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?
Which element is being oxidized and which is being reduced?
1 Answer
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
On the other hand, the oxidation number of nitrogen goes from
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!
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
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#
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
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+)#
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
#{(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))#