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Friday, 9 October

JEE Main 2021 · ChemistryNumerical answerNumerical valueHardCalculation

JEE Main 25 July 2021, Shift 1, Chemistry Q58

Question 58 of 90 in this shift, Chemistry question 28 of 30, Section B.

Consider the cell at 25∘^\circC Zn∣Zn2+(aq),(1 M) ∣∣ Fe3+(aq),Fe2+(aq)∣Pt(s)\mathrm{Zn|Zn^{2+}(aq),(1\ M)\,||\,Fe^{3+}(aq),Fe^{2+}(aq)|Pt(s)} The fraction of total iron present as Fe3+\mathrm{Fe^{3+}} ion at the cell potential of 1.500 V is x×10−2x\times10^{-2}. The value of xx is _________. (Nearest integer) (Given : EFe3+/Fe2+0=0.77E^0_{\mathrm{Fe^{3+}/Fe^{2+}}}=0.77 V, EZn2+/Zn0=−0.76E^0_{\mathrm{Zn^{2+}/Zn}}=-0.76 V)

Official answer

24

NTA final key.