[mous]

C₃H₈O + CrO₃ + H₂SO₄ = Cr₂(SO₄)₃ + C₃H₆O + H₂O

I've balanced this equation for mass and this was the result:

3C₃H₈O + 2CrO₃ + 3H₂SO₄ = Cr₂(SO₄)₃ + 3C₃H₆O + 6H₂O

I was wondering how this would be done using the redox method.
These are the oxidation numbers I found (for the respective elements above):
(+2, +1, -2) + (+6,-2)+ (+1, +6, -2) = (+3, +6, -2) + (-4/3, + 1, -2) + (+1, -2)

Carbon and Chromium are the elements undergoing redox. However, both are gaining electrons, and that does not make sense to me. (Chromium gains 3 from +6 to +3; Carbon gains 3 1/3 from +2 to -4/3) And there is no way to balance the electron transfer since 3 and 3 1/3 have no common multiple.

Is it possible to balance this equation using the redox method?
I was also thinking about the ion electrode method, but as far as I know, that is mostly used if charges are in the equation, and there are none in this equation. Is it valid to balance this equation using the ion electrode method?

[Hunter2]

You have to develop the single Oxidation and Reduction reaction.
Rule: Reduction add H⁺ on Educt side, get water on product side
Oxidation vise versa.

Example MnO₄^- + 8 H⁺ + 5 e^- => Mn²⁺ + 4 H₂O

Try yourself with your given example.

[mous]

You have to develop the single Oxidation and Reduction reaction.
Rule: Reduction add H⁺ on Educt side, get water on product side
Oxidation vise versa.

Example MnO₄^- + 8 H⁺ + 5 e^- => Mn²⁺ + 4 H₂O

Try yourself with your given example.

My result:
C₃H₈O = C₃H₆O + 2H⁺ + 2e^- (x3)

6H⁺ + 6e^- + 3H₂SO₄ + 2CrO₃ = Cr₂(SO₄)₃ + 6H₂O

6H⁺ +6e^- + 3H₂SO₄ + 2CrO₃ + 3C₃H₈O = Cr₂(SO₄)₃ + 6H⁺ + 6H₂O + 3C₃H₆O + 6H⁺ + 6e^-

3H₂SO₄ + 2CrO₃ + 3C₃H₈O = Cr₂(SO₄)₃ + 6H₂O + 3C₃H₆O

[Hunter2]

You had on 6 H⁺ to much.

But the result is correct. Normally the SO₄^2- will not used in the equation, because in sulfuric as well in the chromiumsulfate they are the same.

[mous]

Thanks for catching the typo, and thanks for your help.