How to balance this redox reaction

How to balance this redox reaction (using half reaction method):

$$\ce{Cu + HNO3 -> Cu(NO3)2 + NO + NO2 + H^2O}$$

My attempt:

$$\ce{Cu -> Cu(NO3)2 + 2e-}$$ $$\ce{3e- + HNO3 -> NO}$$ $$\ce{e- + HNO3 -> NO2}$$

So we multiple Cu half reaction by 2 to balance electrons and we get:-

$$\ce{2Cu + 2HNO3 -> 2Cu(NO3)2 + NO + NO2}$$

I would have solved further but the correct answer happens to be different.

Correct answer: $\ce{2Cu + 6HNO3 -> Cu(NO3)2 + NO + NO2 + H2O}$

• Please avoid using Latex in titles due to searching issues. Also, you can get better formatting using the \ce command. Thirdly, what is your understanding of the half reaction method? Your equations don't balance so clearly something is wrong. – bon Nov 2 '15 at 19:03
• Your very first half-reaction looks somewhat suspicious: multiple N and O appear out of nowhere. – Ivan Neretin Nov 2 '15 at 19:55
• Your "correct answer" can't be correct. You have two copper atoms on the left and one on the right. You have 6 H on left and two on the right. And so on... – MaxW Nov 2 '15 at 21:14
• There is no way to "balance" this equation as written. It has hydrogen on the left, but none on the right. – MaxW Nov 2 '15 at 21:23
• The point is that in a half cell reaction all the atoms have to balance. You just to add electrons as a "reactant" or a "product". (e.g. $\ce{Cu_{(s)} -> Cu^{2+}_{(aq)} + 2e^-}$) – MaxW Nov 2 '15 at 21:34

How to balance this redox reaction (using half reaction method):$$\ce{Cu + HNO3 -> Cu(NO3)2 + NO + NO2}$$

You really can't balance this equation for two reasons.

(1) There is hydrogen on the left hand side, but none on the right.

(2) There are two competing redox reactions. One produces $\ce{NO}$ and the other produces $\ce{NO_2}$. The overall reactions are:$$\ce{Cu_{(s)} + 4HNO3_{(aq)} -> Cu(NO3)2_{(aq)} + 2NO2_{(g)} + 2H2O_{(l)}}$$ $$\ce{3Cu_{(s)} + 8HNO3_{(aq)} -> 3Cu(NO3)2_{(aq)} + 2NO_{(g)} + 2H2O_{(l)}}$$

Thus the ration of $\ce{NO2_{(g)}}$ to $\ce{NO_{(g)}}$ is a kinetics problem. In concentrated acid the top reaction predominates and you get mostly the first reaction. In dilute acid the second reaction predominates and you get mostly the second reaction.

Now as an example let's look at the redox reaction of the first reaction.

Looking at the copper first we have: $$\ce{Cu_{(s)} -> Cu^{2+}_{(aq)} + 2e^{-}}$$

Just looking at $\ce{NO3^{-} -> NO2 + e{-} }$ the charge balances, and the N atoms balance, but the O atoms do not balance and hence this is not a valid half cell reaction. The valid reaction is: $$\ce{e^{-} + 2HNO3_{(aq)} -> NO^{2-}_{3} + NO2_{(g)} + H2O_{(l)}}$$

Using "redox" I can simply balance the electrons. So I need the Cu reaction as is, but I need to multiple the nitrite acid reaction by two. $$\ce{Cu_{(s)} -> Cu^{2+}_{(aq)} + 2e^{-}}$$ $$\ce{2e^{-} + 4HNO3_{(aq)} -> 2NO^{2-}_{3} + 2NO2_{(g)} + 2H2O_{(l)}}$$

Now that the electrons match I can add the two equations. $$\ce{Cu_{(s)} + 4HNO3_{(aq)} -> Cu(NO3)2_{(aq)} + 2NO2_{(g)} + 2H2O_{(l)}}$$

• If I added the two competing reactions and divided them by common factor 2 I get $2Cu+6HNO_3 \longrightarrow 2Cu(NO_3)_2 + NO + NO_2 + 3H_2O$. Do you think this is the balanced equation. – Abhishek Mhatre Nov 3 '15 at 3:42
• No. Kinetics adds a whole different level of consideration. That is not a "general solution." It would be true for some very specific reaction conditions. (1) The concentration of acid couldn't change, and (2) the geometry of the Cu would have to sat the same throughout the reaction. So for a copper sphere the geometry of the Cu changes (surface area) as the reaction proceeds. Thus to get your balanced equation the copper would have to be like a film on a glass side so the surface area stays the same. – MaxW Nov 3 '15 at 3:54
• The "wrinkle" with the Cu is caused by the fact that the Cu is a solid and not a liquid. Thus there are two "phases" in the reaction. If the reaction were $\ce{Cu^{+}_{(aq)} -> Cu^{2+}_{(aq)}}$ and then it would be a lot easier to control the Cu. The point is that we want to products to stay the same from the time the reaction starts till it finishes. – MaxW Nov 3 '15 at 4:04
• If you used just enough $\ce{HNO3}$ to dissolve the copper then it would be sort of like 90% of the reaction in strong acid and 10% of the reaction in weak acid. That would really be an interesting case to solve exactly. But again, you'd have to make some assumptions about the Cu geometry. – MaxW Nov 3 '15 at 4:07

To balance a reaction, it's not just the electrons which need to balance, it's all entities - both electrons and atoms.

As Ivan mentions in the comments, your first big problem is that your half reactions aren't balanced with respect to atoms. You need to re-write them such that the number of oxygens/nitrogens/hydrogens on the right match the number of oxygens/nitrogens/hydrogens on the left. This may involve adding additional species (such as $\ce{H2O}$ and $\ce{H+}$ or $\ce{OH-}$ if you're in an aqueous solution). A hint here is that the $\ce{Cu(NO3)2}$ is really a salt, and can be thought of as "equal to" $\ce{Cu^{2+}}$ and two $\ce{NO3-}$

Once you have your half reactions balanced, then you need to balance their combination. This may involve playing around with proportions.

A final note, you may want to double check the "correct" answer. The one you give currently fails that key property of a balanced reaction - that is, having an equal number of each element on both sides. For example, there are two $\ce{Cu}$ on the left, but only one on the right. Likewise, there are 6 $\ce{N}$ on the left, but only 4 total on the right, etc.

First, identify the oxidation and the reduction (which you successfully have done). Metallic copper is oxidized to Cu(II), in the form of a copper dinitrate. Then we have N(V) being reduced to both N(II) and N(IV). Assuming this is carried out in an aqueous solution, we can just work with the nitrate ion, and not the nitric acid molecule, and just work with the $\ce{Cu^2+}$ ion, and not the denitrate.

\begin{align} \textrm{Oxidation: } \ce{Cu^0 &-> Cu^{II}} \\ \textrm{Reduction: } \ce{N^{V}O3^- &-> N^{II}O} \\ \textrm{Reduction: } \ce{N^{V}O3^- &-> N^{IV}O2} \end{align}

Step 1: Now you need to find out how many electrons are needed for each of these reactions, which you have correctly done. Step 2: Then you balance each reaction with respect to both mass and charge. To balance for oxygen, you add water on the oxygen-deficit side. Then you can add protons to balance for hydrogen. If done correctly, your charges will be the same on both sides. Step 3: The last step is to multiply each reaction with a factor so that all three reactions involve the same amounts of electrons. The only way to you can directly add them together, is if each reaction involve the same number of electrons. In other words, find the smallest common multiplum of electrons, and multiply each reaction with the appropriate factor to get to that multiplum.