# Do reduction and oxidation occur together?

Aug 28, 2017

Well, yes.....we conceive of $\text{oxidation}$ as the loss of electrons.....

#### Explanation:

And we conceive of $\text{reduction}$ as the gain of electrons.

Since charge, as well as mass, is CONSERVED in every chemical reaction, for every oxidation there must be a corresponding reduction, and we can assign $\text{oxidation numbers}$ as arbitrary numbers with which we can assess $\text{conceptual electron transfer.....}$

Now ammonia can oxidized to nitrate.......a formal oxidation of $N \left(- I I I\right)$ to $N \left(+ V\right)$, an EIGHT electron transfer.....

And thus for oxidation we write.....

$N {H}_{3} + 3 {H}_{2} O \rightarrow N {O}_{3}^{-} + 9 {H}^{+} + 8 {e}^{-}$ $\left(i\right)$

Mass is balanced, and charge is balanced....so this is a reasonable representation of chemical change. Now the electrons are presumed to GO somewhere on oxidation; i.e. they cause a CORRESPONDING reduction of some other reagent (which of course is the OXIDIZING agent). A typical oxidizing agent is permanganate ion, $M n {O}_{4}^{-}$, which is REDUCED from $M n \left(V I I +\right)$ to colourless $M {n}^{2 +}$, a 5 electron reduction.....

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$ $\left(i i\right)$

Charge and mass are balanced as required. Are they?

To complete the entire redox reaction, we cross multiply to remove the electrons.....$5 \times \left(i\right) + 8 \times \left(i i\right)$:

$8 M n {O}_{4}^{-} + 64 {H}^{+} + 5 N {H}_{3} + 15 {H}_{2} O + 40 {e}^{-} \rightarrow 8 M {n}^{2 +} + 32 {H}_{2} O + 5 N {O}_{3}^{-} + 45 {H}^{+} + 40 {e}^{-}$

And then we cancel out the common reagents......

$8 M n {O}_{4}^{-} + \cancel{64} 19 {H}^{+} + 5 N {H}_{3} + \cancel{15 {H}_{2} O} + \cancel{40 {e}^{-}} \rightarrow 8 M {n}^{2 +} + \cancel{32} 17 {H}_{2} O + 5 N {O}_{3}^{-} + \cancel{45 {H}^{+}} + \cancel{40 {e}^{-}}$

To give finally.......

$8 M n {O}_{4}^{-} + 19 {H}^{+} + 5 N {H}_{3} \rightarrow 8 M {n}^{2 +} + 5 N {O}_{3}^{-} + 17 {H}_{2} O$

Which, despite the whack coefficients, is balanced with respect to mass and charge. And we would see the deep purple colour of permanganate dissipate to give COLOURLESS $M {n}^{2 +}$.