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6 NaI^{-I} + K_{2}Cr_{2}^{VI}O_{7} + 7 H_{2}SO_{4} → 3 Na_{2}SO_{4} + 3 I_{2}^{0} + Cr_{2}^{III}(SO_{4})_{3} + K_{2}SO_{4} + 7 H_{2}O

This is an **oxidation-reduction** (redox) reaction:

6 I^{-I} - 6 e^{-} → 6 I^{0} (oxidation)

2 Cr^{VI} + 6 e^{-} → 2 Cr^{III} (reduction)