The number of moles of KMnO_4 reduced by 1 m | vedclass

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(B) In a basic medium,the reaction between $KI$ and $KMnO_4$ is given by: $I^- + 2MnO_4^- + H_2O \rightarrow IO_3^- + 2MnO_2 + 2OH^-$In this reaction:

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$2$

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(B) In a basic medium,the reaction between $KI$ and $KMnO_4$ is given by: $I^- + 2MnO_4^- + H_2O \rightarrow IO_3^- + 2MnO_2 + 2OH^-$In this reaction:- The oxidation state of $I$ changes from $-1$ to $+5$,which is an increase of $6$ units.- The oxidation state of $Mn$ changes from $+7$ to $+4$,which is a decrease of $3$ units per $Mn$ atom.To balance the electron transfer:$1 \ mol$ of $I^-$ loses $6 \ mol$ of electrons.$1 \ mol$ of $MnO_4^-$ gains $3 \ mol$ of electrons.Therefore,$1 \ mol$ of $I^-$ will reduce $6/3 = 2 \ mol$ of $MnO_4^-$.Thus,$1 \ mol$ of $KI$ reduces $2 \ mol$ of $KMnO_4$.

The number of moles of $KMnO_4$ reduced by $1 \ mol$ of $KI$ in basic medium is .............

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