Balancing redox equations
reactants: Cu, NO3^- (the negative sign is nex to the O, above the 3).
products: Cu^2+, NO2
Medium: Acidic
Cu+ NO3^- ----> Cu^2+ + NO2
Okay, I don't understand how to do this problem...but something someone needs to clear up for me...why does N in NO3^- have an oxidation state of 5 and in then NO2 has an oxidation state of 4?
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.