Answer
$Cr_{2}O_{7}^{2-} (aq) + I^{-} (aq) + 8H^{+} →2Cr^{3+} (aq) + IO_{3}^{-} (aq) + 4H_{2}O$
Oxidizing Agent: $Cr_{2}O_{7}^{2-}$
Reducing Agent: $I^{-}$
Work Step by Step
We add H+ on left hand side and water on right side to balance the equation as it is in acidic medium. Oxidation state of Cr is changed from +6 to +3, hence it is reduced and state of I is changed from -1 to +5, hence it is oxidized.
