    Question

# Balance the following equation by ion electron method: Cr2O2−7+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 Cr2O2−7+3C2H4O+8H+→2Cr3++3C2H4O2+4H2OSince, 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  Suggest Corrections  0      Similar questions  Explore more