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Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Class 12 Chemistry 35 Years JEE Mains &Advance Past yr Paper

JEE : Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

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Integer Value Correct Type

1. All the energy released from the reaction X → Y, Δ_rG° = –193 kJ mol^–1 is used for oxidizing M⁺ as M⁺ → M³⁺ + 2e^–, E° = –0.25 V
Under standard conditions, the number of moles of M⁺ oxidized when one mole of X is converted to Y is [F = 96500 C mol^–1] (JEE Adv. 2015)

Ans: 4

Solution :

Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

Hence DG° for oxidation will be
ΔG° = – nFE°
= –2 × 96500 × (–0.25) = 48250 J = 48.25 kJ
48.25 kJ energy oxidises one mole M⁺

∴ 193 kJ energy oxidises Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev mole M⁺ = 4 mole M⁺

2. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If the difference in their pK_a values, pK_a(HX) – pK_a (HY), is (consider degree of ionization of both acids to be <<1) (JEE Adv. 2015)

Ans: 3

Solution :

Integer Answer Type Questions: Electrochemistry | JEE Advanced Notes | EduRev

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