The correct option is B Cr2O2−7+3C2H4O+8H+→2Cr3++3C2H4O2+4H2O
Since, the reaction is unbalanced,
The oxidation number for all the atoms can be:
Cr2O2−7+C2H4O+H+→2Cr3++C2H4O2+H2O
Oxidation half equation: C2H4O2→C2H4O2
Reduction half equation: Cr2O2−7→2Cr3+
In both the half reactions, hydrogen and oxygen atoms are not balanced
Balance O atoms by adding water molecules on the side containing less number of O atoms.
Balance hydrogen atoms by adding H+ ions to the side containing less number of H- atoms.
Oxidation half equation: C2H4O2+H2O→C2H4O2+2H+
Reduction half equation: Cr2O2−7+14H+→2Cr3++7H2O
Balancing charges by adding number of electrons.
Oxidation half equation: C2H4O+H2O+2e−→C2H4O2+2H+
Reduction half equation: Cr2O2−7+14H+→2Cr3++7H2O+6e−
Multiply the half equation to equalize the number of electrons
in the two half equations
Oxidation half equation: 3C2H4O+3H2O+6e−→3C2H4O2+6H+
Reduction half equation: Cr2O2−7+14H+→2Cr3++7H2O+6e−
Now, adding the two half reactions,
Cr2O2−7+3C2H4O+8H+→2Cr3++3C2H4O2+4H2O