The reaction of aqueous KM nO 4 with H 2 O 2 in acidic conditions givesA- Mn 4- and O 2B- Mn 2- and O 2C- Mn 2- and O 3D- Mn 4- and MnO 2

The correct option is B Mn^2+ and O_2 2KMnO_4+ 3H_2 SO_4+5H_2 O_2 → K_2 SO_4 + 2 MnSO_4 + 8H_2 O + 5O_2 ...

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Byju's Answer

Standard XII

Chemistry

Oxoacids of Halogens

The reaction ...

Question

The reaction of aqueous $K M n O_{4}$ with $H_{2} O_{2}$ in acidic conditions gives

$M n^{4 +}$ and $O_{2}$

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$M n^{2 +}$ and $O_{2}$

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$M n^{2 +}$ and $O_{3}$

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$M n^{4 +}$ and $M n O_{2}$

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Solution

The correct option is B
$M n^{2 +}$ and $O_{2}$

$2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} O_{2} \to K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O + 5 O_{2}$

Suggest Corrections

Similar questions

Q. Give

E_{M n^{7 +} | M n^{2 +}}^{0} = 1.5 V a n d E_{M n^{4 +} | M n^{2 +}}^{0} = 1.2 V, t h e n E_{M n^{7 +} | M n^{4 +}}^{0}

is:

Q. The

E^{o}

for,

M n^{4 +} + 2 e \to M n^{2 +}

is equal to:

Q. The reaction of aqueous

K M n O_{4}

with

H_{2} O_{2}

in acidic conditions gives:

E_{R P}^{o}

and their respective half reactions for some changes are given below:

(a)

M n O_{4}^{-} (a q) + 8 H^{+} + 5 e \to M n^{2 +} (a q) + 4 H_{2} O (l); E^{o} = + 1.52 V

(b)

M n O_{4}^{2 -} (a q) + 4 H^{+} + 2 e \to M n O_{2} (s) + 2 H_{2} O (l); E^{o} = + 2.26 V

(c)

2 M n O_{4}^{-} (a q) + 2 e \to 2 M n O_{4}^{2 -} (a q); E^{o} = + 0.56 V

(acid med.)
(d)

M n^{3 +} + e \to M n^{2 +}; E^{o} = + 1.51 V

(acid med.)
(e)

M n^{4 +} + e \to M n^{3 +}; E^{o} = 0.95 V

(acid med.)

The most stable oxidation state of Mn is :

100

mL of

0.01 M

K M n O_{4}

oxidized

100

mL of

H_{2} O_{2}

in acidic medium. The volume of the same

K M n O_{4}

required in alkaline medium to oxidize

100

mL of the same

H_{2} O_{2}

will be :

(

M n O_{4}^{⊝}

changes to

M n^{2 +}

in acidic medium and to

M n O_{2}

in alkaline medium)

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The reaction of aqueous KMnO4 with H2O2 in acidic conditions gives

The correct option is B Mn2+ and O2 2KMnO4+3H2SO4+5H2O2→K2SO4+2MnSO4+8H2O+5O2

The reaction of aqueous $K M n O_{4}$ with $H_{2} O_{2}$ in acidic conditions gives

The correct option is B
$M n^{2 +}$ and $O_{2}$

$2 K M n O_{4} + 3 H_{2} S O_{4} + 5 H_{2} O_{2} \to K_{2} S O_{4} + 2 M n S O_{4} + 8 H_{2} O + 5 O_{2}$