Question

Balance the following redox equation, identifying the element oxidized and the element reduced. Show all of the work used to solve the problem.

Cr2O7 2- + C2H5OH = Cr3+ + CO2

Answer 1

@sweetburger can you help me?

Answer 2

@tester97

Answer 3

@Carl_Pham

Answer 4

Cr2O7 2- ---> Cr 3+

C2H5OH ---> CO2

First we have to balance all elements excluding O, and H

so Cr2O7 2- ---> 2Cr 3+

C2H5OH ---> 2CO2

Then we balance out the H and O atoms using H2O and H+.

14H+ + Cr2O7 2- ---> 2Cr3+ + 7H2O

3H2O + C2H5OH ---> 2CO2 + 12H+

Then obtain equal charge on both sides of each reaction by the addition of electrons.

6e- + 14H+ + cr2O7 2- ---> 2Cr 3+ + 7H2O

3H2O + C2H5OH ---> 2CO2 + 12H+ +12e-

Because 6e- are transferred in the top half reaction and 12e- are transferred in the bottom reaction we must multiply the entire top reaction by 2.

2(14H+ + Cr2O7 2- ---> 2Cr 3+ + 7H2O) =
28H+ +2Cr2O7 2- ---> 4Cr 3+ + 14H2O


combine both the left sides of the reaction so
12e- + 28H+ + 2Cr2O7 2- + 3H2O + C2H5OH ---> 2Co2 + 12H+ + 4Cr 3+
+ 14H2O + 12e-


Now I will let you complete the last step which involves cancelling each thing that occurs on both sides of the reaction for example cross out the 12e- that occur on both sides.

Answer 5

you might want to check my balancing I did this really fast

Answer 6

also the last step is *combine both the left and right sides of the reaction and then cancel out stuff that reoccurs on both sides of the reaction.