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Oxidising Reagent and Reducing Reagent

Consider the ...

Question

Consider the following statements for the reaction:

$K I O_{3} + 5 K I + 6 H C l \to 3 I_{2} + 6 K C l + 3 H_{2} O$

$1.$ $I^{-}$ is reduced to $I_{2}$
$2.$ ${I O_{3}}^{-}$ is oxidized to $I_{2}$
$3.$ ${I O_{3}}^{-}$ is reduced to $I_{2}$
$4.$ Oxidation number of $I$ increases from $- 1$ (in $K I$ ) to zero (in $I_{2}$ )

The correct statements are:

3

and

4

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1

and

4

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1

3

and

4

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1

2

and

4

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Solution

The correct option is A $3$ and $4$
$K I + 5 O_{3} + 5 K I - 1 + 6 H C l \to 3 I_{2} 0 + 6 K C l + 3 H_{2} O$

$I^{-}$ is oxidized to $I_{2}$ .

$I O_{3}^{-}$ is reduced to $I_{2}$ .

Oxidation number of $I$ increases from $- 1$ (in $K I$ ) to $z e r o$ (in $I_{2}$ ).
Hence, option $A$ is correct.

Suggest Corrections

Similar questions

K I O_{3} + 5 K I + 6 H C l \to 3 I_{2} + 6 K C l + 3 H_{2} O

{I O}_{3}^{-}, {I O}_{4}^{-}, K I

I_{2}

I_{2} + N a O H \to I O_{3} + N a I + 3 H_{2} O

K I O_{3}

K I

I O_{3}^{-} \to I_{2}

I_{2} + S_{2} O_{3}^{2} \to S_{4} O_{6}^{2 -} + I^{-}

=

=

I_{2} =

I O_{3}^{-}

2 \times m e q .

I O_{3}^{-} =

I^{-}

I_{2}

I_{2} \to I O_{3}^{-} + I^{-}

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