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Byju's Answer

Standard XII

Chemistry

Reaction with Metals

MnO4- oxidise...

Question

${M n O_{4}}^{-}$ oxidises $M n^{2 +}$ to $M n O_{2}$ and itself reduces to $x$ in the titration method. How many oxygen atoms does $x$ contain?

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Solution

${M n O_{4}}^{-}$ oxidizes $M n^{2 +}$ to $M n O_{2}$ and itself reduces to $M n O_{2}$ in the titration method. This is an example of comproportionation reaction.
Therefore, $x$ is $M n O_{2}$ and contains $2$ oxygen atoms.

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Similar questions

M n O_{4}^{-} (a q)

can oxidise

I O_{3}^{-} (a q)

I O_{4}^{-} (a q)

and itself gets reduced to

M n O_{2} (s)

. If

x

moles of

I O_{3}^{-} (a q)

can reduce

0.1

mol of

M n O_{4}^{-} (a q)

, then calculate the value of

100 x

{M n O_{4}}^{-}

ions are reduced in acidic condition to

M n^{2 +}

ions whereas they are reduced in neutral condition to

M n O_{2}

. The oxidation of

25 m L

of a solution

X

containing

F e^{2 +}

ions required in acidic condition,

20 m L

of a solution

Y

containing

{M n O_{4}}^{-}

ions. What volume of solution

Y

would be required to oxidise

25 m L

of solution

X

containing

F e^{2 +}

ions in neutral condition?

100

mL of

1 M K M n O_{4}

oxidised

100

mL of

H_{2} O_{2}

in acidic medium (when

{M n O_{4}}^{-}

is reduced to

M n^{2 +}

). Volume of same

K M n O_{4}

required to oxidise

100

mL of

H_{2} O_{2}

in neutral medium (when

{M n O_{4}}^{-}

is reduced to

M n O_{2}

) will be:

Q. In the following reaction:

V O_{3}^{2 -} + M n O_{4}^{-} ⟶ {M n}^{2 +} + V O_{4}^{3 -}

, how many moles of

M n O_{4}

is required to oxidise

1

mol of

V O_{3}^{2 -}

Q. How many Faraday are required to reduce one mol of

{M n O}_{4}^{-}

{M n}^{2 +}

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MnO4− oxidises Mn2+ to MnO2 and itself reduces to x in the titration method. How many oxygen atoms does x contain?

MnO4− oxidizes Mn2+ to MnO2 and itself reduces to MnO2 in the titration method. This is an example of comproportionation reaction.Therefore, x is MnO2 and contains 2 oxygen atoms.

${M n O_{4}}^{-}$ oxidises $M n^{2 +}$ to $M n O_{2}$ and itself reduces to $x$ in the titration method. How many oxygen atoms does $x$ contain?

${M n O_{4}}^{-}$ oxidizes $M n^{2 +}$ to $M n O_{2}$ and itself reduces to $M n O_{2}$ in the titration method. This is an example of comproportionation reaction.
Therefore, $x$ is $M n O_{2}$ and contains $2$ oxygen atoms.