In alkaline condition ${\rm{KMn}}{{\rm{O}}_4}$ reacts as follows :$2{\rm{KMn}}{{\rm{O}}_4} + 2{\rm{KOH}} \to 2{{\rm{K}}_2}{\rm{Mn}}{{\rm{O}}_4} + {{\rm{H}}_2}{\rm{O}} + {\rm{O}}{\rm{.}}$ The eq. wt. of ${\rm{KMn}}{{\rm{O}}_4}$ is :

NEET UG CHEMISTRY MCQ: REDOX REACTIONS

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In alkaline condition

\mathrm{KMn}{\mathrm{O}_4}

reacts as follows :

2\mathrm{KMn}{\mathrm{O}_4} + 2\mathrm{KOH} \to 2{\mathrm{K}_2}\mathrm{Mn}{\mathrm{O}_4} + {\mathrm{H}_2}\mathrm{O} + \mathrm{O}\mathrm{.}

The eq. wt. of

\mathrm{KMn}{\mathrm{O}_4}

is :

View answer and explanation

Correct option: B

158

Option B (158) is correct because in alkaline conditions, KMnO4 is reduced to K2MnO4, involving a change of 1 in oxidation number.

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{I_2}

\mathrm{MnO}_4^ - + {\mathrm{C}_2}\mathrm{O}_4^{2 - } + {\mathrm{H}^ + } \to \mathrm{M}{\mathrm{n}^{2 + }} + \mathrm{C}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O}

\mathrm{MnO}_4^ -

{\mathrm{C}_2}\mathrm{O}_4^{2 - }

{\mathrm{H}^ + }