For a cell reaction, Pb(s) + Hg₂Cl₂(s) → PbCl₂(s) + 2Hg(l), ((∂ Eo)/(∂ T))P is 1.45 × 10⁻⁴ V K⁻¹. The entropy… · ABC26PH0700
Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Nernst Cells · Exam: JAM 2001-2022 · Marks: 2 · Difficulty: Medium
For a cell reaction, $\mathrm{Pb(s) + Hg_2Cl_2(s) \rightarrow PbCl_2(s) + 2Hg}(l)$, $\left(\frac{\partial E^o}{\partial T}\right)_P$ is $1.45 \times 10^{-4}\,\mathrm{V\,K^{-1}}$. The entropy change (in $\mathrm{J\,mol^{-1}\,K^{-1}}$) for the reaction is ____ [Given: 1F = 96500 C $\mathrm{mol^{-1}}$]
Answer
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