Copper wire is placed in a beaker full of nitric acid.
2006-10-31 13:35:39 · 3 answers · asked by j_lyn316 2 in Science & Mathematics ➔ Chemistry
Also, the copper dissolves and nitric oxide is produced. It says to consider the half-reactions.
2006-10-31 14:22:27 · update #1
if this solution is dilluted
3Cu + 8HNO3(l) ---> 3Cu(NO3)2 + 2NO + 4H2O
if this solution is condensed
Cu + 4HNO3(l) ---> Cu(NO3)2 + 2NO2 + 2H2O
2006-10-31 13:46:24 · answer #1 · answered by James Chan 4 · 0⤊ 0⤋
pure trioxonitrate acid will oxidize copper to copper(II)trioxonitrate(V),willitself is reduced to nitrogen(IV)oxide.
Cu+4HNO3-------->Cu(NO3)2+2H2O+2NO2
with dilute trioxonitrate(V)acid(about 50%),copper again is oxidize to copper(II)trioxonitrate(V) while the acid itself is reduced to nitrogen(II)oxide instead of nitrogen(IV)oxide
3Cu+8HNO3------->3Cu(NO3)2+4H2O+2NO
2006-11-01 09:13:00 · answer #2 · answered by jayscanty 2 · 0⤊ 0⤋
Cu + 2HNO3 ---> Cu(NO3)2 + 2H+
2006-10-31 21:45:50 · answer #3 · answered by titanium007 4 · 0⤊ 0⤋