Balancing an oxidation-reduction reaction?
If anyone could help me in breaking this down step by step so I can understand it better, I'd really appreciate it. The way our notes were given makes everything crammed together and nearly impossible to figure out. The steps we were given in balancing an oxidation-reduction reaction like this make sense to me, the example we had is just way too run together.
So this is another problem we were given:
"Balance the following oxidation-reduction reaction that occurs in acidic solution using the half-reaction method:"
(s) = solid (aq) = aqueous (g) = gas
Cu (s) + NO3(-1) (aq) ---> Cu(+2) (aq) + NO (g)
The (-1) would mean the NO3 has a charge of -1.
- PhilLv 51 decade agoFavorite Answer
I'm not in the habit of giving answers, so I'll give you a procedure listed step by step.
First step is to break down the reaction to the oxidation and reduction components. You follow the same set of steps for each half reaction after this.
Balance the elements other than oxygen and hydrogen. There should only be one in each reaction, so this should be fairly simple.
Next, balance the oxygen atoms by adding water to the correct side of the reaction. You're allowed to add water because this is in an aqueous solution.
Then you balance hydrogens using protons, H+. This comes from the acidic conditions. Even if you are working in a basic solution, you should still do this step, and we'll take care of it later.
As the final step for each half reaction, balance the charge by adding electrons. Keep in mind that the electrons should be added to the reagents side for the reduction and to the products side for the oxidation.
Now that you've balanced each half reaction, you combine them by finding the least common multiple of the number of electrons from the two reactions. The goal is to add the reactions together so that the number of electrons is the same on both sides. Then they'll cancel out.
As an example if the reduction reaction takes in 5 electrons and the oxidation releases 2, the least common multiple is 10. You want to scale up each equation so that it uses 10 electrons. That means multiplying all the coefficients of the reduction reaction by 2 and coefficients of the oxidation reaction by 5.
Add the half reactions together.
Once you've done this is to do the final touch ups. Cancel out any excess H+, H2O and electrons. The electrons should all cancel out.
At that point, you're done for acidic conditions. If you're in basic conditions, add enough OH- to each side to neutralize the H+. You'll get more water molecules and may be able to cancel some of them out.
- zanekevin13Lv 41 decade ago
I use a simpler method (half-reaction method is time-consuming) so just read phil's lecture for the half reaction method.
3Cu (s) + 2NO3(-1) (aq) + 2H+ ---> 3Cu(+2) (aq) + 2NO (g) + H2OSource(s): medical technology student