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Common Oxidation States

The product o...

Question

The product of oxidation of $I^{-}$ with $M n O_{4}^{-}$ in alkaline medium is:

A

I O_{3}^{-}

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B

I_{2}

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C

I O^{-}

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D

I O_{4}^{-}

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Solution

The correct option is A $I O_{3}^{-}$
Oxidaiton of $I^{-}$ with $K M n O_{4}$ in alkaline solution is given by
Reaction: $6 M n O_{4}^{-} + I^{-} + 6 O H^{-} \to I O_{3}^{-} + 6 M n O_{4}^{2 -} + 3 H_{2} O$
Here the oxidation state of Mn reduced from +7 to +6 and the oxidation state of I oxidised from -1 to +5.
$6 e^{-}$ is involved in the reaction.
Oxidation product form is $I O_{3}^{-} .$
Hence, option A is correct.

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Q. The product of

I^{⊖}

M n O_{4}^{⊖}

in acidic medium:-

$K M n O_{4}$ acts as an oxidizing agent in alkaline medium. When alkaline $K M n O_{4}$ is treated with KI, iodide ion is oxidized to

K M_{n} O_{4}

acts as an Oxidising agent in alkaline medium. When alkaline

K M_{n} O_{4}

is treated with KI, iodide ion is oxidised to

I^{-}

M n O_{4}^{-}

in alkaline medium,

I^{-}

is converted into

I^{-}

M n O_{4}^{-}

in faintly alkaline medium,

I^{-}

is convered into ?

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Common Oxidation States

Standard XII Chemistry

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