Question

Identify , which atom is oxidized and which atom is reduced in following unbalanced redox reaction :-
IO3- and IO- gives rise to i- and I3-

Answer 1

The first, you need to find oxidation state for I. In reaction with $\mathrm{IO}_3^-$ and $\mathrm{IO}^-$:

I in $\mathrm{IO}_3^-$ has: $(3*(-2))_{\mathrm{oxygen}} + (-1)_{\mathrm{extra}} = -7$, so I is $+7$

Then it becomes: $(-2)_{\mathrm{oxygen}} + (-1)_{\mathrm{extra}} = -3$ so I is $+3$

Simplified reaction:

$$\mathrm{I}^{+7} + 5\mathrm{e} \rightarrow \mathrm{I}^{+3}$$

Using next statements:

Oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom, or ion.

Reduction is the gain of electrons or a decrease in oxidation state by a molecule, atom, or ion.

$$\mathrm{I}^{+7} + 5\mathrm{e} \rightarrow \mathrm{I}^{+3}$$

So $\mathrm{I}^{+7}$ is reduced

I suppose that reaction between $\mathrm{I}^-$ and $\mathrm{I}_3^-$ is nor redox reaction, cause there is no changing in oxidation state:

$$\mathrm{I}^- + \mathrm{I}_2 = \mathrm{I}_3^- (\mathrm{I}_2^* \mathrm{I}^-) \text{ - complex}$$