Balance the following redox reaction occurring in acidic solution
Cu(s) + NO3-(aq)--> Cu2+(aq) +NO2(g)
This is what I have done so far
Cu-->Cu2+ + 2e-
2H + NO-3 --> NO2 + H2O
Need help please
Cu ==> Cu^2+ + 2e is correct.
Stepwise: acid medium.
NO3^- ==> NO2
N changes from +5 on the left to +4 on the right so it gained 1 e.
NO3^- + e ==> NO2
Count the charge. -2 on the left; zero on the right. Add H^+ to balance the charge.
NO3^- + e + 2H^+ ==> NO2
Add water to balance the H^+
NO3^- + e + 2H^+ ==> NO2 + H2O
It should be balanced but I like to check them to see. 1N on both sides. 3O on both sides. 2 H on both sides. zero charge on both sides. N changed from +5 + e to +4. Everything ok.
To balance this redox reaction in acidic solution, follow these steps:
Step 1: Assign oxidation numbers to all elements in the equation.
Cu(s) → Cu2+(aq) (Copper changes from 0 to +2)
NO3-(aq) → NO2(g) (Nitrogen changes from +5 to +4)
Step 2: Separate the reaction into two half-reactions, one for the oxidation and one for the reduction.
Oxidation half-reaction: Cu(s) → Cu2+(aq)
Reduction half-reaction: NO3-(aq) → NO2(g)
Step 3: Balance the atoms except hydrogen and oxygen in each half-reaction.
Oxidation: Cu(s) → Cu2+(aq) + 2e-
Reduction: NO3-(aq) + 2H+(aq) → NO2(g) + H2O(l)
Step 4: Balance the electrons in each half-reaction.
Oxidation: Cu(s) → Cu2+(aq) + 2e-
Reduction: NO3-(aq) + 2H+(aq) + 2e- → NO2(g) + H2O(l)
Step 5: Equalize the number of electrons in both half-reactions. Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 1.
2Cu(s) → 2Cu2+(aq) + 4e-
2NO3-(aq) + 4H+(aq) + 4e- → 2NO2(g) + 2H2O(l)
Step 6: Add both half-reactions together and eliminate common species.
2Cu(s) + 2NO3-(aq) + 4H+(aq) → 2Cu2+(aq) + 2NO2(g) + 2H2O(l)
Step 7: Simplify the equation by dividing all coefficients by the greatest common divisor, if necessary.
Cu(s) + NO3-(aq) + 2H+(aq) → Cu2+(aq) + NO2(g) + H2O(l)
The balanced redox reaction in acidic solution is:
Cu(s) + NO3-(aq) + 2H+(aq) → Cu2+(aq) + NO2(g) + H2O(l)
To balance a redox reaction in acidic solution, you need to follow these steps:
1. Assign oxidation numbers to all elements in the reaction.
Cu(s) + NO3-(aq) --> Cu2+(aq) + NO2(g)
2. Identify the element that is undergoing oxidation and the element that is undergoing reduction.
In this case, Cu is being oxidized from 0 to +2, and NO3- is being reduced from +5 to +4.
3. Write the two half-reactions for oxidation and reduction.
Oxidation half-reaction: Cu(s) --> Cu2+(aq)
Reduction half-reaction: NO3-(aq) --> NO2(g)
4. Balance the number of atoms in each half-reaction, starting with the non-oxygen and non-hydrogen atoms.
Oxidation half-reaction: Cu(s) --> Cu2+(aq) + 2e-
Reduction half-reaction: 2NO3-(aq) --> 2NO2(g)
5. Balance the charges by adding electrons to the side that has a higher charge. In this case, add 2 electrons to the left side of the oxidation half-reaction.
Oxidation half-reaction: Cu(s) --> Cu2+(aq) + 2e-
Reduction half-reaction: 2NO3-(aq) +2e- --> 2NO2(g)
6. Balance the oxygen atoms by adding water molecules to the side that needs more oxygen. In this case, add 2 water molecules to the right side of the reduction half-reaction.
Oxidation half-reaction: Cu(s) --> Cu2+(aq) + 2e-
Reduction half-reaction: 2NO3-(aq) +2e- --> 2NO2(g) + 2H2O
7. Balance the hydrogen atoms by adding H+ ions to the side that needs more hydrogen. In this case, add 4 hydrogen ions (H+) to the left side of the reduction half-reaction to balance hydrogens.
Oxidation half-reaction: Cu(s) --> Cu2+(aq) + 2e-
Reduction half-reaction: 2NO3-(aq) +2e- + 4H+ --> 2NO2(g) + 2H2O
8. Multiply the oxidation half-reaction by the appropriate factor so that the number of electrons transferred is the same for both half-reactions.
In this case, the factor of 2 can be used to multiply the oxidation half-reaction:
2Cu(s) --> 2Cu2+(aq) + 4e-
9. Combine the two half-reactions, canceling out the electrons.
2Cu(s) + 2NO3-(aq) + 4H+(aq) --> 2Cu2+(aq) + 4NO2(g) + 2H2O(l)
10. Simplify the equation if possible.
Cu(s) + NO3-(aq) + 2H+(aq) --> Cu2+(aq) + 2NO2(g) + H2O(l)
And that's how you balance the redox reaction occurring in an acidic solution.