Balancing Redox Reactions?

Question

Balance the following equation by the half-reaction method in an acidic or basic solution as indicated.


Cr2O72- + C2H5OH --> Cr3+ + CO2 ( acidic )


You must enter the correct coefficients of the above reactants and products as a single group of integers.


Example:

C2H5OH(aq) + Cr2O72-(aq) --> CH3CO2H(aq) + Cr3+(aq)



the balanced equation is

3C2H5OH(aq) + 2Cr2O72-(aq) + 16H3O+(aq) --> 3CH3CO2H(aq) + 4Cr3+(aq) + 27H2O(l)



you would enter 3234 for an answer.

Answer 1

Cr2O7 2- + 14H+ + 6e- >> 2Cr3+ + 7H2O
C2H5OH + 3H2O >> 2 CO2 + 12H+ + 12e-
Now I have to multiply the first half-reaction for 2

2Cr2O7 2- + 28H+ + 12e- >> 4Cr3+ + 14H2O
C2H5OH + 3H2O >> 2CO2 + 12H+ + 12e-

Now I add up thehalf-reactions and I cancel any duplication of species on the left and right sides

2Cr2O7 2- + C2H5OH + 16H+>>4Cr3+ +2CO2+11H2O