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