How would you classify the reaction #I_2(aq)+2S_2O""_3^(2-) (aq) -> 2I^(-) (aq) + S_4O""_6^(2-)(aq)#?

I won't add state symbols because the reaction occurs in an aqueous solution, but you can verify that this is what's happening by giving the elements involved in the reaction oxidation numbers.

The fact that the tetrathionate anion contains sulfur with a fractional oxidation number should not be taken as a clue; rather, it indicates that not every sulfur atom in the anion has the same oxidation state.

Iodine is reduced to iodide anions in the reduction half-reaction.

Iodine gains two electrons, which causes it to become reduced.

The thiosulfate anion is oxidized to tetrathionate anion in the oxidation half-reaction.

Thus, you are in fact dealing with a redox reaction since the sulfur loses electrons and the iodine gains them.

This Socratic answer provides further information on the fractional oxidation number of sulfur in the tetrathionate anion.

The reaction between I2(aq) and 2S2O3^2-(aq) to form 2I^-(aq) and S4O6^2-(aq) is a redox reaction, specifically a reduction-oxidation or redox reaction.

The reaction between ( \text{I}_2 ) and ( \text{S}_2\text{O}_3^{2-} ) to form ( \text{I}^- ) and ( \text{S}_4\text{O}_6^{2-} ) is a redox reaction. Specifically, it's a redox reaction because iodine undergoes a change in oxidation state from ( 0 ) in ( \text{I}_2 ) to ( -1 ) in ( \text{I}^- ), while sulfur undergoes a change in oxidation state from ( +2 ) in ( \text{S}_2\text{O}_3^{2-} ) to ( +5 ) in ( \text{S}_4\text{O}_6^{2-} ).