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NEET CHEMISTRYMedium

For the reduction of silver ions with copper metal, the standard cell potential was found to be +0.46 V+0.46 \text{ V} at 25∘C25^\circ\text{C}. The value of standard Gibbs energy, ΔG∘\Delta G^\circ will be: (F=96500 C mol−1F = 96500 \text{ C mol}^{-1})

A

−89.0 kJ-89.0 \text{ kJ}

B

−89.0 J-89.0 \text{ J}

C

−44.5 kJ-44.5 \text{ kJ}

D

−98.0 kJ-98.0 \text{ kJ}

Step-by-Step Solution

  1. Identify the Cell Reaction: The reduction of silver ions (Ag+Ag^+) by copper metal (CuCu) is represented by the balanced chemical equation: Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) In this reaction, Copper loses 2 electrons and two Silver ions gain 1 electron each. Therefore, the number of electrons transferred, n=2n = 2 .
  2. Formula: The relationship between standard Gibbs energy change (ΔrG∘\Delta_r G^\circ) and standard cell potential (Ecell∘E^\circ_{cell}) is given by: ΔrG∘=−nFEcell∘\Delta_r G^\circ = -nFE^\circ_{cell} Where FF is the Faraday constant .
  3. Calculation: n=2n = 2 F=96500 C mol−1F = 96500 \text{ C mol}^{-1}
  • Ecell∘=+0.46 VE^\circ_{cell} = +0.46 \text{ V} ΔrG∘=−2×96500 C mol−1×0.46 V\Delta_r G^\circ = -2 \times 96500 \text{ C mol}^{-1} \times 0.46 \text{ V} ΔrG∘=−193000×0.46 J mol−1\Delta_r G^\circ = -193000 \times 0.46 \text{ J mol}^{-1} ΔrG∘=−88780 J mol−1\Delta_r G^\circ = -88780 \text{ J mol}^{-1} Converting to kJ: ΔrG∘≈−88.8 kJ\Delta_r G^\circ \approx -88.8 \text{ kJ} Rounding to the nearest option gives −89.0 kJ-89.0 \text{ kJ}.
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