Sciencemadness Discussion Board

A Redox Look at Cu/HNO3 Reaction

AJKOER - 6-2-2018 at 09:04

To begin, I examine a redox reaction that first involves NO2- . This may provide a starting point for an analysis of possible reaction mechanics between copper metal and nitric acid along with any pure radical based contributions. Note the following redox half cell from a source (see https://answers.yahoo.com/question/index?qid=20110810234717A... ):

Half reaction 1: NO2- + 2OH- = NO3- + H2O + 2e-

Half reaction 2: Fe3+ + e- = Fe2+

Net cell: .NO2- + 2Fe3+ + 2 OH- = NO3- + 2Fe2+ + H2O

Upon adding 2 H+ + NO3- to each side of the first half cell reaction, which I assume can be driven in either direction:

(2 H+ + NO2- + NO3-) + 2 OH- = 2 NO3- + H2O + 2 H+ + 2 e-

As 2 .NO2 + H2O = 2 H+ + NO2- + NO3- , upon substituting and cancelling H2O from each side:

Revised Half Cell 1: 2 .NO2 + 2 OH- = 2 NO3- + 2 H+ + 2 e

I will rewrite the second half cell replacing ferric/ferrous with cuprous/copper:

Revised Half Cell 2: 2 Cu(+) + 2 e- = 2 Cu

Implied net reaction upon dividing through by 2:

Net: Cu(+) + .NO2 + OH- = Cu + H+ + NO3-

Or, upon reversing the reaction(into the direction it apparently spontaneously proceeds) and adding H+ + NO3- to each side:

Cu + 2 H+ + 2 NO3- = Cu(+) + NO3- + H2O + .NO2

Adding the following likely redox reaction with more HNO3 forming the hydroxyl radical (source: see "Fenton chemistry in biology and medicine*" by Josef Prousek, to quote reaction (15) on page 2330: "For Fe(II) and Cu(I), this situation can be generally depicted as follows [20,39], Fe2+/Cu+ + HOX → Fe3+/Cu2+ + .OH + X- (15) where X = Cl, ONO, and SCN. " ):

Cu(+) + HNO3 → Cu(2+) + .OH + NO2-

From the last two reactions:

Cu + 3 HNO3 = Cu(2+) + NO2- + NO3- + H2O + .OH + .NO2

Adding a final 2HNO3 to each side:

Cu + 5 HNO3 = Cu(NO3)2 + (2 H+ + NO2- + NO3-) + H2O + .OH + NO2

and applying again, 2 .NO2 + H2O = 2 H+ + NO2- + NO3- , we have:

Cu + 5 HNO3 = Cu(NO3)2 + 3 .NO2 + 2 H2O + .OH

But, a radical reaction is possible: .OH + .NO2 + M --> HNO3 + M (see https://pubs.acs.org/doi/abs/10.1021/j100488a024 and https://pubs.acs.org/doi/abs/10.1021/jp9939928 )

Which would imply a final equation of:

Cu + 4 HNO3 = Cu(NO3)2 + 2 NO2 + 2 H2O

which agrees with the usual cited result (see, for example, http://m.wolframalpha.com/input/?i=Cu+%2B+HNO3+-%3E+Cu(NO3)2+%2B+NO2+%2B+H2O ).
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Alternate little shorter workup making an assumption:

Cu + HNO3 = Cu(+) + .NO2 + OH- (from reversing previously cited redox above: Cu(+) + .NO2 + OH- = Cu + H+ + NO3- )

Cu + HNO3 → Cu(+) + .OH + NO2- (assuming a questionable application of a prior referenced reaction: Cu+ + HOX → Cu2+ + .OH + X-, with X = NO2 and lowering the valent states of copper)

The last two reactions above imply the following valid (and I would argue reversible) radical reaction occurring upon mixing of the NO2 in an alkaline solution :

.NO2 + OH- = .OH + NO2- (source: see, for example, Box 1, reaction (R3a) at https://www.envchemgroup.com/anna-jones.html )

which likely could be driven in the reverse direction if the NO2 gas is vented from say a hot solution with added acid (removing the OH- also). Note, this scenario is in line with the overall claimed reaction of Cu with at least 4 HNO3.

I suspect many have have held the view that the action of nitric acid on copper metal may proceed via redox and radical pathways, and I hope the above exercise has presented some education and support for that view.

