Question

Iodate ion, ${I{O}_3}^{-}$, oxidises ${S{O}_3}^{2-}$ to ${S{O}_4}^{2-}$ in acidic medium. If 100 mL sample of solution containing 2.14 g of $KI{O}_3$ reacts with 60 mL of 0.5 M $N{a}_{2}S{O}_3$ solution, then final oxidation state of Iodine is $-x$. The value of $x$ is :

Answer 1

The oxidation state of S changes from +4 to +6.
Thus, the oxidation of 1 mole of sulphite to sulphate involves 2 moles of electrons.
60 mL of 0.5 M sodium sulphitesolution contains $\frac{60}{1000}\times 0.5=0.03$ moles of sulphite ions.
The oxidation of 0.03 moles of sulphite ions will involve $2\times 0.03=0.06$ moles of electrons.
The molar mass of $KI{O}_3$ is 214 g/mol.
The number of moles of $KI{O}_3$ are $\frac{2.14}{214}=0.01$
Now, 0.01 mole of $KI{O}_3$ requires 0.06 moles of electrons.
1 mole of $KI{O}_3$ will require 6 moles of electrons.
The oxidation number of iodine in $KI{O}_3$ is +5. It will change to $+5-6=-1$ which is equal to -x. hence, $x=1$