Question

In strongly alkaline solution containing excess of barium ions, a solution containing 0.01587 g of $I^{-}$ was treated with 0.1 molar ${MnO}_4^{-}$ until a pink colour presisted in the solution : 10.0 mL of the ${MnO}_4^{-}$ solution were required. Under these conditions ${MnO}_4^{-}$ is converted into the sparingly soluble ${BaMnO}_4$. The product of the oxidation of iodide is $I{O}_4^{-}$
if true enter 1, else enter 0.

Answer 1

$BaMn{O}_4=B{a}^{2+}+Mn{O}_4^{2-}$
In this reaction, $\underset{(+7)}{Mn{O}_4^{-}}$ is reduced to $\underset{(+6)}{Mn{O}_4^{2-}}$
The oxidation number changes from 7 to 6. The change in oxidation number is 1 unit
$\frac{10X0.1}{1000}=0.001$ Moles of $Mn{O}_4^{-}$ are used
Equivalent of $Mn{O}_4^{-}$ used = 0.001
Let $I^{-}$ be oxidised to species with oxidation number x
$\underset{-1}{I}⟶(I{)}_x^{x+}$
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 $I{O}_4^{-}$