Question #2fc11

The first thing to do here is assign oxidation numbers to the atoms that take part in the reaction.

#stackrel(color(blue)(0))("Cu")_ ((s)) + stackrel(color(blue)(+5))("N") stackrel(color(blue)(-2))("O"_ 3^(-)) ""_ ((aq)) + stackrel(color(blue)(+1))("H"^(+))""_ ((aq)) -> stackrel(color(blue)(+2))("Cu"^(2+))""_ ((aq)) + stackrel(color(blue)(+4))("N") stackrel(color(blue)(-2))("O")_ (2(g)) + stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(-2))("O")_ ((l))#

Notice that the oxidation state of nitrogen goes from #color(blue)(+5)# on the reactants' side to #color(blue)(+4)# on the products' side. Because its oxidation state decreased, you can say that nitrogen is being reduced.

On the other hand, the oxidation state of copper went from #color(blue)(0)# on the reactants' side to #color(blue)(+2)# on the products' side. Because its oxidation state increased, you can say that copper is being oxidized.

The oxidation half-reaction looks like this

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

Here each copper atom is losing #2# electrons.

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

#stackrel(color(blue)(+5))("N")"O"_3^(-) + "e"^(-) -> stackrel(color(blue)(+4))("N")"O"_2#

Here each nitrogen atom gains #1# electron.

Now, notice that the reactants' side contains hydrogen ions, #"H"^(+)#. This means that the reaction is taking place in acidic solution.

As a consequence, you can use water molecules and protons, #"H"^(+)#, to balance the atoms of oxygen and of hydrogen.

To balance the oxygen atoms, add water to the side that needs oxygen. To balance the hydrogen atoms ,add protons to the side that needs hydrogen.

In this case, you have #3# atoms of oxygen on the reactants' side and #2# on the products' side, so add one molecule of water to that side to get

#stackrel(color(blue)(+5))("N")"O"_3^(-) + "e"^(-) -> stackrel(color(blue)(+4))("N")"O"_2 + "H"_2"O"#

Now you need to add #2# protons to the reactants' side to balance the hydrogen atoms.

#2"H"^(+) + stackrel(color(blue)(+5))("N")"O"_3^(-) + "e"^(-) -> stackrel(color(blue)(+4))("N")"O"_2 + "H"_2"O"#

In a redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

To balance out the electrons, multiply the reduction half-reaction by #2# and add the two half-reactions

# {(color(white)(aaaaaaaaaaaaa)stackrel(color(blue)(0))("Cu") -> stackrel(color(blue)(+2))("Cu"^(2+)) + 2"e"^(-)), (2"H"^(+) + stackrel(color(blue)(+5))("N")"O"_3^(-) + "e"^(-) -> stackrel(color(blue)(+4))("N")"O"_2 + "H"_2"O" color(white)(aaa)|xx 2) :}#
#color(white)(aaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#

#"Cu"_ ((s)) + 4"H"_ ((aq))^(+) + 2"NO"_ (3(aq))^(-) + color(red)(cancel(color(black)(2"e"^(-)))) -> "Cu"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"e"^(-)))) + 2"NO"_ (2(g)) + 2"H"_ 2"O"_ ((l))#

This will be equivalent to

#color(green)(|bar(ul(color(white)(a/a)color(black)("Cu"_ ((s)) + 4"H"_ ((aq))^(+) + 2"NO"_ (3(aq))^(-) -> "Cu"_ ((aq))^(2+) + 2"NO"_ (2(g)) + 2"H"_ 2"O"_ ((l)))color(white)(a/a)|)))#