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

1 Answer
Mar 22, 2016

Answer:

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?