2 KMnO4 (aq) + 10 NaI (aq) + 8 H2SO4 (aq) → 5 I2 (s) + 2 MnSO4 (aq) + 5 Na2SO4 (aq) + K2SO4 (aq) + 8 H2O (l)
This is an oxidation-reduction (redox) reaction:
2 MnVII + 10 e- → 2 MnII (reduction)
10 I-I - 10 e- → 10 I0 (oxidation)
KMnO4 is an oxidizing agent, NaI is a reducing agent.
This is a precipitation reaction: I2 is the formed precipitate.