Answer:
2 NaI (aq) + Br2 (aq) → 2 NaBr (aq) + I2 (s)
This is an oxidation-reduction (redox) reaction:
2 Br0 + 2 e- → 2 Br-I
(reduction)
2 I-I - 2 e- → 2 I0
(oxidation)
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