Answer
$2Cr_{2}O_{7}^{2-} (aq) + 3CH_{3}OH (aq) + 16H^{+} → 3HCO_{2}H (aq) + 4Cr^{3+} (aq) +11H_{2}O (l)$
Oxidizing Agent: $2Cr_{2}O_{7}^{2-}$
Reducing Agent: $CH_{3}OH$
Work Step by Step
We add H to left side and water to right side to balance the equation as it is in acidic medium. Oxidation state of S is changed from -2 to +2, hence it is oxidized and state of Cr is changed from +6 to +3, hence it is reduced.