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## Q. 4.8

Balancing Simple Oxidation–Reduction Reactions by the Half-Reaction Method

Consider a more difficult problem, the combination (oxidation–reduction) reaction of magnesium metal and nitrogen gas:

$Mg(s) + N_2(g) → Mg_3N_2(s)$

Apply the half-reaction method to balance this equation.

PROBLEM STRATEGY

Start by identifying the species undergoing oxidation and reduction and assigning oxidation numbers. Then write the two balanced half reactions, keeping in mind that you add the electrons to the more positive side of the reaction. Next, multiply each of the half-reactions by a whole number that will cancel the electrons on each side of the equation. Finally, add the half-reactions together to yield the balanced equation.

## Verified Solution

Identify the oxidation states of the elements:

$\overset{0}{Mg}(s) + \overset{0}{N_2}(g) \longrightarrow \overset{+2}{Mg_3} \overset{-3}{N_2} (s)$

In this problem, a molecular compound, nitrogen ($N_2$), is undergoing reduction.  when a species undergoing reduction or oxidation is a molecule, write the formula of the molecule (do not split it up) in the half-reaction. Also, make sure that both the mass and the charge are balanced. (Note the $6e^-$ required to balance the charge due to the $2N^{3-}$.)

$Mg → Mg^{2+} + 2e^-$  (balanced oxidation half-reaction)

$N_2 + 6e^- → 2N^{3-}$  (balanced reduction half-reaction)

We now need to multiply each half-reaction by a factor that will cancel the electrons.

$3 \times (Mg\longrightarrow Mg^{2+} + 2e^-)\\\underline{1 \times (N_2 + 6e^-\longrightarrow 2N^{3-})} \\3Mg + N_2 + \cancel{6e^-} \rightarrow 3Mg^{2+} + 2N^{3-} + \cancel{6e^-}$

Therefore, the balanced combination (oxidation–reduction) reaction is

$3Mg + N_2 \rightarrow 3Mg^{2+} + 2N^{3-}$

From inspecting the coefficients in this reaction, looking at the original equation, and knowing that $Mg^{2+} and N^{3-}$ will combine to form an ionic compound ($Mg_3N_2$), we can rewrite the equation in the following form:

$3Mg(s) + N_2(g) \rightarrow Mg_3N_2(s)$