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Byju's Answer

Standard XII

Chemistry

Oxidation Number Method

0.1 M of Mn...

Question

$0.1 M$ of $M n O_{4}^{-}$ (in acidic medium) can not oxidise completely:
(Assume the volume of reducing agent is same as of the $M n O_{4}^{-}$ solution)

0.25 M C_{2} O_{4}^{2 -}

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0.5 M F e^{2 +}

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0.166 M F e C_{2} O_{4}

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0.6 M C r_{2} O_{7}^{2 -}

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Solution

The correct option is C $0.6 M C r_{2} O_{7}^{2 -}$
Both $K M n O_{4}$ and $K_{2} C r_{2} O_{7}$ are strong oxidizing agents. In both these compounds, the central metal atom is present in the highest oxidation state and it cannot be oxidized further. Hence, $K M n O_{4}$ cannot completely oxidize $K_{2} C r_{2} O_{7}$ .

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Similar questions

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:

F e^{2 +}

as well as

H N O_{2}

can be used to estimate

M n O_{4}^{-}

in acidic medium. If 100 mL of 1 M

F e^{2 +}

can be oxidised by

V

mL of 1 M

M n O_{4}^{-} / H^{+}

solution, calculate volume of 1 M

H N O_{2}

that can be oxidises by

V

mL of 1 M

M n O_{4}^{-} / H^{+}

solution.

Q. Potassium acid oxalate

K_{2} C_{2} O_{4} .3 H_{2} C_{2} O_{4} .4 H_{2} O

can be oxidized by

M n O_{4}^{-}

in acidic medium. Calculate the volume of

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

reacting in acid solution with

10.16 g

of the acid oxalate.
(Molar mass of potassium acid oxalate is

508 g m o l^{- 1}

Q. Assertion :

1

mol of

F e C_{2} O_{4}

is oxidised by

0.6

mol of

M n O_{4}^{-}

in acidic medium. Reason:

M n O_{4}^{-}

oxidises both

{F e}^{2 +}

as well as

C_{2} O_{4}^{2 -}

{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?

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0.1M of MnO−4 (in acidic medium) can not oxidise completely:(Assume the volume of reducing agent is same as of the MnO−4 solution)

The correct option is C 0.6MCr2O2−7Both KMnO4 and K2Cr2O7 are strong oxidizing agents. In both these compounds, the central metal atom is present in the highest oxidation state and it cannot be oxidized further. Hence, KMnO4 cannot completely oxidize K2Cr2O7.

$0.1 M$ of $M n O_{4}^{-}$ (in acidic medium) can not oxidise completely:
(Assume the volume of reducing agent is same as of the $M n O_{4}^{-}$ solution)