Complete and balance the following chemical equation :

$M n O_{4}^{-} + H_{2} O + I^{-} \to$

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Solution

The complete, unbalanced chemical equation is :

$M n O_{4}^{-} + H_{2} O + I^{-} \to M n O_{2} + I O_{3}^{-}$

I atoms and Mn atoms are balanced.
The oxidation number of iodine increases from $- 1$ (in $I^{-}$ ) to $+ 5$ (in $I O_{3}^{-}$ ). Total increase in oxidation number for iodine is $+ 5 - (- 1) = 6$ .
The oxidation number of Mn decreases from $+ 7$ (in $M n O_{4}^{-}$ ) to $+ 4$ (in $M n O_{2}$ ). Total decrease in oxidation number for Mn is $7 - 4 = 3$ .
To balance the total increase in oxidation number with the total decrease in oxidation number, add coefficient 2 to Mn species.
$2 M n O_{4}^{-} + H_{2} O + I^{-} \to 2 M n O_{2} + I O_{3}^{-}$

Balance O atoms by adding 2 water molecules to the products side.
$2 M n O_{4}^{-} + H_{2} O + I^{-} \to 2 M n O_{2} + I O_{3}^{-} + 2 H_{2} O$

One water molecule cancels on each side.
$2 M n O_{4}^{-} + I^{-} \to 2 M n O_{2} + I O_{3}^{-} + H_{2} O$
Balance H atoms by adding 2 protons to the reactants side.
$2 M n O_{4}^{-} + I^{-} + 2 H^{+} \to 2 M n O_{2} + I O_{3}^{-} + H_{2} O$

This is the complete balanced chemical equation.

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