1
$\begingroup$

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: $\ce{C6H12O6}$?

$\endgroup$
0
$\begingroup$

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):

$$\ce{2H+ + 2OH- + 2CO2 + 2e- -> H2C2O4 + 2OH-}$$ $$\ce{2H2O + 2CO2 + 2e- -> H2C2O4 + 2OH-}$$

$\endgroup$
  • $\begingroup$ 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) ? $\endgroup$ – Anis Souames Sep 5 '16 at 19:56
  • $\begingroup$ @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. $\endgroup$ – DHMO Sep 5 '16 at 19:59
  • $\begingroup$ @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? $\endgroup$ – Dr. J. Jun 10 '18 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.