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ChemistryNumerical+3Question 32 of 54

JEE Advanced 2018 · Paper 2, Chemistry question 14

Consider an electrochemical cell: A(s) ∣ An+(aq, 2 M) ∣∣ B2n+(aq, 1 M) ∣ B(s)\mathrm{A(s)\,|\,A^{n+}(aq,\ 2\ M)\,||\,B^{2n+}(aq,\ 1\ M)\,|\,B(s)}. The value of ΔH⊖\Delta H^\ominus for the cell reaction is twice that of ΔG⊖\Delta G^\ominus at 300 K. If the emf of the cell is zero, the ΔS⊖\Delta S^\ominus (in J K−1^{-1} mol−1^{-1}) of the cell reaction per mole of B formed at 300 K is ____.

(Given: ln(2) = 0.7, RR (universal gas constant) = 8.3 J K−1^{-1} mol−1^{-1}. HH, SS and GG are enthalpy, entropy and Gibbs energy, respectively.)

Answer and worked solution

Official answer: -11.82

Cell reaction, per mole of B:
2 A+BX2n+→2 AXn++B(2n electrons).\ce{2A + B^{2n+} -> 2A^{n+} + B}\qquad(2n\text{ electrons}).

Zero emf. E=0E=0 means ΔG=0\Delta G=0, so
ΔG∘=−RTln⁡Q=−RTln⁡[AXn+]2[BX2n+]=−RTln⁡4.\Delta G^\circ=-RT\ln Q=-RT\ln\frac{[\ce{A^{n+}}]^2}{[\ce{B^{2n+}}]}=-RT\ln4.

Entropy. ΔH∘=2ΔG∘\Delta H^\circ=2\Delta G^\circ, so
ΔS∘=ΔH∘−ΔG∘T=ΔG∘T=−Rln⁡4=−8.3×1.4=−11.62 J K−1 mol−1.\Delta S^\circ=\frac{\Delta H^\circ-\Delta G^\circ}T=\frac{\Delta G^\circ}T=-R\ln4=-8.3\times1.4=-11.62\ \mathrm{J\,K^{-1}\,mol^{-1}}.

Answer: −11.62 (key: −11.82 to −11.42)

Topics:Gibbs energy and EMFNernst equationGibbs energy and spontaneity
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