# How to write half equations based on oxidizing/reducing agents pairs?

In most videos, people use oxidation numbers to find the oxidizing agent and the reducing agent, It's a great way to use, but our teacher never used it before he would give us a reaction and ask us directly to write the half reactions and what are the oxidizing and reducing agent on each. Using oxidization numbers I can easily find the oxidizing/reducing agents and write half reaction for each.

However, sometimes we don't have the final reaction, we only have oxidizing/reducing agents and are asked to write half equations. For example: Given the pair ($\ce{CO2/H2C2O4}$) as an example, should I write the half equation for the $\ce{CO2}$ to turn into $\ce{H2C2O4}$, or the inverse?

Another question is how can I find oxidation numbers on compound elements such as: $\ce{C6H12O6}$?

Firstly, you need to understand that there is a maximum and minimum oxidation number.

For metals, the minimum is usually $0$ while the maximum is usually the number of their outermost electrons. For example, $\ce{Na}$ which has $1$ outermost shell electron would have a minimum of $0$ and a maximum of $1$.

For non-metals, the minimum is usually the number of their outermost shell electrons minus $8$, while the maximum is usually the number of their outermost shell electrons. For example, $\ce{S}$ which has 6 outermost shell electrons would have a minimum of $-2$ and a maximum of $6$.

For the pair $\ce{CO2}$ and $\ce{H2C2O4}$:

Firstly, note that the most electronegative element in each compound is $\ce{O}$. You can learn more about electronegativity here.

Therefore, it takes the minimum, which is $-2$.

Secondly, usually hydrogen is $1$ (in most covalent compounds) or $-1$ (in most ionic compounds). In this case, hydrogen is $1$.

It follows from calculation (that the sum of the electronegativity of each element in a compound must be zero) that the $\ce{C}$ in the first compound has oxidation number $4$ while that in the second compound has oxidation number $3$.

Since the oxidation number has dropped, this is a reduction reaction.

We write the unbalanced equation as a starter:

$$\ce{2CO2 -> H2C2O4}$$

Balancing the number of $\ce{C}$.

Now, we are in an acidic environment, so we add $\ce{H2O}$ to balance the oxygen and $\ce{H+} to balance the hydrogen. The oxygen is already balanced, so: $$\ce{2H+ + 2CO2 -> H2C2O4}$$ Note that the charges are still not balanced while the elements are balanced, so we add electrons: $$\ce{2H+ + 2CO2 + 2e- -> H2C2O4}$$ Thus our half equation is complete. The method to find the oxidation number of each element in$\ce{C6H12O6}$is detailed above, and I leave this for you as an exercise. PS: in an alkaline environment, we add$\ce{OH-}$to both sides of the equation to eliminate the$\ce{H+}$, i.e. (the following is wrong because we are in an acidic environment thanks to the acid$\ce{H2C2O4}$(acetic acid): $$\ce{2H+ + 2OH- + 2CO2 + 2e- -> H2C2O4 + 2OH-}$$ $$\ce{2H2O + 2CO2 + 2e- -> H2C2O4 + 2OH-}$$ • Great answer, but what I can't get into my mind is that why we wrote a half reaction for CO2 to turn to C2H2O4 instead of writing a half reaction to turn the C2H2O4 to CO2 ? My basic question is that how do we know which element turn to which other element given the agent pair (ox/red) ? – Anis Souames Sep 5 '16 at 19:56 • @AnisSouames You can also write a half reaction to turn$\ce{C2H2O4}$into$\ce{CO2}$: $$\ce{C2H2O4 -> 2H+ + 2CO2 + 2e-}$$ So when to use which equation? Well, when you are converting$\ce{CO2}$to$\ce{C2H2O4}\$, use the first equation. If vice versa, use this equation. – DHMO Sep 5 '16 at 19:59
• @DHMO. I believe when you wrote "It follows from calculation (that the sum of the electronegativity of each element in a compound must be zero)", you meant that the sum of the oxidation numbers ...must be zero. Yes? – Dr. J. Jun 10 '18 at 11:34