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The number of moles of $\mathrm{KMnO}_4$ reduced by one mole of $\mathrm{KI}$ is alkaline medium is:
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Verified Answer
The correct answer is:
two
In alkaline medium:
Related Theory
In acidic medium: $\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow$ $\mathrm{Mn}_2+4 \mathrm{H}_2 \mathrm{O}$
In basic medium: $\mathrm{MnO}_4^{-}+e^{-} \rightarrow \mathrm{MnO}_4^{2-}$
In neutral medium: $\mathrm{MnO}_4^{-}+4 \mathrm{H}^{+}+3 e^{-} \longrightarrow$ $\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$
Related Theory
In acidic medium: $\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow$ $\mathrm{Mn}_2+4 \mathrm{H}_2 \mathrm{O}$
In basic medium: $\mathrm{MnO}_4^{-}+e^{-} \rightarrow \mathrm{MnO}_4^{2-}$
In neutral medium: $\mathrm{MnO}_4^{-}+4 \mathrm{H}^{+}+3 e^{-} \longrightarrow$ $\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$
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