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The standard cell potential of the following cell Zn | Zn^2+( aq) | Fe^2+( aq) mid Fe is 0.32 V . Calculate the standard Gibbs energy change for the…

NEET UG 2024 — Chemistry Physical Chemistry

2024mcqmedium

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}Zn​Zn2+(aq)​Fe2+(aq)∣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})Zn(s)+Fe2+(aq)→Zn2+(aq)+Fe(s) (Given : 1 F=96487C1 \mathrm{~F}=96487 \mathrm{C}1 F=96487C )

Official previous-year question

Held on 30 Apr 2024 · Verified 9 Jul 2026.

Options

  1. A

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

  2. B

    +5.006 kJ mol−1+5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}+5.006 kJ mol−1

  3. C

    −5.006 kJ mol−1-5.006 \mathrm{~kJ} \mathrm{~mol}^{-1}−5.006 kJ mol−1

  4. D

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

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