And so...
#Cu(s) rarr Cu^(2+) + 2e^(-)# #(i)#
And nitrate is REDUCED to nitrous oxide..#N_2O#...this beast has formal CHARGES as shown...#stackrel(0)N-=stackrel(+)N-O^(-)#...AN AVERAGE oxidation state with respect to nitrogen of #+1/2#...and these are REDUCED from #stackrel(+V)N#:
#2NO_3^(-) +10H^+ +8e^(-)rarr N_2O+5H_2O# #(ii)#
For both equations, charge and mass are balanced (are they?)...and they must be if we purport to represent chemical reality. And so we adds the half equations in such a manner as to ELIMINATE the electrons...#4xx(i)+(ii)=#
#4Cu(s)+2NO_3^(-) +10H^+ +8e^(-)rarr 4Cu^(2+) + 8e^(-)+N_2O+5H_2O#
And so we make cancellations....
#4Cu(s)+2NO_3^(-) +10H^+ rarr 4Cu^(2+) +N_2O+5H_2O#
And for completeness we could add #8xxNO_3^(-)# to EACH side (and after all nitric acid is the source of the hydronium ion) to give..
#4Cu+10HNO_3 rarr 4Cu(NO_3)_2 +N_2O+5H_2O#
The which again is balanced with respect to mass and charge. Do I win five pounds?
#N_2O# (the old laffing gas that was sometime used as an anaesthetic in dentistry) is often air-oxidized up to #NO_2#..
#N_2O +3/2O_2rarr 2NO_2#
