Help with balancing redox reaction

Question

C4H8 (2-methylpropene) KMnO4 H2SO4 ----> C3H6O (propanone) K2SO4 MnSO4 CO2 H2O

I know that C is the the reducing agent and that Mn is the oxidizing agent. I can't figure out how to balance this redox reaction. Can anyone help me?

Also, can you tell me how many electrons receives/gives C and Mn ?

I don't know how to balance the reaction because in the first substance, the oxidation number of C is 0 (if I calculated well), in CO2 the oxidation number of C is 4 and in propanone the oxidation number of C is 2. In KMnO4, the oxidation number of Mn is 7 and in MnSO4 the oxidation number of Mn is 2. So, Mn gives 5 electrons. What about C?

Answer 1

Sometimes it's better not to worry about all of the different oxidation states, particularly when an organic is involved. Just set up the half-reactions with MnO4- => Mn2+ and C4H8 => C3H6O + CO2. I also left out the spectator ions like K+ and SO42-.

As for the first one, you get

MnO4-  +  8H+ + 5e-   =>  Mn2+ + 4H2O

In the organic one, the C is already balanced (4 on each side). Add 3 H2O on the left to balance O, and add 8H+ on the right to balance H. When you do that, you'll have a charge of 0 on the left and +8 on the right, so add 8e- to the right. You end up with

C4H8  + 3H2O    =>   C3H6O  +  CO2  +  8H+  +  8e-

You can add the two half reactions together after you multiply the reduction reaction by 8 (to give 40e- on the left) and the oxidation reaction by 5 (to give 40e- on the right).

8MnO4- + 64H+ + 40e- + 5C4H8 + 15H2O => 8Mn2+ + 32H2O + 5C3H6O + 5CO2 + 40H+ + 40e-

Crossing out the extra H+ and H2O gives

8MnO4- + 24H+ + 5C4H8 => 8Mn2+ + 17H2O + 5C3H6O + 5CO2

Th elements and charge on bith sides are balanced! If you want to put the spectator ions back in, put K+ with MnO4- and SO42- with H+ and Mn2+. You need to add 4K2SO4 on the right to balance the spectators.

8KMnO4 + 12H2SO4 + 5C4H8 => 8MnSO4 + 17H2O + 5C3H6O + 5CO2 + 4K2SO4

I hope this helps.