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Friday, 9 October

JEE Main 2020 · ChemistryNumerical answerNumerical valueMediumCalculation

JEE Main 8 January 2020, Shift 1, Chemistry Q48

Question 48 of 75 in this shift, Chemistry question 23 of 25, Section B.

What would be the electrode potential for the given half cell reaction at pH=5 ? ______. 2H2O→O2+4H⊕+4e−2\mathrm{H_2O} \rightarrow \mathrm{O_2} + 4\mathrm{H}^{\oplus} + 4e^{-} ; Ered0=1.23 VE^0_{red} = 1.23\ \mathrm{V} (R=8.314 J mol−1 K−1R = 8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}; Temp =298 K= 298\ \mathrm{K}; oxygen under std. atm. pressure of 1 bar)

Official answer

-0.93 to -0.94

NTA final key (Jan 2020). Any value in the range scored.

Chapter
Electrochemistry
Topic
Nernst Equation
Idea tested
pH Dependence of Electrode Potential

Same idea in other shifts

Asked 2× in all
  1. 8 Apr 2025, Shift 2 · Q75Consider 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…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.