UPSC Combined Geo-Scientist, Mains · 2026 · Mains Paper II
Q13(a). ΔᵣG, ΔᵣH and ΔᵣS of the cell reaction
Question
At 298 K, the galvanic cell: (Pt) H₂(g, 1 atm) | HCl(aq, 0.1 M) | AgCl | Ag, has the EMF (E) =0.35 V and ((dE)/(dT))_P=-18×10⁻⁴ V K⁻¹. Determine ΔᵣG, ΔᵣH and ΔᵣS of the cell reaction for the production of 1 faraday electricity.
Answer in brief
ΔᵣG=-33.8 kJ, ΔᵣS=-173.7 J K⁻¹, ΔᵣH=-85.5 kJ (per mole of electrons, n=1).
Key points
- ΔᵣG=-nFE=-(1)(96485)(0.35)=-3.38×10⁴ J=-33.8 kJ
- ΔᵣS=nF((dE)/(dT))_P=(96485)(-18×10⁻⁴)=-173.7 J K⁻¹
- ΔᵣH=ΔᵣG+TΔᵣS=-33.77+(298)(-0.1737)=-33.77-51.75=-85.5 kJ
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
n=1 faraday: 12H₂(g)+AgCl(s)(s)+H+(aq)+Cl-(aq).
- ΔᵣG=-nFE=-(1)(96485)(0.35)=-3.38×10⁴ J=-33.8 kJ
- ΔᵣS=nF((dE)/(dT))_P=(96485)(-18×10⁻⁴)=-173.7 J K⁻¹
- ΔᵣH=ΔᵣG+TΔᵣS=-33.77+(298)(-0.1737)=-33.77-51.75=-85.5 kJ
The heat exchanged when the cell works reversibly is qᵣₑᵥ=TΔᵣS=-51.8 kJ: the cell gives out heat, and the maximum electrical work (33.8 kJ) is less than -ΔᵣH (85.5 kJ).
Coversthermodynamics of cellsΔᵣGΔᵣH and ΔᵣS of the cell reaction
All parts of Question 13
- 13(a) ΔᵣG, ΔᵣH and ΔᵣS of the cell reaction10 marks
- 13(b) Buffer capacity of an acetic acid–acetate buffer10 marks