# Building an incomplete redox half reaction/reduction

I can't seem to figure out this redox reaction.

$$\ce{CH3COO- + CO2 -> CH3CH2COO-}$$

Is the charge on the carbon in the $\ce{CO2}$ supposed to match the charge on the carbon in the $\ce{CH3COO-}$? I am supposed to build a half reaction using $\ce{H2O}$ and $\ce{H+}$ as needed, but I can't seem to figure out how many electrons are actually transferred. What are the charges on the carbon?

It helps to use "fully compressed" formulae so you can easily count atoms and figure out how to balance carbons, oxygens, hydrogens, and charges. That is, instead of

• $\ce{CH3COO- +CO2 -> CH3CH2COO-}$

write it this way:

• $\ce{C2H3O2- +CO2 -> C3H5O2-}$

Then it's easy to see that this is not a balanced equation. Carbon atoms are the same on each side, but not oxygen or hydrogen atoms. You need two more $\ce{O}$ atoms on the products side. Those have to come from water since that is the only oxygen-containing species you are allowed to use.

• $\ce{C2H3O2- +CO2 -> C3H5O2- + 2H2O}$

Now oxygen is balanced, but hydrogen is still imbalanced. That can be rectified by adding $\ce{H+}$ into the equation as appropriate.

• $\ce{C2H3O2- +CO2 +6H+ -> C3H5O2- + 2H2O}$

Now only charge is imbalanced, which can be fixed by adding electrons to complete the balanced half-reaction:

• $\ce{C2H3O2- +CO2 +6H+ +6e- -> C3H5O2- + 2H2O}$