# Question 1d773

Oct 15, 2015

$16 {\text{H"_2"O"_text((l]) + "P"_text(4(s]) -> 4"H"_2"PO"_text(4(aq])^(-) + 20e^(-) + 24"H}}_{\textrm{\left(a q\right]}}^{+}$

#### Explanation:

Start by assigning oxidation numbers to the atoms that take part in this half-reaction

${\stackrel{\textcolor{b l u e}{0}}{\text{P")_text(4(s]) -> stackrel(color(blue)(+1))("H")_2stackrel(color(blue)(+5))("P")stackrel(color(blue)(-2))("O}}}_{\textrm{2 \left(a q\right]}}^{-}$

Notice that you're dealing with an oxidation half-reaction, since phosphorus' oxidation state changes from $\textcolor{b l u e}{0}$ on the reactants' side, to $\textcolor{b l u e}{+ 5}$ on the products' side.

To balance this half-reaction, start by balancing the phosphorus atoms

stackrel(color(blue)(0))("P")_4 -> 4"H"_2stackrel(color(blue)(+5))("P")"O"_4^(-)

So, each phosphorus atom will lose five electrons, so four phosphorus atoms will lose a total of

$4 \times 5 {e}^{-} = 20 {e}^{-}$

stackrel(color(blue)(0))("P")_4 -> 4"H"_2stackrel(color(blue)(+5))("P")"O"_4^(-) + 20e^(-)#

Now balance the oxygen and hydrogen atoms. Assuming that the reaction takes place in acidic medium, you can use water molecules to balance the oxygen and protons, ${\text{H}}^{+}$, to balance the hydrogen.

So, you have zero oxygen atoms on the reactants' side and a total of 16 on the products' side, which means that you're going to need 16 water molecules on the reactants' side.

$16 {\text{H"_2"O" + stackrel(color(blue)(0))("P")_4 -> 4"H"_2stackrel(color(blue)(+5))("P")"O}}_{4}^{-} + 20 {e}^{-}$

Now focus on the hydrogen atoms. You now have a total of 32 hydrogen atoms on the reactants' side, and a total of 8 hydrogen atoms on the products' side.

${\text{no. of protons" = 32 - 8 = 24 "H}}^{+}$

to the products' side to balance the hydrogen atoms.

$16 {\text{H"_2"O" + stackrel(color(blue)(0))("P")_4 -> 4"H"_2stackrel(color(blue)(+5))("P")"O"_4^(-) + 20e^(-) + 24"H}}^{+}$

And there you have it, this oxidation half-reaction is balanced.

$16 {\text{H"_2"O"_text((l]) + "P"_text(4(s]) -> 4"H"_2"PO"_text(4(aq])^(-) + 20e^(-) + 24"H}}_{\textrm{\left(a q\right]}}^{+}$

You would now go on to do the same for the reduction half-reaction, make sure that the number of electrons lost in the oxidation half-reaction is equal to the number of electronsgained in the reduction half-reaction, and balance the overall equation.