How many faraday are needed to reduce a mole | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The reduction of $1 \, 	ext{mole}$ of $MnO_4^-$ to $Mn^{2+}$ is represented by the following half-reaction:$MnO_4^- + 8H^+ + 5e^- ightarrow Mn^

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The reduction of $1 \, 	ext{mole}$ of $MnO_4^-$ to $Mn^{2+}$ is represented by the following half-reaction:$MnO_4^- + 8H^+ + 5e^- ightarrow Mn^{2+} + 4H_2O$In $MnO_4^-$,the oxidation state of $Mn$ is $+7$.In $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.The change in oxidation state is $7 - 2 = 5$.Since $5 \, 	ext{moles}$ of electrons are required for the reduction of $1 \, 	ext{mole}$ of $MnO_4^-$,the amount of electricity required is $5 \, 	ext{Faraday}$.

How many faraday are needed to reduce a mole of $MnO_4^-$ to $Mn^{2+}$?

Similar Questions

In the electrolysis of an aqueous $NiI_2$ solution using $Ni$ electrodes,what happens to the mass of the cathode upon passing $1$ equivalent of charge? (Given: atomic mass of $Ni = 59$)

The value of current required to deposit $0.972 \ g$ of chromium in $3 \ hours$ if the $E.C.E.$ of chromium is $0.00018 \ g \ C^{-1}$,is ............ $A$.

How many total Faraday currents will be required to obtain $1 \ mole$ of $Ag$,$Mg$,and $Al$ respectively?

The relation between Faraday constant $(F)$,chemical equivalent $(E)$,and electrochemical equivalent $(Z)$ is

$A$ certain current liberated $0.504 \ g$ of hydrogen in $2 \ hours$. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module