# What is the first step to balance a redox equation using the half-reaction method?

Mar 30, 2017

Assignment of $\text{oxidation numbers...........}$

#### Explanation:

Redox reactions are conceived to occur on the basis of electron transfer, which requires the assignment of formal oxidation states. The oxidation state is the charge left on the central atom when all the bonding pairs of electrons are distributed on the basis of a few predetermined rules.

And thus for, say $M n {O}_{4}^{-}$, a potent oxidizing agent, we require that the sum of the oxidation numbers equals the charge on the ion, which is here $- 1$. The oxidation number of oxygen is usually $- I I$ and it is here, which leaves the oxidation number of $M n$ at $+ V I I$. When manganese is reduced, it is conceived to accept electrons, i.e. $\text{formal reduction}$, and it is typically reduced to $M {n}^{2 +}$, so it is conceived to have accepted 5 electrons:

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

And once you have assigned the oxidation numbers, the difference between such numbers is the number of electrons transferred during the redox equation.

Are charge and mass balanced here?

And for every reduction, there is a corresponding oxidation:

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

And in the overall redox reaction, we cross-multiply the individual redox equations to eliminate the electrons:

And thus $4 \times \left(i\right) + 5 \times \left(i i\right) =$

$4 M n {O}_{4}^{-} + \cancel{32} 12 {H}^{+} + \cancel{20 {e}^{-}} + 5 R C {H}_{2} O H + \cancel{5 {H}_{2} O} \rightarrow 4 M {n}^{2 +} + \cancel{16} 11 {H}_{2} O + 5 R C {O}_{2} H + \cancel{20 {H}^{+}} + \cancel{20 {e}^{-}}$

.........to give finally,

$4 M n {O}_{4}^{-} + 5 R C {H}_{2} O H + 12 {H}^{+} \rightarrow 4 M {n}^{2 +} + 5 R C {O}_{2} H + 11 {H}_{2} O$

Which is, as far as I know, balanced with respect to mass and charge, as indeed it must be if we purport to model a chemical process.

What we would see in this reaction, if we mixed stoichiometric reagents, was the disappearance of the deep purple colour of potassium permanganate to give almost colourless $M {n}^{2 +}$ ion.

What are the oxidation numbers of the central atom in $C {r}_{2} {O}_{7}^{2 -}$, and $C r {O}_{4}^{2 -}$, and $C l {O}_{4}^{-}$?