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Friday, 9 October

JEE Main 2025 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 8 April 2025, Shift 2, Chemistry Q75

Question 75 of 75 in this shift, Chemistry question 25 of 25, Section B.

Consider the following half cell reaction Cr2O72−(aq)+6e−+14H+(aq)⟶2Cr3+(aq)+7H2O(l)\mathrm{Cr_2O_7^{2-}(aq) + 6e^- + 14H^+(aq) \longrightarrow 2Cr^{3+}(aq) + 7H_2O(l)} The reaction was conducted with the ratio of [Cr3+]2[Cr2O72−]=10−6\frac{[\mathrm{Cr^{3+}}]^2}{[\mathrm{Cr_2O_7^{2-}}]} = 10^{-6}. The pH value at which the EMF of the half cell will become zero is __________. (nearest integer value) [Given : standard half cell reduction potential ECr2O72−,H+/Cr3+∘=1.33E^\circ_{\mathrm{Cr_2O_7^{2-},H^+/Cr^{3+}}} = 1.33V, 2.303RTF=0.059\frac{2.303RT}{F} = 0.059V.]

Official answer

10

NTA final key.

Same idea in other shifts

Asked 2× in all
  1. 8 Jan 2020, Shift 1 · Q48What would be the electrode potential for the given half cell reaction at pH=5 ? ______.…MediumNumerical value

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 42 of the 174 JEE Main shifts held from 2017 to 2026.