# What is oxidation number? Write its uses.

Dec 10, 2017

Well we could write an essay....

#### Explanation:

Oxidation number is the charge left on the central atom when all the bonding pairs of electrons are broken with the charge assigned to the most electronegative atom...

Note that this is a fictitious exercise, but we use it to help us balance oxidation/reduction reactions....where the oxidized species is conceived to lose electrons, and the reduced species is conceived to gain electrons...See this old link and for formal rules of assignment see here..

And for an example, we often use high valent manganese or chromium oxides to effect oxidation...we would write the reduction reaction as....

$C r \left(V I +\right) \rightarrow C r \left(I I I +\right)$ $\left(i\right)$
$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O$

And...the corresponding oxidation:

$C \left(- I\right) \rightarrow C \left(+ I I I\right)$ $\left(i i\right)$
${H}_{3} C - C {H}_{2} O H + {H}_{2} O \rightarrow {H}_{3} C - C {O}_{2} H + 4 {H}^{+} + 4 {e}^{-}$

We add $2 \times \left(i\right) + + 3 \times \left(i i\right)$ to remove the electrons, virtual particles of convenience....

$2 C {r}_{2} {O}_{7}^{2 -} + \cancel{28} 16 {H}^{+} + 3 {H}_{3} C - C {H}_{2} O H + \cancel{3 {H}_{2} O + 12 {e}^{-}} \rightarrow 4 C {r}^{3 +} + 3 {H}_{3} C - C {O}_{2} H + \cancel{12 {H}^{+}} + \cancel{12 {e}^{-}} + 11 \cancel{3} {H}_{2} O$

...to give finally...

$2 C {r}_{2} {O}_{7}^{2 -} + 16 {H}^{+} + 3 {H}_{3} {\text{CCH"_2OHrarr 4Cr^(3+)+3H_3"CCO}}_{2} H + 11 {H}_{2} O$

...which I think is balanced with respect to mass and charge if I have done my sums correctly....