Redox reaction help please

H (aq) + Cr2O7^2- (aq) + C2H5OH(l) = Cr^3+ (aq) + CO2 (g) + H2O(l)

I know that oxidation is the loss of electrons and reduction is the gain or electrons, also I know that in this example Cr is reduced. What are the half cell reactions and what is oxidised.

P.S. Don't tell me to balance it as I need to know the half cell reactions to do this. Please help I asked this question but got a terrible response.
Thanks in advance

H+ (aq) + Cr2O7^2- (aq) + C2H5OH(l) = Cr^3+ (aq) + CO2 (g) + H2O(l)

reduction of Cr's : (Cr2O7)2- with Cr's at +6 each take 6 electreons ---> 2 Cr+3
oxidation of C's : C2H5OH with 2C's at -2 each --> lose 12 electrons becoming C+4's in 2 CO2's

simplifies to
reduction of Cr's : (Cr2O7)2- & 6 e- ---> 2 Cr+3
oxidation of C's : C2H5OH --> 2 CO2 & 12 e-


elecrons balance as
reduction: 2 (Cr2O7)2- & 12 e- ---> 4 Cr+3
oxidation: C2H5OH --> 2 CO2 & 12 e-

giving us:
2 (Cr2O7)2- & C2H5OH --> 2 CO2 & 4 Cr+3

making H2O's to balance O's:
2 (Cr2O7)2- & C2H5OH --> 2 CO2 & 4 Cr+3 & 11 H2O

adding H's to balance H:
16 H+ & 2 (Cr2O7)2- & C2H5OH --> 2 CO2 & 4 Cr+3 & 11 H2O

even the charges of the ions are balanced
12 + --> 12 +