The value of n in${\rm{MnO}}_4^ - + 8{{\rm{H}}^ + } + n{e^ - } \to {\rm{M}}{{\rm{n}}^{2 + }} + 4{{\rm{H}}_2}{\rm{O}}$ is

JEE CHEMISTRY MCQ: REDOX REACTIONS

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The value of n in

\mathrm{MnO}_4^ - + 8{\mathrm{H}^ + } + n{e^ - } \to \mathrm{M}{\mathrm{n}^{2 + }} + 4{\mathrm{H}_2}\mathrm{O}

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Correct option: A

Option A (5) is correct because the balanced half-reaction shows that 5 electrons are required to reduce MnO₄⁻ to Mn²⁺.

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\mathrm{H}

2 \mathrm{KClO}_3 \longrightarrow 2 \mathrm{KCl}+3 \mathrm{O}_2

\mathrm{HNO}_3

\mathrm{HNO}_3

\mathrm{MnO}_4^{-}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{MnO}_2(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})

\mathrm{Br}

-1

+5.