Balancing reaction between copper and nitric acid given the ratio of formed nitrogen oxides

Question

Problem

$\ce{Cu}$ reacts with $\ce{HNO3}$ according to the equation

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

If $\ce{NO}$ and $\ce{NO2}$ are formed in 2:3 ratio, what is coefficient of $\ce{Cu}$ when equation is balanced with simplest whole numbers?

Answer

9

Question

I am getting the coefficient of $\ce{Cu}$ as $2,$ which contradicts the answer:

$$\ce{2 Cu + 9 HNO3 + 5 H+ -> 2 Cu(NO3)2 + 2 NO + 3 NO2 + 7 H2O}$$

Answer 1

The final arbiter of formal correctness of chemical reaction enumeration are laws of mass and charge conservation. If total counts of charge and atoms of every element are not the same on each side, the equation is wrong. Reaction enumeration by following these laws may be troublesome, as general solution leads to resolving a set of linear equations. That can be useful for educational purposes, or rather to implant terror and despair in eyes of young chemistry students.

Chemists do not go that way. They usually use various heuristics to speed up the process, enumerating equations in several steps. For redox reactions, the heuristic is inventory of exchanged electrons, which is tracked by formal oxidation numbers (ON). These numbers are rather just arbitrary convention, but extremely useful.

For enumeration of your reaction ( with expressed ONs of relevant atoms):

$$\ce{Cu^{0} + HN^{+V}O3 -> Cu^{+II}(N^{+V}O3)2 + N^{+II}O + N^{+IV}O2 + H2O}$$

the first step is matching components undergoing redox change.

• Copper changes ON from 0 to +II.i.e. by 2.
• Nitrogen changes ON from +V to +IV ( $\ce{NO2}$), or from +V to +II($\ce{NO}$), i.e. by 1 resp. 3.
• There is given $\ce{NO/NO2}$ ratio 2:3, so there would be as products $\ce{2 NO + 3 NO2}$, or their multiples.
• $\ce{2 NO}$ exchange $2 \cdot 3 = 6$ electrons.
• $\ce{3 NO2}$ exchange $3 \cdot 1 = 3$ electrons.

For simplicity, we can take formally $\ce{2 NO + 3 NO2}$ as a single entity, created from $\ce{HNO3}$ by exchanging totally 9 electrons.

Now comes the "magical" formula $9 \cdot 2 = 2 \cdot 9$.

9 Cu exchanging 2 electrons each gives 18 electrons totally.
2 ( $\ce{2 NO + 3 NO2}$ ) exchanging 9 electrons each has accepted 18 electrons.

So the intermediate result is: $$\ce{9 Cu + HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + H2O}$$

This is enough for the question answer, but we have to finish the equation. It may sometimes happen some coefficients would get fractional, so all would get multiplied to have all integer.

We need supply enough nitrogen for the salt and oxides: $2\cdot 9 + 4 + 6 = 28$ $$\ce{9 Cu + 28 HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + H2O}$$

We need to create water from hydrogens: $$\ce{9 Cu + 28 HNO3 -> 9 Cu(NO3)2 + 4 NO + 6 NO2 + 14 H2O}$$

And we are done. We can formally check the item conservation:

$\ce{Cu}$: $9=9$
$\ce{H}$: $28=28$
$\ce{N}$: $28=28$
$\ce{O}$: $3\cdot 28=84=54+4+12+14=84$
charge: $0=0$

For a quick check,we may just check items we were not enumerating before, like just oxygen and charge.

Answer 2

As @Poutnik has already very well described the intuitive way to solve such problem so I will try to describe a standard method for such problems.

First analyse the reaction and find out the species which are being oxidised and reduced by finding the change in their oxidation state. Initially ignore all other ions which are not included in the redox reaction.

In this reaction copper is being oxidised from its $0$ state to $+2$ state. Whereas the nitrate ion is being reduced to it's $+4$ and $+2$ states from its $+5$ state.

Now try to form half cell reactions for the oxidation and reduction parts of the reaction.

In this example oxidation of copper and reduction of nitrate can be described as follows:

$$ \ce{Cu -> Cu^{2+} + 2e-}\\ \ce{3(NO3- + e^-) + 2(NO3- + 3e^-) + 14H+ -> 3NO2 + 2NO + 7H2O} $$

Note, I have used coefficients $3$ and $2$ in the reduction half cell as the question clearly states that $\ce{NO2}$ and $\ce{NO}$ are produced in the ratio of $3:2$.

Now add these half reactions by multiplying them with coefficients in such a way that there is no net electron gain or deficit.

In this case the coefficients or as @Poutnik says magic numbers/components of magic equation are $9$ and $2$.

Multiplying the half reactions with "magic numbers" and than adding them gives the following balanced ionic reaction:

$$\ce{9Cu + 10NO3- + 28H+ -> 9Cu^{2+} + 6NO2 + 4NO + 14H2O}$$

Now to get the molecular reaction we can add $\ce{18NO3-}$ ions to the reaction to form $\ce{HNO3}$ and $\ce{Cu(NO3)2}$ respectively.

This gives us the final equation:

$$\ce{9Cu + 28HNO3 -> 9Cu(NO3)2 + 6NO2 + 4NO + 14H2O}$$