Which element is reduced in this reaction? 16h++2cr2o72-+c2h5oh→4cr3++11h2o+2co2?

The equation, for better presentation, is

6H⁺ + 2 Cr₂O₇²⁻ + C₂H₅OH → 4 Cr³⁺ + 11 H₂O + 2 CO₂

To tell which element is reduced, let's solve the oxidation numbers of derived compounds.

H⁺ becomes H₂O. The oxidation number of H⁺ is ⁺1, while the oxidation number of H in H₂O is:

2x - 2 = 0

x = ⁺1

Hence, the oxidation number remained the same.

Next, Cr₂O₇²⁻ became Cr³⁺. The oxidation number of Cr in Cr³⁺ is ⁺3. The oxidation number of Cr in Cr₂O₇²⁻ isL

2x + 7 (-2) = - 2

x = 6

Thus, the oxidation from ⁺6 became ⁺3. The oxidation number reduced. So, the element reduced is Cr or Chromium.