[Edited on 7-2-2018 by AJKOER]

AJKOER - 6-2-2018 at 15:20

Well, my analysis may, in itself, could be reflective of at least one opinion as to the underlying products produced, which, per a source, is not primarily NO2. Per “New stoichiometry for copper dissolution in nitric acid”, by William D Hill, in J. Chem. Educ., 1987, 64 (12), p 1069, DOI: 10.1021/ed064p1069.4, to quote:

“ To the Editor:
It was indicated in the reaction described by F. M. ElCheikh, S. A. Khalil, M. A. El-Manguch, and Hadi A. Omar [1985, 69, 7611 that copper(Il) nitrate, nitrous acid, nitric oxide, and water are the products of the copper-nitric acid reaction with the indication that nitrogen dioxide does not appear to be a primary product in this reaction. “

Link: https://pubs.acs.org/doi/pdf/10.1021/ed064p1069.4

My translation, touted net summary reaction of:

Cu + 4 HNO3 = Cu(NO3)2 + 2 NO2 + 2 H2O

is likely a too simplistic rendition of underlying reactions.
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Another comment, per “Stoichiometry for copper dissolution in nitric acid: A comment”, by James D. Carr, in J. Chem. Educ., 1990, 67 (2), p 183, DOI: 10.1021/ed067p183.2 . To quote:

“As El-Cheikh et al. state, many sets of coefficients will serve to balance eq 1. The reason is that more than one independent reaction is disguised within a single equation. “

Link: https://pubs.acs.org/doi/pdf/10.1021/ed067p183.2
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Lastly, “Copper dissolution in nitric acid” by Robert A. Stairs in J. Chem. Educ., 1990, 67 (2), p 184, DOI: 10.1021/ed067p184.3 , to quote:

“The correspondence from W. D. Hill [1987,64,1069] and F. M. El-Cheikh et al. [1987,64,1070], on the stoichiometry of the dissolution of copper in nitric acid, is confused as a result of the attempt to write a single equation. The phenomenon is better treated, as far as the stoichiometry is concerned at least, as two or more parallel reactions. “

Link: https://pubs.acs.org/doi/pdf/10.1021/ed067p184.3


[Edited on 6-2-2018 by AJKOER]

Sulaiman - 6-2-2018 at 16:28

I've done no significant copper chemistry yet,
but I have done a little chemistry with its neigbour below it in the periodic table, silver.

When reacting silver with nitric acid the concentration of the acid determines whether NO (dilute) or NO2 is produced by the reaction.
ref. : the Synthesis section of https://en.wikipedia.org/wiki/Silver_nitrate

so Cu + 4 HNO3 = Cu(NO3)2 + 2 NO2 + 2 H2O
may not be such a simple matter.

But then I do not understand why AgNO3 yet Cu(NO3)2
when both copper and silver have full electron shells and one outer s electron,
so I'm clearly ignor.ant in this area

AJKOER - 6-2-2018 at 17:33

My understanding is that there is a variation in the gas produced (varying generally from N2O, NO and NO2, but apparently, even NH3 with Mg and Mn) which is dependent on pH and the particular metal (Cu, Fe, Zn, Co, Ni, Ag, As, ...) .

The underlying mechanics as to the particular gas is likely a sequence of radical based and standard chemical reactions. For example:

H+ + e-(aq) = .H

.H + NO3- = OH- + .NO2

2 .NO2 + H2O = 2 H+ + NO2- + NO3-

.H + NO2- = OH- + .NO

.H + .NO = HNO

2 HNO --> H2N2O2 --> H2O + N2O

[Edited on 7-2-2018 by AJKOER]

DraconicAcid - 6-2-2018 at 17:45

The product being NO2 or NO depends on the concentration of the acid (NO2 when more concentrated, NO when more dilute). The exact pathway obviously involves redox chemistry, and since it produces radical species such as NO and NO2 obviously means that radicals are involved somewhere. So I'm not sure what is the point of saying "I suspect many have have held the view that the action of nitric acid on copper metal may proceed via redox and radical pathways, and I hope the above exercise has presented some education and support for that view. "

AJKOER - 6-2-2018 at 18:22

DraconicAcid: My point is that the underlying mechanism is complex via multiple pathways.

If I was teaching chemistry, I could understand how it is more convenient not to introduce more material leading to more student questions. As such, this may foster a degree of reductionism even on more inherently complex reactions.

Also, read the links provided on educators objecting to the simple reaction summary (with apparently the body of the educational community on board with the single reaction format), with some arguing that there is at least two underlying reactions in play for the copper nitric acid reaction.

[Edited on 7-2-2018 by AJKOER]

DraconicAcid - 6-2-2018 at 21:05

Even first year texts say that there is more than one reaction going on. Your speculation is hardly insightful.

aga - 7-2-2018 at 02:24

Well, at least I found it interesting, having recently part-processed a pile of gold waste slime using this reaction.

