# What is the basic formula of a redox reaction?

Jul 18, 2018

Well, we conceive of electrons, and oxidation numbers...

#### Explanation:

We conceive of $\text{oxidation}$ as the loss of electrons, and when an oxidation number, an imaginary number, or rather a conceptual number, goes up, this corresponds to the loss of electrons. And please note that this is an entirely CONCEPTUAL exercise....

Oxygen is the quintessential oxidant .... it is conceived to GAIN electrons to form oxide ions...as shown....

$\frac{1}{2} {\stackrel{0}{O}}_{2} \left(g\right) + 2 {e}^{-} \rightarrow {O}^{2 -}$ $\left(i\right)$

...dioxygen gas is said to be $\text{REDUCED}$; it has gained electrons...and its oxidation number has been reduced from $0$ to $- I I$; the $- I I$ is the charge on the oxide anion; and again if mass and charge are not balanced, if stoichiometry is NOT preserved, then the reaction cannot be accepted as a representation of chemical reality.

In the given reduction equation, the which represents the reduction of dioxygen gas, the electrons are conceived to come from somewhere. Some other species MUST LOSE electrons, and this process is dubbed $\text{oxidation}$...

For example, we could write the oxidation of dihydrogen gas as shown....

${\stackrel{0}{H}}_{2} \left(g\right) \rightarrow 2 {H}^{+} + 2 {e}^{-}$ $\left(i i\right)$

We adds $\left(i\right)$ and $\left(i i\right)$ in such a way that the electrons are removed from the final equation.....i.e. $\left(i\right) + \left(i i\right)$ gives...

${H}_{2} + \frac{1}{2} {O}_{2} \rightarrow {\underbrace{2 {H}^{+} + {O}^{2 -}}}_{\text{one water molecule}}$ or....

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(l\right)$

And this is a simple example of a redox equation. Confused yet? But all I have done is assign some oxidation numbers, and conserved mass and conserved charge....

Formal rules of assignment of oxidation numbers are given here....

$1.$ $\text{The oxidation number of a free element is always 0.}$

$2.$ $\text{The oxidation number of a mono-atomic ion is equal}$ $\text{to the charge of the ion.}$

$3.$ $\text{For a given bond, X-Y, the bond is split to give } {X}^{+}$ $\text{and}$ ${Y}^{-}$, $\text{where Y is more electronegative than X.}$

$4.$ $\text{The oxidation number of H is +1, but it is -1 in when}$ $\text{combined with less electronegative elements.}$

$5.$ $\text{The oxidation number of O in its}$ compounds $\text{is usually -2, but it is -1 in peroxides.}$

$6.$ $\text{The oxidation number of a Group 1 element}$ $\text{in a compound is +1.}$

$7.$ $\text{The oxidation number of a Group 2 element in}$ $\text{a compound is +2.}$

$8.$ $\text{The oxidation number of a Group 17 element in a binary compound is -1.}$

$9.$ $\text{The sum of the oxidation numbers of all of the atoms}$ $\text{in a neutral compound is 0.}$

$10.$ $\text{The sum of the oxidation numbers in a polyatomic ion}$ $\text{is equal to the charge of the ion.}$

Note that these are for reference...and not for learning off by heart....The best way is to learn these is to apply the given rigmarole to actual equations...for instance the oxidation of ammonia to nitrate...i.e. an eight electron oxidation with respect to nitrogen...

$\stackrel{- I I I}{N} {H}_{3} + 3 {H}_{2} O \rightarrow \stackrel{+ V}{N} {O}_{3}^{-} + 9 {H}^{+} + 8 {e}^{-}$