6 NaI + MnO2 + 4 H2SO4 → MnSO2 + 3 Na2SO4 + 3 I2 + 4 H2O
This is an oxidation-reduction (redox) reaction:
MnIV + 2 e- → MnII (reduction)
SVI + 4 e- → SII (reduction)
6 I-I - 6 e- → 6 I0 (oxidation)
MnO2 is an oxidizing agent, H2SO4 is an oxidizing agent, NaI is a reducing agent.