# How do you use the half-equation to represent oxidation-reduction reactions?

Mar 22, 2016

The half equation method features the loss or or gain of electrons as ACTUAL particles in order to balance redox equations.

#### Explanation:

Let's represent the oxidation of aluminum to give alumina, $A {l}_{2} {O}_{3}$.

This ionic species features $A {l}^{3 +}$ and ${O}^{2 -}$ ions. Since the charge on the elemental ion is the oxidation state, we can speak of $A l \left(I I {I}^{+}\right)$ and $O \left(- I I\right)$.

$\text{Oxidation (electron loss from zerovalent aluminum):}$

$A {l}^{0} \rightarrow A {l}^{3 +} + 3 {e}^{-}$ $\left(i\right)$

$\text{Reduction (electron gain to zerovalent oxygen):}$

${O}_{2}^{0} + 4 {e}^{-} \rightarrow 2 {O}^{2 -}$ $\left(i i\right)$

In each redox equation, charge and mass is conserved, as it must be.

In the final equation, we wish to eliminate electrons, because these are hypothetical particles: so we cross-multiply, $2 \times \left(i\right) + 3 \times \left(i i\right) =$

$2 A l + 3 {O}_{2} \rightarrow 2 A {l}_{2} {O}_{3}$

Now this was simply a combustion reaction, which is intuitively easy to solve. Try the oxidation of elemental sulfur to $S \left(V I\right)$, as in $S {O}_{4}^{2 -}$. Sulfur loses 6 electrons.

$S \rightarrow S {O}_{4}^{2 -} + 6 {e}^{-}$

Balance the oxygens (using water)

$S + 4 {H}_{2} O \rightarrow S {O}_{4}^{2 -} + 6 {e}^{-}$

Now balance the hydrogens:

$S + 4 {H}_{2} O \rightarrow S {O}_{4}^{2 -} + 8 {H}^{+} + 6 {e}^{-}$

Again, both mass and CHARGE are balanced, as required. We need a corresponding reduction reaction to ACCEPT the electrons.

Another reaction, oxidation of ammonia to nitric oxide:

${N}^{- I I I} {H}_{3} + 2 {H}_{2} O \rightarrow {N}^{+ I V} {O}_{2} + 7 {H}^{+} + 7 {e}^{-}$

Is this reaction balanced with respect to mass and charge? Why?