JEE Main 2021 · ChemistryNumerical answerNumerical valueEasyCalculation

JEE Main 26 February 2021, Shift 1, Chemistry Q56

Question 56 of 90 in this shift, Chemistry question 26 of 30, Section B.

Consider the following reaction

\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{+2} + 4H_2O}

\mathrm{E^\circ} = 1.51

V. The quantity of electricity required in Faraday to reduce five moles of

\mathrm{MnO_4^-}

is __________. (Integer answer)

Official answer

NTA final key.

Chapter: Electrochemistry
Topic: Faraday's Laws of Electrolysis
Idea tested: Charge-Mole Relationship

Same idea in other shifts

Asked 4× in all

31 Jan 2024, Shift 1 · Q85One Faraday of electricity liberates $x\times10^{-1}$ gram atom of copper from copper sulphate. $x$ is _________.EasyNumerical value
26 Jul 2022, Shift 1 · Q87The amount of charge in F (Faraday) required to obtain one mole of iron from $\mathrm{Fe_3O_4}$ is ____________. (Nearest Integer)MediumNumerical value
28 Jun 2022, Shift 1 · Q87The quantity of electricity in Faraday needed to reduce 1 mol of $\mathrm{Cr_2O_7^{2-}}$ to $\mathrm{Cr^{3+}}$ is ________EasyNumerical 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 108 of the 174 JEE Main shifts held from 2017 to 2026.