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Oxyacids of Phosphorus

Equivalent ma...

Question

Equivalent mass of $K M n O_{4}$ in alkaline medium is $\frac{M}{x}$ . The value of $x$ is :

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Solution

In a strongly basic solution, permanganate (VII) is reduced to the green $+ 6$ oxidation state of the manganate ion, $M n O_{4}^{2}$ .
$M n O_{4}^{-} + e^{-} \to M n O_{4}^{2 -}$
As we know,
Equivalent weight $= \frac{m o l a r w t .}{n - f a c t o r} = \frac{M}{1} = M$ .
So, the value of $x$ is $1$ .

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Similar questions

Q. Calculate the equivalent mass of

K M n O_{4}

in the below reaction.

M n O_{4}^{-} (+ 7) \to M n O_{4}^{- 2} (+ 6)

(alkaline medium)

The equivalent mass of $K M n O_{4}$ in acidic medium is (molar mass of $K M n O_{4}$ = 158 g/mol)

Q. In an alkaline medium,

K M n O_{4}

reacts as follows (At wt. of

K = 39.09

M n = 54.94

O = 16.0

2 K M n O_{4} + 2 K O H \to 2 K_{2} {M n O}_{4} + H_{2} O + [O]

i) The equivalent weight of

K M n O_{4}

in alkaline medium is 158.
ii) The number of electrons gained by one molecule of

K M n O_{4}

in the alkaline medium is 1.
iii) In the alkaline medium,

K M n O_{4}

acts as a reducing agent.

Select the correct option.

Q. Assertion :Equivalent mass of

K M n O_{4}

is equal to one-third of its molecular mass when it acts as an oxidising agent in an alkaline medium. Reason: Oxidation number of

M n

K M n O_{4}

.

Q. Calculate the equivalent mass of

K M n O_{4}

in the below reaction.

M n O_{4}^{-} (+ 7) \to M n O_{4}^{- 2} (+ 6)

(alkaline medium)

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