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Saturday, 31 October

JEE Main 2021 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 3 August 2021, Shift 1, Chemistry Q56

Question 56 of 90 in this shift, Chemistry question 26 of 30, Section B.

The logarithm of equilibrium constant (log⁡10\log_{10}K) for the reaction Pd(aq)2++4Cl(aq)−⇌PdCl42−(aq)\mathrm{Pd^{2+}_{(aq)} + 4Cl^{-}_{(aq)} \rightleftharpoons PdCl_4^{2-}{}_{(aq)}} is xx × 10−110^{-1}. The value of xx is ________. (Nearest integer) [Given emf for the half-cell at 25ºC Pd(aq)2++2e−⟶Pd(s)\mathrm{Pd^{2+}_{(aq)} + 2e^{-} \longrightarrow Pd_{(s)}} Eº = 0.83V PdCl42−(aq)+2e−⟶Pd(s)+4Cl−\mathrm{PdCl_4^{2-}{}_{(aq)} + 2e^{-} \longrightarrow Pd(s) + 4Cl^{-}} Eº = 0.64V]

Official answer

64

NTA final key.