stackrel(0)Cu+Hstackrel(+5)NO_3->stackrel(+2)Cu(NO_3)_2+stackrel(+2)NO+H_2O0Cu+H+5NO3→+2Cu(NO3)2++2NO+H2O
Change in Oxidation Number
Cu(0)->Cu(+2)=>"2 unit oxidation"Cu(0)→Cu(+2)⇒2 unit oxidation
N(+5)->N(+2)=>"3 unit reduction"N(+5)→N(+2)⇒3 unit reduction
"Reacting Ratio of " Cu(+2):N(+5)=3:2Reacting Ratio of Cu(+2):N(+5)=3:2
3mol Cu(NO_3)_2Cu(NO3)2 formed from 3mol Cu requires extra 6 mol HNO_3HNO3 as salt former which is added in LHS making total 8 mol HNO_3HNO3 in reactant side.The RHS is then adjusted as required.
3stackrel(0)Cu+8Hstackrel(+5)NO_3->3stackrel(+2)Cu(NO_3)_2+2stackrel(+2)NO+4H_2O30Cu+8H+5NO3→3+2Cu(NO3)2+2+2NO+4H2O
Another Way
(1) First write the balanced equation for decomposition reaction of
HNO_3HNO3 forming the oxide of Nitrogen produced, water and oxygen as follows
2HNO_3->H_2O+2NO+3O2HNO3→H2O+2NO+3O
(2) Adjust the O-atom produced in step (1) combining with required proportion of Cu atoms as follows
3Cu+3O->3CuO3Cu+3O→3CuO
(3) Write balanced equation of reaction of 3CuO with HNO_3HNO3 forming salt and water
3CuO+6HNO_3->3Cu(NO_3)_2+3H_2O3CuO+6HNO3→3Cu(NO3)2+3H2O
(4) Finally add those equations to have the required balance equation.
2HNO_3->H_2O+2NO+cancel(3O)
3Cu+cancel(3O)->cancel(3CuO)
cancel(3CuO)+6HNO_3->3Cu(NO_3)_2+3H_2O
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3Cu+8HNO_3->3Cu(NO_3)_2+2NO+4H_2O
Shortcut tips for balancing the equations of reactions between metal and nitric acids in various cases where oxides of nitrogen are produced.
(1) Let moles Cu be x and moles of HNO_3 be y in the balanced equation of a reaction.
Then x=5-"ON of N in its oxide formed"
As for above reaction x=5-2=3
(2) Then y=x xxv+v,
"where" v = "valency of metal or OS of metal in its nitrate salt "
So for above reaction v=2
and y=3xx2+2=8
(3) Knowing the value of x and y we can easily balance the equation.
