Balancing redox equations

IT has a lower oxidation state because it was oxidized. Nitrogen can have several states.

Cu+NO3- + 2H+ >> Cu-2 + NO2 + H20

acidic medium means Hydrogen ion is present.

so, b/c it is acidic, there is always an 2H and H2O even before I balance it?

To balance the redox equation, we need to first determine the oxidation states of each element involved.

In the reactant species, nitrogen in NO3^- has an oxidation state of +5 because oxygen usually has an oxidation state of -2, and there are three oxygen atoms in NO3^- which results in a total charge of -6. Since nitrate (NO3^-) has an overall charge of -1, nitrogen must have an oxidation state of +5 to balance the charges.

In the product species, nitrogen in NO2 has an oxidation state of +4. This is because oxygen usually has an oxidation state of -2 and there are two oxygen atoms in NO2 which results in a total charge of -4. Since nitrogen must balance out the charge of 2+, it must have an oxidation state of +4.

Now, let's balance the redox equation step by step:

1. Assign oxidation numbers:
Cu^0 + NO3^- (+5) -> Cu^2+ + NO2 (+4)

2. Separate the half-reactions:
Cu^0 -> Cu^2+
NO3^- (+5) -> NO2 (+4)

3. Balance the atoms other than oxygen and hydrogen:
Cu^0 -> Cu^2+
2NO3^- (+5) -> 2NO2 (+4)

4. Balance oxygen atoms by adding water molecules (H2O):
Cu^0 -> Cu^2+
2NO3^- (+5) -> 2NO2 (+4) + 3H2O

5. Balance hydrogen atoms by adding hydrogen ions (H+):
Cu^0 -> Cu^2+
2NO3^- (+5) + 6H+ -> 2NO2 (+4) + 3H2O

6. Balance charges by adding electrons (e-):
Cu^0 -> Cu^2+ + 2e-
2NO3^- (+5) + 6H+ + 2e- -> 2NO2 (+4) + 3H2O

7. Make the number of electrons equal in both half-reactions by multiplying them:
Cu^0 -> Cu^2+ + 2e-
2NO3^- (+5) + 6H+ + 10e- -> 2NO2 (+4) + 3H2O

8. Combine the half-reactions, cancelling out the electrons:
Cu^0 + 2NO3^- (+5) + 6H+ -> Cu^2+ + 2NO2 (+4) + 3H2O

Now, the equation is balanced in an acidic medium.