Question

Balance the following equation by ion electron method in acidic medium
a) Cr2O7²- +C2H4O→ C2H4O2 +Cr³+​

Answer 1

Answer:

here your answee

Explanation:

Cr2O72- + C2H2O- → C2H2O + Cr3+ ...(i) Here Cr2O72- → 2Cr3+ i.e., Cr2O72- → 2Cr3+ + 7H2O 6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O [H2O + C2H4O- → C2H4O2 + 2H+ + 3e-] x 2 ...(ii) 10H+ + Cr2O72- + 2C2H4O- → 2C2H4O2 + 2Cr3+ + 5H2ORead more on Sarthaks.com - https://www.sarthaks.com/608497/cr2o7-2-c2h4o-c2h4o2-cr-3-in-acid-solution-by-ion-electron-method

Answer 2

✎Answer

The unbalanced redox equation is as follows:

Cr2O72−+C2H4O+H+→C2H4O2+Cr3+

Balance all  atoms other than H and O.

Cr2O72−+C2H4O+H+→C2H4O2+2Cr3+

The oxidation number of Cr changes from 6 to 3. So the change in oxidation number of one Cr atom is 3.

For 2 Cr atoms, the oxidation number changes by 6.

The oxidation number of C changes from -1 to 0. So the change in oxidation number of 1 C atom is 1.

For 2 C atoms, the oxidation number changes by 2.

The increase in the oxidation number is balanced with decrease in the oxidation number by  multiplying

 C2H4O and C2H4O2  with 3.

Cr2O72−+3C2H4O+H+→3C2H4O2+2Cr3+

O atoms are balanced by adding 4 water molecules on RHS.

Cr2