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Saturday, 31 October

JEE Main 2021 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 31 August 2021, Shift 1, Chemistry Q55

Question 55 of 90 in this shift, Chemistry question 25 of 30, Section B.

Consider the following cell reaction Cd(s)+Hg2SO4(s)+95H2O(l)⇌CdSO4⋅95H2O(s)+2Hg(l)\mathrm{Cd_{(s)}+Hg_2SO_{4(s)}+\frac{9}{5}H_2O_{(\mathit{l})}\rightleftharpoons CdSO_4\cdot\frac{9}{5}H_2O_{(s)}+2Hg_{(\mathit{l})}}. The value of Ecell0\mathrm{E^0_{cell}} is 4.315 V at 25°C. If ΔH∘=−825.2\Delta\mathrm{H}^\circ=-825.2 kJ mol−1^{-1}, the standard entropy change ΔS∘\Delta\mathrm{S}^\circ in J K−1^{-1} is __________. (Nearest integer) [Given : Faraday constant=96487=96487 C mol−1^{-1}]

Official answer

25

NTA final key.

Same idea in other shifts

Asked 2× in all
  1. 26 Jun 2022, Shift 1 · Q65The (∂E∂T)P\left(\frac{\partial E}{\partial T}\right)_P of different types of half cells are as follows : A: 1×10−41\times10^{-4} B:…MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.