Charge required for reduction of 0.01 mol of | vedclass

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(D) In a neutral medium,the reduction of permanganate ion $(MnO_4^-)$ to manganese dioxide $(MnO_2)$ is represented by the following half-reaction: $M

Vedclass · Accepted Answer

$2895$

Vedclass · Answer

(D) In a neutral medium,the reduction of permanganate ion $(MnO_4^-)$ to manganese dioxide $(MnO_2)$ is represented by the following half-reaction: $MnO_4^- + 2H_2O + 3e^- ightarrow MnO_2 + 4OH^-$. From the stoichiometry of the reaction,$1 \ mol$ of $MnO_4^-$ requires $3 \ mol$ of electrons for reduction. Therefore,$0.01 \ mol$ of $MnO_4^-$ requires $0.01 	imes 3 = 0.03 \ mol$ of electrons. The charge $Q$ is given by $Q = n 	imes F$,where $n$ is the number of moles of electrons and $F$ is Faraday's constant $(96500 \ C/mol)$. $Q = 0.03 \ mol 	imes 96500 \ C/mol = 2895 \ C$.

Charge required for reduction of $0.01 \ mol$ of $MnO_4^-$ in neutral medium is ............ $C$.

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