When ${\rm{KMn}}{{\rm{O}}_4}$ as oxidising agent and ultimately forms ${\rm{MnO}}_4^{2 - }, \, {\text{MnO}}_{2}, \, {\rm{M}}{{\rm{n}}_2}{{\rm{O}}_3} \, {\rm{and M}}{{\rm{n}}^{2 + }},$ the number of electrons transferred per mole of ${\rm{KMn}}{{\rm{O}}_4}$ each case respectively is:

JEE CHEMISTRY MCQ: REDOX REACTIONS

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Application +4 | -1

When

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

as oxidising agent and ultimately forms

\mathrm{MnO}_4^{2 - }, \, {\text{MnO}}_{2}, \, \mathrm{M}{\mathrm{n}_2}{\mathrm{O}_3} \, \mathrm{and M}{\mathrm{n}^{2 + }},

the number of electrons transferred per mole of

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

each case respectively is:

View answer and explanation

Correct option: C

1, 3, 4, 5

Option C (1, 3, 4, 5) is correct because it represents the number of electrons transferred per mole of KMnO4 in each reduction to MnO4^2-, MnO2, Mn2O3, and Mn²⁺ respectively.

More Questions from REDOX REACTIONS

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The oxidation number of

\mathrm{H}

in sodium hydride is A reagent, which can decrease the oxidation number of an element, is called Consider the following decomposition reaction

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

In the above reaction, Assertion (A):

\mathrm{HNO}_3

acts as oxidizing agent.

Reason (R): Oxidation no. of nitrogen is +5, no increase in oxidation no. beyond +5 can occur. The oxidation no. of

\mathrm{HNO}_3

can only decrease. Assertion :- Bromide ion is serving as a reducing agent in the reaction.

\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})

Reason :- Oxidation number of

\mathrm{Br}

increases from

-1

+5.

A substance, that by its sharp colour change indicates the completion of reaction is known as :