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NEET 2026
Exam
NEET
Year
2026
Subject
Chemistry
Chapter
Chapter

NEET 2026 · Chemistry · Chapter

Consider the following electrochemical cell reaction: Cr(s) + 3Ag^...

Consider the following electrochemical cell reaction:
Cr(s)+3Ag+(aq)⟶Cr3+(aq)+3Ag(s)\mathrm{Cr}(\mathrm{s}) + 3\mathrm{Ag}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cr}^{3+}(\mathrm{aq}) + 3\mathrm{Ag}(\mathrm{s})
If the standard cell potential (Ecell∘E^{\circ}_{\text{cell}}) is 1.54 V1.54 \text{ V}, determine the standard Gibbs free energy change (ΔG∘\Delta G^{\circ}) for the process.
(Given: 1 F=96487 C mol−11 \text{ F} = 96487 \text{ C mol}^{-1})

Options

  1. A−148.59 kJ mol−1-148.59 \text{ kJ mol}^{-1}
  2. B−445.77 J mol−1-445.77 \text{ J mol}^{-1}
  3. C−445.77 kJ mol−1-445.77 \text{ kJ mol}^{-1}
  4. D+445.77 kJ mol−1+445.77 \text{ kJ mol}^{-1}

Correct answer

C. −445.77 kJ mol−1-445.77 \text{ kJ mol}^{-1}

Step-by-step solution

The standard Gibbs free energy change is given by the relation:

ΔG∘=−nFEcell∘\Delta G^{\circ} = -nFE^{\circ}_{\text{cell}}

For the given reaction, the number of electrons transferred is n=3n = 3.

Substituting the given values:

ΔG∘=−3×96487 C mol−1×1.54 V\Delta G^{\circ} = -3 \times 96487 \text{ C mol}^{-1} \times 1.54 \text{ V}

ΔG∘=−445769.94 J mol−1\Delta G^{\circ} = -445769.94 \text{ J mol}^{-1}

ΔG∘≈−445.77 kJ mol−1\Delta G^{\circ} \approx -445.77 \text{ kJ mol}^{-1}

Answer: −445.77 kJ mol−1-445.77 \text{ kJ mol}^{-1}
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