What is the balanced redox reaction between lead and ammonia?

#PbO"(s)"+NH_3 "(g)" -> N_2"(g)"+H_2O"(l)"+Pb"(s)"#

Which element is being oxidized and which is being reduced?

1 Answer
Dec 20, 2017

Answer:

#3"PbO"_ ((s)) + 2"NH"_ (3(g)) ->"N"_ (2(g)) + 3"H"_ 2"O"_ ((l)) + 3"Pb"_ ((s))#

Explanation:

In this redox reaction, lead(II) oxide is being reduced to lead metal and ammonia is being oxidized to nitrogen gas.

#stackrel(color(blue)(+2))("Pb")"O"_ ((s)) + stackrel(color(blue)(-3))("N")"H"_ (3(g)) -> stackrel(color(blue)(0))("N") _ (2(g)) + "H"_ 2"O"_ ((l)) + stackrel(color(blue)(0))("Pb")_ ((s))#

More specifically, the oxidation number of lead goes from #color(blue)(+2)# on the reactants' side to #color(blue)(0)# on the products' side, which implies that lead is being reduced, i.e. its oxidation number is decreasing.

On the other hand, the oxidation number of nitrogen goes from #color(blue)(-3)# on the reactants' side to #color(blue)(0)# on the products' side, which implies that nitrogen is being oxidized, i.e. its oxidation number is increasing.

Now, start by writing the half-reactions for this redox reaction. To make the balancing easier, you can assume that the reaction is taking place in acidic solution--keep in mind that it's not taking place in acidic solution!

#color(white)(a)#
The reduction half-reaction looks like this

#stackrel(color(blue)(+2))("Pb")"O" + 2"e"^(-) -> stackrel(color(blue)(0))("Pb")#

Here every lead(II) cation is taking in #2# electrons to become lead metal. To balance out the atoms of oxygen, add water molecules to the side that needs oxygen and protons, #"H"^(+)#, to the side that needs hydrogen.

You will have

#2"H"^(+) + stackrel(color(blue)(+2))("Pb")"O" + 2"e"^(-) -> stackrel(color(blue)(0))("Pb") + "H"_ 2"O"#

Notice that the half-reaction is balanced in terms of charge because you have

#2 xx (1+) + 0 + 2 xx (1-) = 0 + 0#

#color(white)(a)#
The oxidation half-reaction looks like this

#2stackrel(color(blue)(-3))("N")"H"_ 3 -> stackrel(color(blue)(0))("N")_ 2 + 6"e"^(-)#

Here each nitrogen atom is losing #3# electrons, so two nitrogen atoms will lose a total of #6# electrons. To balance out the atoms of hydrogen, add protons to the side that needs hydrogen.

You will have

#2stackrel(color(blue)(-3))("N")"H"_ 3 -> stackrel(color(blue)(0))("N")_ 2 + 6"e"^(-) + 6"H"^(+)#

Notice that the half-reaction is balanced in terms of charge because you have

#2 xx 0 = 0 + 6 xx (1-) + 6 xx (1+)#

#color(white)(a)#
Now, in any redox reaction, the number of electrons lost in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction.

To balance out the number of electrons transferred between the two elements, multiply the reduction half-reaction by #3#

#{(2"H"^(+) + stackrel(color(blue)(+2))("Pb")"O" + 2"e"^(-) -> stackrel(color(blue)(0))("Pb") + "H"_ 2"O"" " " " " "|xx 3), (color(white)(aaaaaaaaaaa)2stackrel(color(blue)(-3))("N")"H"_ 3 -> stackrel(color(blue)(0))("N")_ 2 + 6"e"^(-) + 6"H"^(+)) :}#

and add the two half-reactions to get

#{(6"H"^(+) + 3stackrel(color(blue)(+2))("Pb")"O" + 6"e"^(-) -> 3stackrel(color(blue)(0))("Pb") + 3"H"_ 2"O"), (color(white)(aaaaaaaaaaaa)2stackrel(color(blue)(-3))("N")"H"_ 3 -> stackrel(color(blue)(0))("N")_ 2 + 6"e"^(-) + 6"H"^(+)) :}#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#
#color(red)(cancel(color(black)(6"H"^(+)))) + 3stackrel(color(blue)(+2))("Pb")"O" + color(red)(cancel(color(black)(6"e"^(-)))) + 2stackrel(color(blue)(-3))("N")"H"_ 3 -> stackrel(color(blue)(0))("N")_ 2 + 3stackrel(color(blue)(0))("Pb") + color(red)(cancel(color(black)(6"e"^(-)))) + 3"H"_ 2"O" + color(red)(cancel(color(black)(6"H"^(+))))#

Notice that the protons we've used to balance the half-reactions cancel each other out, which is what confirms that the reaction is not taking place in acidic solution.

The balanced chemical equation that describes this redox reaction will look like this

#3"PbO"_ ((s)) + 2"NH"_ (3(g)) ->"N"_ (2(g)) + 3"H"_ 2"O"_ ((l)) + 3"Pb"_ ((s))#