# Are there alternative ways to find n or number of e- in problems like the one pictured? [closed]

I need help understanding how we are to come up with 12 e- because according to the directions, we are supposed to focus on any element, but it doesn't tell us what to look for. I see how carbon has a -2 value in the reactants side and a +4 on the products side. the difference is 6 e- and we are suppose to multiply that by the number of carbons.

But is there another way to calcualte to find 12 e- ?

Couldn't we add the charges from the reactants and products after multiplying them respectively with the subscripts? For instance: -2 X 2 = -4 or 4 e- and +4 X 2 = 8 or 8 e- and so 4 e- + 8 e- = 12 e- ?

Isn't this a valid substitute than the explanation below? Because the directions below ask us to look for the difference in any of the elements. But this doesn't explain how to calulcate for hydrogen. But in the explanation I prepared above does, doesn't it?

But then again, this doesn't explain how we can calculate for oxygen, which has a 0 charge in it's own gas phase and a -2 in water and -2 in carbon dioxide.

So what would be another way to calculate for -12 e- ? In this case, we've to calculate the oxidation states of carbon in the reactant and product side separately, the difference between those two and finally multiply it with the number of carbon atoms. While this is may be a seemlingly simple calculation, things can get uber-tricky because of the mammoth number of exceptional cases. Let's start with the basics; here's a gift for you:

$${\begin{array}{|l|c|} \hline \pu{Element} & \pu{Oxidation Number} & \pu{Remarks} \\\hline \pu{Alkali Metals} & \pu{+1} & \pu{Always} \\ \pu{Alkali Earth Metals} & \pu{+2} & \pu{Always} \\ \pu{Flourine} & \pu{-1} & \pu{Always} \\ \pu{Halogen Family} & \pu{-1} & \pu{Generally} \\ \pu{Oxygen Family} & \pu{-2} & \pu{Generally} \\ \pu{Hydrogen} & \pu{+1} & \pu{Generally} \\\hline \end{array}}$$

Whoa! I see three exceptional cases already (remarked "generally" in the table) and things haven't started yet. What's up with them?

Oxygen: Oxygen doesn't always exist as an oxide ion $(\ce{O^2-})$ in its compounds, it can often-times exist as

• Peroxide $(\ce{O_2^2-})$: Oxidation state in this case is $-1$. Popular examples include $\ce{H2O2}$, $\ce{Na2O2}$, $\ce{BaO2}$, etc.
• Superoxide $(\ce{O_2^-})$: Oxidation state is $-0.5$. Common examples $\ce{KO2}$, $\ce{RbO2}$, $\ce{CrO2}$
• Ozonide ($\ce{O3-}$): $\frac{-1}{3}$ oxidation state; for instance in $\ce{KO3}$.

Halogen atoms (except flourine): When $\ce{Cl, Br or I}$ combine with elements of higher electronegativity ($\ce{O and F}$), then these halogen atoms can show positive oxidation numbers; commonly $\pu{+1, +3, +5, +7}$. In $\ce{BrF5}$, that's an inter-halogen compound, $\ce{Br}$ is in $+5$ oxidation state.

Hydrogen: In case of metal hydrides, it can show $-1$ oxidation state. As in $\ce{NaH}$.

Phew, er any more exceptions? You dare ask? Metals. Wait! Aren't metals electro-positive!? Yes, they are, but in some cases they can show zero or even negative oxidation numbers. For example in $\ce{CsAu}$, gold shows a $-1$ oxidation state. Let's bring in coordination-compounds: nickel in $\ce{[Ni(CO)4]}$ finds itself in zero oxidation number.

There are plenty of them, I'm saving them for another day.

Let's try calculating things you've asked.

For ethanol ($\ce{C2H6O}$) we don't know the oxidation state of the carbon atom. Why don't we assume it to be $x$. Now, the oxidation state of a compound is either zero (if it's a neutral species) or it's equal to the charge it bears. Is ethanol neutral? Yes it is. So the individual oxidation numbers of elements (times the number of elements bearing that oxidation state) should sum up to be zero.

$$2x+6(+1)+1(-2)=0$$

Solving, we get $x=-2$. Doing the same thing for carbon-dioxide, we can get $x=+4$. So a carbon atom changed its oxidation state from $-2$ to $+4$, that's a jump of $+6$. And how many carbon atoms were doing this? Two carbons. That gives us $n=12$.