Redox mass balancing trouble

Question

I'm a university student. I've been given the following redox to study and balance using ionic-electronic method:

$$CuS_{(aq)} + HNO_{3(aq)} \longrightarrow Cu(NO_3)_{2(aq)} + H_2O_{(l)} + SO_{2(g)} + NO_{2(g)}$$

Now, I've bene having some trouble understanding what's going on. First things first: it is written $CuS_{(aq)}$, however $CuS$ doesn't dissolve in water. Isn't the subscript ${(aq)}$ wrong? Besides that, the equation can be rewritten in it's ionic form:

$$CuS + H^+ + NO_3^- \longrightarrow Cu^{2+} + 2NO_3^- + H_2O + SO_2 + NO_2$$

Now I have some doubts: how do I determine wether the ambient is acidic or basic? My line of thinking is the following: $HNO_3$ is a strong acid, henceforth the ambient should be acidic. On the other hand, by using the reaction of ionization of water

$$H_2O \longrightarrow H^+ + OH^-$$

$H^+$ cancels out with the one produced by the strong acid, thus only $OH^-$ remains, making the ambient basic. What Is the right interpretation? Should the water be considered ionized?

Moreover, when removing spectator ions from the ionic equation, after finding the oxidation Number of each element, it seems that the oxidation equation is

$$ CuS \longrightarrow SO_2 $$

(I've written $CuS$ and not $S^{2-}, as it does not dissolve in water) where the sulfur changes from -2 to +4

The species which reduct is the nitrogen. The strange things is that, once spectator ions have been removed, the reduction of nitrogen happens between $NO_3^-$ and $NO_2$, which are both products. At this point I asked myself if the elimination of spectator ions Is even correct.

One more thing. It seems impossible to balance the oxidation reaction of sulfur. If it is correct to write $CuS$ instead of $S^{2-}$, then the balancing should look like this:

$$CuS \longrightarrow SO_2 + 6e^- + 6H^+$$

if the ambient is acidic, while for a basic ambient, it should look like this:

$$CuS + 6OH^- \longrightarrow SO_2 + 6e^- $$

In both cases, it is impossible to balance the mass using water molecules. I've also tried separating $Cu^{2+}$ and $S^{2-}$, obtaining the following:

$$S^{2-} + 2H_2O \longrightarrow SO_2 + 6e^- + 4H^+$$

if the ambient is acidic

$$S^{2-} + 2H_2O + 4OH^- \longrightarrow SO_2 + 6e^-$$

if the ambient is basic. The other equation results in:

$$NO_3^- + e^- + 2H^+ \longrightarrow NO_2 + H_2O$$

if the ambient is acidic

$$NO_3^- + e^- \longrightarrow NO_2 + H_2O + 2OH^-$$

if the ambient is basic. Now combining the two (and multiplying the second by 6), One should obtain the balanced ionic equation. However when using only $S^{2-}$ instead of $CuS$, the final equation is:

$$S^{2-} + 6NO_3^- + 8H^+ \longrightarrow SO_2 + 6NO_2 + 4H_2O$$

if the ambient is acidic. Which should result in the following molecular form:

$$CuS + 6HNO_3 + 2H^+ \longrightarrow Cu(NO_3)_2 + SO_2 +6NO_2 + 4H_2O $$

It is possibile to see that $NO_3^-$ is not balanced. The same thing happens if the ambient is basic (except for the fact that there are no free $OH^-$, as instead is happening with $H^+$ ions for the acidic ambient$

To sum up, these are my questions:

1. Is it right to wirite $CuS_{(aq)}?

2. Has $CuS$ have to be considered in its non ionized form?

3. Has the water in the original reaction have to be separated into $H^+$ and $OH^-$

4. Have spectator ions have to be canceled out?

5. If yes, how Is It possibile that the oxidation happens between two products?

6. Is the ambient acidic or basic?

7. If $CuS$ must not be separated, how is it possibile that the oxidation reaction can't be balanced in mass using water?

8. If $CuS$ has to be separated, how Is It possibile that the final molecular form of the reaction is not balanced?

Thanks in Advance for tour kind help!

Answer 1

I was on a Navy ship with salt water ballasted (a foolish idea) JP-4 fuel in Monel metal tanks. We were anchored in estuarial water and we noticed that the jet fuel started to have a slight copper corrosion. One day the crew chief brought me lots of a black, granular solid saying "I think we broke a carbon filter" Lets check! Thinking that carbon would react slowly with nitric acid I placed a large spatula full in some water and swirled the mix and got some sort of suspension We went to the hood and added some concentrated nitric acid. The result was spectacular!! (luckily i had used a large Erlenmeyer flask) The reaction foamed, emitted a huge amount of red brown gas and formed a granular yellow solid and a blue-green solution. More complete analysis revealed the solid was copper sulfide. Investigation revealed anaerobic bacteria reduced sulfate to $\ce{H2S}$. The $\ce{H2S}$ reacted with the copper in the Monel metal. Net result Nitric acid does react with $\ce{CuS}$ in this case giving molecular sulfur. $\ce{S}$ can be oxidized further to $\ce{SO2}$ with proper conditions.

$\ce{CuS}$ suspended in water can loosely be considered aqueous; requires some imagination that should not be lacking in a chemist. There is a suspension of $\ce{CuS}$ particles in water. $\ce{CuS}$ is an ionic compound with considerable charge feedback from $\ce{S^{2-}}$ to $\ce{Cu^{2+}}$, a possible cause of the black coloration. We can think it as an ionic lattice of $\ce{Cu^{2+}}$ and $\ce{S^{2-}}$ ions; this gives us an oxidation half reaction:

$\ce{CuS + 2H2O <=> Cu^{2+} + SO2 + 4H+ + 6e-}$ [notice acid is formed here]

Reduction: nitric acid, ie. nitrate ion, is reduced to $\ce{NO2}$

$\ce{NO3- + H2O + 1e- <=> NO2 + 2OH- }$

Notice $\ce{OH-}$ is formed here; this is another way of saying that $\ce{H+}$ is consumed. Since the reaction is run in aqueous nitric acid lets write it the second way:

$\ce{NO3- +2H+ +e- <=> NO2 + H2O }$ [acid is consumed here]

The reduction reaction is multiplied by 6 to equalize oxidation and reduction and the equations added.

$\ce{CuS +2H2O + 6NO3- + 12H+ +6e- <=> Cu^{2+} + SO2 + 4H+ +6e- + 6NO2 + 6H2O}$

Remove redundancies:

$\ce{CuS + 6NO3- + 8H+ <=> Cu^{2+} + SO2 + 6NO2 +4H2O}$

There are 2 extra $\ce{H+}$ they are accompanied by two $\ce{NO3-}$ to make one $\ce{Cu[NO3]2}$. These two extra acids needn't be nitric they can be any available acid.

The purpose of the initial anecdote besides not jumping to conclusions was to point out with multiple oxidation states possible different conditions can give different mechanisms and reactions. I think that the sulfur will be oxidized to $\ce{SO2}$ only when nitrate is reduced further to $\ce{NO}$ or even $\ce{N2O}$ and $\ce{N2}$ and the $\ce{NO2}$ is contained in the reaction with careful control of the acid concentration.