Question #1d773
1 Answer
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
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,
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.