Question #1d773

1 Answer
Oct 15, 2015

Answer:

#16"H"_2"O"_text((l]) + "P"_text(4(s]) -> 4"H"_2"PO"_text(4(aq])^(-) + 20e^(-) + 24"H"_text((aq])^(+)#

Explanation:

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

#stackrel(color(blue)(0))("P")_text(4(s]) -> stackrel(color(blue)(+1))("H")_2stackrel(color(blue)(+5))("P")stackrel(color(blue)(-2))("O")_text(2(aq])^(-)#

Notice that you're dealing with an oxidation half-reaction, since phosphorus' oxidation state changes from #color(blue)(0)# on the reactants' side, to #color(blue)(+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 xx 5e^(-) = 20e^(-)#

#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, #"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"H"_2"O" + stackrel(color(blue)(0))("P")_4 -> 4"H"_2stackrel(color(blue)(+5))("P")"O"_4^(-) + 20e^(-)#

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.

You thus need to add

#"no. of protons" = 32 - 8 = 24 "H"^(+)#

to the products' side to balance the hydrogen atoms.

#16"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"H"_2"O"_text((l]) + "P"_text(4(s]) -> 4"H"_2"PO"_text(4(aq])^(-) + 20e^(-) + 24"H"_text((aq])^(+)#

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.