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Saturday, 31 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.