The material was 'unclean' to say the least and a lot of frothing happened.

While the larger froth bubbles formed the NO gas was transparent.

As soon as the bubble popped it instantly oxidised to become the famous dark orange/brown NO2 gas.

AJKOER - 7-2-2018 at 06:26

Quote: Originally posted by DraconicAcid  
The product being NO2 or NO depends on the concentration of the acid (NO2 when more concentrated, NO when more dilute)...


Interestingly, at some point with many metals in addition to copper, the amount of NxOy (see https://books.google.com/books?id=SPxJAQAAMAAJ&pg=PA1403... ) and the very vigor of the action of nitric acid on the metal can actually decline with increasing HNO3 concentration.

This may be due to the formation of an insoluble layer of the nitrate (passivation, see https://www.researchgate.net/publication/215895254_REACTION_... ), or I do believe I recall an experiment arguing for the required presence of some HNO2. So a uniform increase in reaction rate with concentration of HNO3 is not always observed.

Here are some dated but supportive extracts from Atomistry.com (http://nitrogen.atomistry.com/nitric_acid_and_metals.html) on the need for nitrous acid:

"Divers assumed that a very small quantity of nitrous acid is necessary to initiate the reaction and functions in a catalytic manner......
The second class of metals includes zinc, magnesium, aluminium, cadmium, tin, lead, iron, and the alkali metals, and no nitrous acid is required to start their reaction with nitric acid. According to Divers, nitrous acid is not produced in appreciable amounts, because further reduction occurs which is due to the action of nascent hydrogen:

HNO3 + 2H = HNO2 + H2O; (Nitrous acid.)
2HNO3 + 8H = H2N2O2 + 4H2O; (Hyponitrous acid.)
HNO3 + 6H = NH2OH + 2H2O; (Hydroxylamine.)
HNO3 + 8H = NH3 + 3H2O. (Ammonia.) "

Modern science suggests replacing 'nascent hydrogen' with the hydrogen atom radical created from:

M(n) --> M(n+1) + e-
e- + kH2O = e-(aq)
H+ + e- (aq) = .H

[Edited on 7-2-2018 by AJKOER]

MrHomeScientist - 7-2-2018 at 07:31

Quote: Originally posted by AJKOER  
My understanding is that there is a variation in the gas produced (varying generally from N2O, NO and NO2, but apparently, even NH3 with Mg and Mn)

Are you saying that reacting magnesium with nitric acid produces ammonia? Do you have a reference for that claim?

AJKOER - 7-2-2018 at 10:26

I can across it recently in an old text. I will see if I can find it.

However, see my edit to my response above to DraconicAcid, where apparently highly anodic metals (like Mg, or better, Mg+NaCl) can introduce solvated electrons which in the presence of H+ forms the hydrogen atom radical which acting on HNO3 creates products via:

.H + NO3- = OH- + .NO2

2 .NO2 + H2O = HNO2 + HNO3

.H + NO2- = OH- + .NO

.H + .NO = HNO

2 HNO --> H2N2O2 --> H2O + N2O

and further to NH3.

Interestingly, apparently also highly anodic aluminum in HNO3 plus NaCl can create the hydrogen atom reducing radical (see “REACTION OF ALUMINIUM WITH DILUTE NITRIC ACID CONTAINING DISSOLVED SODIUM CHLORIDE: ON THE NATURE OF GASEOUS PRODUCT”, by Vladimir Petrusevski, et al, at Chemistry Education in New Zealand, May 2011, p. 7., at https://www.researchgate.net/publication/215904100_REACTION_...). Nitric acid with Al and NaCl is claimed to be reduced down to mainly N2O and some NO2 (possibly from presence of copper ions+Al-->Cu + HNO3-->NO + O2-->NO2).

[Edited on 8-2-2018 by AJKOER]

MrHomeScientist - 7-2-2018 at 11:48

That seems very unlikely to me; I'd be interested to read the reference.

If it does happen the ammonia would immediately react with the rest of the nitric acid, so practically speaking you wouldn't be able to detect it. I suppose it could be tested for by calculating how much the pH should change by reacting a known amount of Mg; if it is greater than expected, that would suggest something else had neutralized some of the acid. Or evaporate to dryness and test the residue for ammonium ions.

AJKOER - 7-2-2018 at 14:15

My Atomistry reference above does list magnesium and as possible final reaction product, ammonia.

