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

Question

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: $\ce{C6H12O6}$?

Answer 1

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 $6$.

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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.

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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.

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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.

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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):

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