Question

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

Answer 1

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, `Al_2O_3`.

This ionic species features `Al^(3+)` and `O^(2-)` ions. Since the charge on the elemental ion is the oxidation state, we can speak of `Al(III^+)` and `O(-II)`.

`"Oxidation (electron loss from zerovalent aluminum):"`

`Al^(0) rarr Al^(3+) + 3e^-` `(i)`

`"Reduction (electron gain to zerovalent oxygen):"`

`O_2^(0) +4e^(-) rarr 2O^(2-)` `(ii)`

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, `2xx(i)+3xx(ii)=`

`2Al+3O_2 rarr 2Al_2O_3`

Now this was simply a combustion reaction, which is intuitively easy to solve. Try the oxidation of elemental sulfur to `S(VI)`, as in `SO_4^(2-)`. Sulfur loses 6 electrons.

`S rarr SO_4^(2-) + 6e^-`

Balance the oxygens (using water)

`S + 4H_2O rarr SO_4^(2-) + 6e^-`

Now balance the hydrogens:

`S + 4H_2O rarr SO_4^(2-) + 8H^+ + 6e^-`

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^(-III)H_3 + 2H_2O rarr N^(+IV)O_2 + 7H^(+) +7e^-`

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