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JEE Main 202126 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
26 February
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2021 · 26 February · Morning Shift · Chemistry · Electrochemistry

Consider the following reaction MnO _ 4 ^ - + 8 H ^ + + 5 e ^ - →...

Consider the following reaction

MnO4−+8H++5e−→Mn+2+4H2O,E°=1.51V{MnO}_{4}^{-} + 8{H}^{+} + 5{e}^{-} \to {Mn}^{+ 2} + 4{H}_{2}O, E ° = 1 . 51V

The quantity of electricity required in Faraday to reduce five moles of MnO4−{MnO}_{4}^{-} is

Correct answer

25

Step-by-step solution

For this balanced reaction, the change in oxidation number of the metal ion is Mn+7→Mn+2{Mn}^{+ 7} \to {Mn}^{+ 2}.

Therefore one mole MnMn required 55 Faraday

Hence, for five mole MnMn required 2525 Faraday.

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