2 NaI (aq) + H2O2 (aq) → I2 (s) + 2 NaOH (aq)
This is an oxidation-reduction (redox) reaction:
2 I-I - 2 e- → 2 I0 (oxidation)
2 O-I + 2 e- → 2 O-II (reduction)
NaI is a reducing agent, H2O2 is an oxidizing agent.
This is a precipitation reaction: I2 is the formed precipitate.