cu}\left|\mathrm{Cu}^{2+}(1 \mathrm{M})\right|\left|\mathrm{Ag}^{+}(1 \mathrm{M})\right| \mathrm{Ag}

Step-by-Step Verified Solution

∵ΔG∘=−nFE∘ Option (1)E∘=0.8+0.76=1.56 V∴ΔG∘=−2×F×1.56=−3.12 V Option (2) E∘=−2.37+0.76=−1.61 V∴ΔG∘=−2×F×(−1.61)=+3.22 V Option (3) E∘=−2.37−0.8=−3.17 V∴ΔG∘=−2=F×(−3.17)=+6.34 Option (4)E∘=0.8−0.34=0.46 VΔG∘=−2×F×0.46=−0.92 V\begin{gathered}\because \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\\text { Option }(1) \mathrm{E}^{\circ}=0.8+0.76 \\=1.56 \mathrm{~V} \\\therefore \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times 1.56 \\=-3.12 \mathrm{~V} \\\text { Option (2) } \mathrm{E}^{\circ}=-2.37+0.76 \\=-1.61 \mathrm{~V} \\\therefore \Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times(-1.61) \\=+3.22 \mathrm{~V} \\\text { Option (3) } \mathrm{E}^{\circ}=-2.37-0.8 \\=-3.17 \mathrm{~V} \\\begin{array}{c}\therefore \Delta \mathrm{G}^{\circ}=-2\end{array} \\=\mathrm{F} \times(-3.17) \\=+6.34 \\\text { Option }(4) \mathrm{E}^{\circ}=0.8-0.34 \\=0.46 \mathrm{~V} \\\Delta \mathrm{G}^{\circ}=-2 \times \mathrm{F} \times 0.46 \\=-0.92 \mathrm{~V}\end{gathered}

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