Question

For a particular redox reaction NO is oxidized to NO3^- and Cu^2+ is reduced to Cu^+ . Complete and balance the equation for this reaction in basic solution.

We are given this as a guide to start

NO + Cu^2+ --> NO3^- + Cu^+

I started with:

Cu^2+ + 2e- --> Cu
No + 2H2O --> NO3^- + 4H^+ + 3e-

Combined and balanced to get
3Cu^2+ + 2NO + 4H2O --> 3Cu + 2NO3^- + 8H^+

Apparently it is wrong and i don't know why.

Answer 1

Cu^2+ --> Cu^+

To balance that out you don't need to add 2e^-

Answer 2

In which case the only one you would need to multiply to get en equal amount of e^- would have to be the half reaction of Cu^2+

Answer 3

Also, I will add that this is a basic solution. A basic solution cannot have an excess of H+. That would make it acidic so you have to add OH^- to both sides of the final equation.

Answer 4

So would this be right then

NO + 2H2O + 3Cu^2+ + OH^- ---> 3Cu^+ + NO3^- + 4H^+ +OH^-

Answer 5

Close. You see how you have 4H^+? To get rid of that you need to add 4OH^-. That will make it into 4H2O. But you have to add 4OH^- to both sides of the equation. Then the H2O will cancel out leaving the OH^- on the left and 2H2O on the right.

Answer 6

The final answer I got was
NO + 3Cu^2+ + 4OH^- --> NO3- + 3Cu^+ + 2H2O

Which was right. Thank you very much for your help.:)

Answer 7

You're welcome.