I'm stuck on this:

Cu + HNO3 ------> CuOH + NO2

Here's my attempt:

Copper is oxidised: Cu ---> Cu2+ + 2OH-
Nitric Acid is reduced: HNO3 ---> NO2 + O2-

It looks crap I know..

Here's what I did again:

Cu ---> Cu2+ + 2OH- + 2e-
HNO3 ---> NO2+ + HO-

Any better?

here's my solution:

Cu + HNO3 ---> CuOH + NO2
CU + 2HNO3 ---> Cu(OH)2 + 2NO2
Cu ---> Cu2+ + 2e- : Oxidation
2OH- + 2e- ---> (OH)2 : Reduction

But what happens to the NO3/NO2? Do I have to include that in the half equation? If so, how do I do it?:confused:

Cu: current oxidation state of 0. However as an ion it has an ox-state of +2
HNO3: current oxidation state of 0 (H:+1, NO3:-1).

The reaction goes as planned: Acid + Metal ----> Metal Salt + Hydrogen Gas.

Hydrogen gas = H2, so the equation is not balanced.

Chemical equation now = Cu (s) + 2HNO3 (aq) ----> CuNO3 (aq) + H2 (g)

Therefore the ionic equation must be:

Cu2+2 + 2NO3-2 -----> CuNO3

I ANSWERED MY QUESTION MOTHERFAC*ERS!!

Hey evil dead sorry I didn't answer, I haven't done half equations for many years :)

Lol no problem. I want to become a Chemical Engineer or some sort of Physicist, so I had to make sure this question be answered.