UPSC Combined Geo-Scientist, Mains · 2023 · Mains Paper II
Q1(i). E° of Na⁺/Na from ΔrH° and ΔrS°
Question
Calculate the standard reduction potential at 298 K for the reaction Na⁺(aq) + e⁻ → Na(s) with respect to standard hydrogen electrode. For the overall reaction ΔrH° = 240.1 kJ mol⁻¹; ΔrS° = −73.1 J K⁻¹ mol⁻¹.
Answer in brief
ΔrG° = +261.9 kJ mol⁻¹, so E°(Na⁺/Na) = −2.71 V (table value −2.71 V).
Key points
- Overall reaction against the SHE (n = 1): Na⁺(aq) + ½H₂(g) → Na(s) + H⁺(aq).
- ΔrG° = ΔrH° − TΔrS° = 240.1 − (298)(−0.0731) = +261.88 kJ mol⁻¹.
- E°SHE = 0, so E° = −ΔrG°/nF = −261 884/(1)(96 485) = −2.714 V.
- ΔrG° > 0, E° < 0: H₂ cannot reduce Na⁺; Na is a strong reducing agent, made by electrolysis of fused NaCl (Downs cell).
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversstandard reduction potentialGibbs energy and emfΔG = −nFEsodium electrodestandard hydrogen electrode
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