Q19 / 150 · MHT CET 2025 (21 Apr Shift 2) · Chemistry
Which of the following relations about $\mathrm{E}^{\circ}$ cell is false?
- A$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{(\text {cathode })}^{\circ}-\mathrm{E}_{(\text {anode })}^{\circ}$
- B$\mathrm{E}_{\text {cell }}^{\circ}=\frac{0.0592}{\mathrm{n}} \times \log _{10} \mathrm{~K}$
- Correct answerC$\mathrm{E}_{\text {cell }}^{\circ}=\frac{\Delta \mathrm{G}^{\circ}}{\mathrm{nF}}$
- D$\mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {(anode) }}^{\circ}+\mathrm{E}_{\text {(cathode) }}^{\circ}$
Solution
Determining the false electrochemical relation:
The standard formula for cell potential is $E_\text{cell}^\circ = E_\text{cathode}^\circ - E_\text{anode}^\circ$ where $E_\text{cathode}^\circ$ and $E_\text{anode}^\circ$ represent standard reduction potentials, so option A is correct.
At 298 K, the relation $E_\text{cell}^\circ = \frac{0.0592}{n} \times \log_{10} K$ derives from the Nernst equation at equilibrium and thermodynamic principles, making option B valid.
The thermodynamic relation between Gibbs free energy and cell potential is $\Delta G^\circ = -nFE_\text{cell}^\circ$. Rearranged, $E_\text{cell}^\circ = -\frac{\Delta G^\circ}{nF}$. Option C omits the negative sign and is therefore incorrect.
Option D represents $E_\text{cell}^\circ = E_\text{anode}^\circ + E_\text{cathode}^\circ$ where $E_\text{anode}^\circ$ is interpreted as oxidation potential. Since oxidation potential equals the negative of reduction potential, this simplifies to $E_\text{cathode}^\circ - E_\text{anode}^\circ$, matching option A and making D valid.
Only option C contains a fundamental error in sign convention.
The false relation is $\boxed{C}$.
Hence, the correct answer is option: 'C'.