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NEET 2024
Exam
NEET
Year
2024
Subject
Chemistry
Chapter
Electrochemistry

NEET 2024 · Chemistry · Electrochemistry

The standard cell potential of the following cell Zn|Zn^ 2+ (aq)|...

The standard cell potential of the following cell Zn∣Zn2+(aq)∣Fe2+(aq)∣Fe\mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq})\right| \mathrm{Fe}^{2+}(\mathrm{aq}) \mid \mathrm{Fe} is 0.32 V . Calculate the standard Gibbs energy change for the reaction : Zn(s)+Fe2+(aq)→Zn2+(aq)+Fe(s)\mathrm{Zn}(\mathrm{s})+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Fe}(\mathrm{s}) (Given : 1 F=96487C1 \mathrm{~F}=96487 \mathrm{C} )

Options

  1. A−61.75 kJ mol−1-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}
  2. B+5.006 kJ mol−1+5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}
  3. C−5.006 kJ mol−1-5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}
  4. D+61.75 kJ mol−1+61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}

Correct answer

A. −61.75 kJ mol−1-61.75 \mathrm{~kJ} \mathrm{~mol}^{-1}

Step-by-step solution

ΔrG⊖=−nFEcell ⊖\Delta_r G^{\ominus}=-n F E_{\text{cell }}^{\ominus} For the given reaction, n=2∴ΔrG⊖=−2×96487×0.32=−61751.68 J mol−1=−61.751 kJ mol−1n=2 \therefore \quad \Delta_{\mathrm{r}} \mathrm{G}^{\ominus}=-2 \times 96487 \times 0.32 =-61751.68 \mathrm{~J} \mathrm{~mol}^{-1}=-61.751 \mathrm{~kJ} \mathrm{~mol}^{-1}
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