Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Consider the following reaction <img alt="MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{+2}+4H_{2}O, E^{o}=1.51 V." src="https://entrancecorner.codecogs.com/gif.latex?MnO_%7B4%7D%5E%7B-%7D&plus;8H%5E%7B&plus;%7D&plus;5e%5E%7B-%7D%5Crightarrow%20

Consider the following reaction MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{+2}+4H_{2}O, E^{o}=1.51 V. The quantity of electricity required in Faraday to reduce five moles of MnO_{4}^{-} is _________.(Integer answer)
Option: 1 25
Option: 2 -
Option: 3 -
Option: 4 -

Answers (1)

best_answer

\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{+2}+4 \mathrm{H}_{2} \mathrm{O}

1 mole of \mathrm{MnO}_{4}^{-} require 5 faraday charge
5 moles of \mathrm{MnO}_{4}^{-} will require 25 faraday charge.

Posted by

sudhir.kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 withheld results declared; NTA issues show cause notices ahead of action
anam-khan

JCECEB 2025 registration for BTech admission through JEE Main rank begins jceceb.jharkhand.gov.in
anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective

Latest Question