Answer
$I_{2} (s) + 5OCl^{-} (aq) +H_{2}O (l) → 2IO_{3}^{-} (aq) + 5Cl^{-} (aq)+ 2H^{+}$
Oxidizing agent: OCl-
Reducing Agent: I2
Work Step by Step
We add water on left hand side and H+ on right side to balance the equation as it is in acidic medium. Oxidation state of I is changed from 0 to +5, hence it is oxidized and state of Cl is changed from +1 to -1, hence it is reduced.