2Li^++2I^-==>2Li+I2

would 2I^- be oxidised and 2Li be reduced reaction?

Write the half reactions so you can tell which is losing and which gaining electrons.

Li^+ + e ==> Li
2I^- ==> I2 + 2e
Li is gaining electrons which makes it reduced (and the oxidizing agent).
I^- loses electrons which makes it oxidized (and the reducing agent).

Yes, in the given reaction, 2I^- ions are being oxidized to I2 and 2Li ions are being reduced to 2Li+ ions.

To determine whether a species is being oxidized or reduced in a chemical reaction, you need to compare the oxidation states (also known as oxidation numbers) of the atoms before and after the reaction.

In this reaction, the oxidation state of iodine (I) changes from -1 in I^- to 0 in I2. Therefore, iodine is being oxidized, as its oxidation state increases from -1 to 0.

On the other hand, the oxidation state of lithium (Li) changes from +1 in Li^+ to 0 in Li. Thus, lithium is being reduced, as its oxidation state decreases from +1 to 0.

Remember that oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. In this reaction, iodine gains electrons and goes from -1 to 0 (reduction), while lithium loses electrons and goes from +1 to 0 (oxidation).