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Introduction to Oxidation and Reduction

In strongly a...

Question

In strongly alkaline solution containing excess of barium ions, a solution containing 0.01587 g of $I^{-}$ was treated with 0.1 molar ${M n O}_{4}^{-}$ until a pink colour presisted in the solution : 10.0 mL of the ${M n O}_{4}^{-}$ solution were required. Under these conditions ${M n O}_{4}^{-}$ is converted into the sparingly soluble ${B a M n O}_{4}$ . The product of the oxidation of iodide is $I O_{4}^{-}$
if true enter 1, else enter 0.

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Solution

$B a M n O_{4} = B a^{2 +} + M n O_{4}^{2 -}$
In this reaction, $M n O_{4}^{-} (+ 7)$ is reduced to $M n O_{4}^{2 -} (+ 6)$
The oxidation number changes from 7 to 6. The change in oxidation number is 1 unit
$\frac{10 X 0.1}{1000} = 0.001$ Moles of $M n O_{4}^{-}$ are used
Equivalent of $M n O_{4}^{-}$ used = 0.001
Let $I^{-}$ be oxidised to species with oxidation number x
$I - 1 ⟶ (I)_{x}^{x +}$
Change in oxidation number = x-(-1)
Equivalent of $I^{-}$ = $\frac{0.01587}{127}$ = (x+1)
Hence, $\frac{0.01587 (x + 1)}{127} = 0.001$
$x = 7$
Thus, iodide is oxidised to $I O_{4}^{-}$

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