dilute H2SO4+Cu

when it is an OA , It is in liquid state.
it means that it is a molecule. So, only H2SO4 can act as a OA but not SO4 2-.
Both of them are redox reaction, not neuturalization.

NO3-(aq) + 4 H+(aq) + 3 e- -----> NO(g) + 2 H2O E=+0.96
NO3-(aq) + 2 H+(aq) + e- -----> NO2(g) + H2O (for concentrated) E=>+0.96
Cu2+(aq) + 2 e- -----> Cu(s) E=+0.34
2NO + O2 ----->NO2 ( which is redox too, you may write the half eqt for both reactant)
dilute
Ecell = 0.96-0.34=0.62V
conc.
Ecell >0.62V
so NO3- directly convert to NO2 instead of oxidation with O2.
However, it is my guess. If i am wrong, hope anyone can correct it



2012-03-18 15:02:18 補充：
it should be no reaction because it is a typical acid in dilute

2012-03-18 15:58:19 補充：
since HNO3(aq) is an oxidizing agent already
HNO3(l) is also a oxidizing agent
The difference is ,high concentration, most of it exists in molecule.
If you want the reason why HNO3 has oxidizing property, I hope other advanced users can anwser you

conc. H2SO4 react with Cu by redox reaction
conc HNO3 react with metal , evolving NO2
dilute HNO3 react with metal producing NO

Is there any reaction between dilute sulphuric acid and copper metal?
No.