# Question #8d261

Jun 8, 2015

Start by assigning oxidation numbers to the atoms that take part in the reaction.

$\stackrel{\textcolor{b l u e}{0}}{{P}_{4}} + \stackrel{\textcolor{b l u e}{+ 5}}{N} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{3}^{-}} \to \stackrel{\textcolor{b l u e}{+ 1}}{{H}_{2}} \stackrel{\textcolor{b l u e}{+ 5}}{P} \stackrel{\textcolor{b l u e}{- 2}}{{O}_{4}^{-}} + \stackrel{\textcolor{b l u e}{+ 2}}{N} \stackrel{\textcolor{b l u e}{- 2}}{O}$

Notice that the oxidation state of phosphorus goes from 0 on the reactants' side, to +5 on the products' side. This means that phosphorus is being oxidized.

On the other hand, nitrogen's oxidation number goes from +5 on the reactants' side, to +2 on the products' side, which means that it is being reduced.

Write the oxidation and reduction half-reaction.

• Oxidation half-reaction

$\stackrel{\textcolor{b l u e}{0}}{{P}_{4}} \to {H}_{2} \stackrel{\textcolor{b l u e}{+ 5}}{P} {O}_{4}^{-}$

Notice that you have 4 phosphorus atoms on the lefthand side, and only 1 on the righthand side of the equation. Multiply the latter by 4 to get

$\stackrel{\textcolor{b l u e}{0}}{{P}_{4}} \to 4 {H}_{2} \stackrel{\textcolor{b l u e}{+ 5}}{P} {O}_{4}^{-} + 20 {e}^{-}$

Each phosphorus atom loses 5 electrons, so the total number of electrons lost is equal to 20.

Since you're in an acidic medium, you can balance the oxygen and the hydrogen atoms by adding water to the side that lacks oxygen and protons, ${H}^{+}$, to the side that lacks hydrogen.

Notice that you need 16 oxygen on the lefthand side, so add 16 water molecules on that side of the equation. This will get you 32 hydrogens on the reactants' side, and only 8 on the products' side $\to$ add 24 protons.

$16 {H}_{2} O + \stackrel{\textcolor{b l u e}{0}}{{P}_{4}} \to 4 {H}_{2} \stackrel{\textcolor{b l u e}{+ 5}}{P} {O}_{4}^{-} + 20 {e}^{-} + 24 {H}^{+}$

• Reduction half-reaction

$\stackrel{\textcolor{b l u e}{+ 5}}{N} {O}_{3}^{-} + 3 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 2}}{N} O$

Add water and protons to balance the oxygen and hydrogen atoms.

$4 {H}^{+} + \stackrel{\textcolor{b l u e}{+ 5}}{N} {O}_{3}^{-} + 3 {e}^{-} \to \stackrel{\textcolor{b l u e}{+ 2}}{N} O + 2 {H}_{2} O$

In a redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

This means that you need to multiply the oxidation half-reaction by 3 and the reduction half-reaction by 20, for a total number of 60 electrons transferred.

This will get you

$\left\{\begin{matrix}48 {H}_{2} O + 3 {P}_{4} \to 12 {H}_{2} P {O}_{4}^{-} + 60 {e}^{-} + 72 {H}^{+} \\ 80 {H}^{+} + 20 N {O}_{3}^{-} + 60 {e}^{-} \to 20 N O + 40 {H}_{2} O\end{matrix}\right.$

Add the two half-reactions to get

$48 {H}_{2} O + 3 {P}_{4} + 80 {H}^{+} + 20 N {O}_{3}^{-} + \cancel{60 {e}^{-}} \to 12 {H}_{2} P {O}_{4}^{-} + 20 N O + \cancel{60 {e}^{-}} + 40 {H}_{2} O + 72 {H}^{+}$

Remove the water and the protons present on both sides of the equation to get the balanced chemical equation

$8 {H}_{2} O + 8 {H}^{+} + 3 {P}_{4} + 20 N {O}_{3}^{-} \to 12 {H}_{2} P {O}_{4}^{-} + 20 N O$