Your reaction has dilute nitric acid reacting with copper metal to produce nitric oxide and #Cu^(2+)# ions. The reaction will take place in acidic solution, which means you can add water and #H^(+)# to balance oxygen and hydrogen.
So,
#stackrel(color(blue)(+5))(N)# #stackrel(color(blue)(-2))(O_(3(aq))^(-))# # + stackrel(color(blue)(0))(Cu_((s)))# # -> stackrel(color(blue)(+2))(N)# #stackrel(color(blue)(-2))(O_((g))# # + Cu_((aq))^(2+)#
Notice that nitrogen is reduced from a +5 to a +2 oxidation state, while copper is oxidized from a 0 to a +2 oxidation state.
#stackrel(color(blue)(0))(Cu) -> stackrel(color(bue)(+2))(Cu) + 2e^(-)# #| * 3# #-># oxidation
#stackrel(color(blue)(+5))(N) + 3e^(-) -> stackrel(color(blue)(+2))(N)# #| *2# #-># reduction
The balanced half-reactions will be
#3stackrel(color(blue)(0))(Cu) -> 3stackrel(color(bue)(+2))(Cu) + 6e^(-)#
#2stackrel(color(blue)(+5))(N) + 6e^(-) -> 2stackrel(color(blue)(+2))(N)#
Before balancing the oxygen and hydrogen atoms ,the equation will look like this
#2stackrel(color(blue)(+5))(N)# #stackrel(color(blue)(-2))(O_(3(aq))^(-))# # + 3stackrel(color(blue)(0))(Cu_((s)))# # -> 2stackrel(color(blue)(2))(N)# #stackrel(color(blue)(-2))(O_((g))# # + 3Cu_((aq))^(2+)#
SIDE NOTE Here's how you'd go about balancing the chemical equation. First, balance the oxygen atoms. Notice you have 6 oxygen atoms on the ractants' side and only 2 on the products' side #-># add 4 water molecules on the products' side.
#2stackrel(color(blue)(+5))(N)# #stackrel(color(blue)(-2))(O_(3(aq))^(-))# # + 3stackrel(color(blue)(0))(Cu_((s)))# # -> 2stackrel(color(blue)(2))(N)# #stackrel(color(blue)(-2))(O_((g))# # + 3Cu_((aq))^(2+) + 4H_2O_((l))#
FInally, balance the hydrogen atoms by adding 8 #H^(+)# on the reactants' side
#8H_((aq))^(+) + 2stackrel(color(blue)(+5))(N)# #stackrel(color(blue)(-2))(O_(3(aq))^(-))# # + 3stackrel(color(blue)(0))(Cu_((s)))# # -> 2stackrel(color(blue)(2))(N)# #stackrel(color(blue)(-2))(O_((g))# # + 3Cu_((aq))^(2+) + 4H_2O_((l))#
