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Question

Balance the following equation by ion electron method:
Cr2O27+C2H4O+H+2Cr3++C2H4O2+H2O

A
Cr2O27+3C2H4O+12H+2Cr3++3C2H4O2+6H2O
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B
Cr2O27+3C2H4O+8H+2Cr3++3C2H4O2+4H2O
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C
Cr2O27+3C2H4O+4H+2Cr3++3C2H4O2+2H2O
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D
2Cr2O27+3C2H4O+8H+4Cr3++3C2H4O2+4H2O
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Solution

The correct option is B Cr2O27+3C2H4O+8H+2Cr3++3C2H4O2+4H2O
Since, the reaction is unbalanced,
The oxidation number for all the atoms can be:
Cr2O27+C2H4O+H+2Cr3++C2H4O2+H2O
Oxidation half equation: C2H4O2C2H4O2
Reduction half equation: Cr2O272Cr3+
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+H2OC2H4O2+2H+
Reduction half equation: Cr2O27+14H+2Cr3++7H2O
Balancing charges by adding number of electrons.
Oxidation half equation: C2H4O+H2O+2eC2H4O2+2H+
Reduction half equation: Cr2O27+14H+2Cr3++7H2O+6e
Multiply the half equation to equalize the number of electrons
in the two half equations
Oxidation half equation: 3C2H4O+3H2O+6e3C2H4O2+6H+
Reduction half equation: Cr2O27+14H+2Cr3++7H2O+6e
Now, adding the two half reactions,
Cr2O27+3C2H4O+8H+2Cr3++3C2H4O2+4H2O

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