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Friday, 9 October

JEE Main 2025 · ChemistryNumerical answerNumerical valueHardMulti-step

JEE Main 2 April 2025, Shift 1, Chemistry Q71

Question 71 of 75 in this shift, Chemistry question 21 of 25, Section B.

Consider the following electrochemical cell at standard condition. Au(s)∣QH2,Q∣NH4X(0.01 M)∣∣Ag+(1 M)∣Ag(s)\mathrm{Au(s)|QH_2,Q|NH_4X(0.01\ M)||Ag^+(1\ M)|Ag(s)} Ecell=+0.4E_{cell}=+0.4 V The couple QH2/Q\mathrm{QH_2/Q} represents quinhydrone electrode, the half cell reaction is given below : [structure: Q (p-benzoquinone)] +2e−+2H+→+2e^-+2H^+\rightarrow [structure: QH2\mathrm{QH_2} (hydroquinone)] EQ/QH2∘=+0.7E^\circ_{Q/QH_2}=+0.7 V [Given : EAg+/Ag∘=+0.8E^\circ_{Ag^+/Ag}=+0.8 V and 2.303RTF=0.06\frac{2.303RT}{F}=0.06 V] The pKb\mathrm{pK_b} value of the ammonium halide salt (NH4X\mathrm{NH_4X}) used here is ___________. (nearest integer)

The figure, in words

Half-cell reaction: p-benzoquinone (Q, six-membered ring with two C=O at para positions) + 2e- + 2H+ -> hydroquinone (QH2, benzene ring with two OH at para positions); E°(Q/QH2) = +0.7 V.

Official answer

6

NTA final key.

Chapter
Electrochemistry
Topic
Nernst Equation
Idea tested
Quinhydrone Electrode for pH

Same topic in other shifts

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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. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.