Note, per another reference on aluminum/nitric acid provided above, the presence of NaCl (and also Cu(ll)) act as a catalyst, which I will expect likely for magnesium.

Looking for my original reference. [EDIT] Found an alternate reference, to quote:

"The amount of ammonia produced in the reaction between magnesium and nitric acid increases with the concentration of the latter until 40 per cent acid is reached, the quantity then decreases. Much hydrogen is formed; the mixture of this gas and nitric oxide liberated by 13 per cent, acid may be exploded by an electric spark."

Source: Journal of the Chemical Society, Volume 62, Part 2
By Chemical Society (Great Britain), page 1403.

Link: https://books.google.com/books?pg=PA1403&lpg=PA1403&...
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Per a more recent source, to quote an answer to the question "Why does nitric acid release hydrogen ONLY when reacting with Mg and Mn?"

In my opinion, the best answer is:

"Actually, it usually doesn't. It produces all types of products ranging from NO2 to NH4NO3 with exact products depending on concentration and conditions."

Link: https://chemistry.stackexchange.com/questions/27357/why-does...

The implication being that any NH3 formed could be converted into NH4NO3. I would add the following possible source of hydrogen:

.H + .H --> H2

[Edited on 7-2-2018 by AJKOER]

aga - 7-2-2018 at 14:24

I kinda lost interest in chemistry aged 13 when the teacher swore blind that FeCl3 could not possibly dissolve copper, yet i had already used it to etch copper-clad circuit boards.

That was 38 years ago, so i guess Scientific Knowledge has advanced since then, the knowledge being inexistent back when she was learning.

As far as i can tell, very few, if any reactions proceed by 'simple' mechanisms.

In this respect OC is a bit of an eye-opener.

AJKOER - 8-2-2018 at 06:59

Quote: Originally posted by aga  
Well, at least I found it interesting, having recently part-processed a pile of gold waste slime using this reaction.

The material was 'unclean' to say the least and a lot of frothing happened.

While the larger froth bubbles formed the NO gas was transparent.

As soon as the bubble popped it instantly oxidised to become the famous dark orange/brown NO2 gas.


Thanks for your comment which is supportive of educators who decide to demonstrate the action of copper on nitric acid to their students and feel a multiple equation presentation (of some form) more closely fits reality.
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Interestingly, came across another reference citing the need for HNO2 to start a silver/nitric acid reaction (see https://www.jstor.org/stable/115052?seq=1#page_scan_tab_cont... ).

I have two proposed explanation. First, per the 3rd quoted reaction in the opening of this thread:

"Net cell: .NO2- + 2Fe3+ + 2 OH- = NO3- + 2Fe2+ + H2O "

So, assuming the metal to be attacked by HNO3 has an oxide (or sulfide...) coating, then in the presence of NO2- (from HNO2), the above reaction provides a path to lower the valence state (like elemental silver) to start things off.

The second explanation is based on the reaction:

2 H+ + NO2- + NO3- = 2 .NO2 + H2O

where if there is a passivating layer based on a nitrate formation, the HNO2 in the presence of a strong acid may institute a NO2 gas formation clearing the surface. Or, there just are little to no solvated electrons created by the particular metal per reaction R0 below, and as such reaction R1 is not as much in play, and the above reaction supplants reaction R2 in forming NO2.

I would also suggest trying to employ dissolved NaCl based on its effects with aluminum as referenced above.
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A final observation for the system that helps explain the change in gas compositions:

R0: M(n) --> M(n+1) + e-

R1. H+ + e-(aq) = .H

R2. .H + NO3- = OH- + .NO2

R3. 2 .NO2 + H2O = 2 H+ + NO2- + NO3-

R4. .H + NO2- = OH- + .NO

R5. .H + .NO = HNO

R6. 2 HNO --> H2N2O2 --> H2O + N2O

First point, reactions R1, R2 and R4 are promoted in the presence of acid.

Second, copper alloys may be able to form a galvanic cell liberating electrons into solution and thereby are less reliant on reaction R0 to create solvated electrons for R1.

Third, with limited .H, reactions R2, R3 and R4 imply that increasing dilution promotes NO formation at the expense of NO3-, as is seen in the case of copper metal.

Also, applying heat advances the decomposition of HNO2 in R3 via 3 HNO2 = HNO3 + 2 NO + H2O, promoting NO again in a new path.

Finally, the highly anodic metals (like Mg, Al and Zn) create more of the solvated electrons, which are generally in short supply (and not the H+ ion). As such, as long as there is some water for R3 (that is, not addressing the case of fuming nitric acid), the reactions R2, R4 and R5 consume the .H and lead to N2O or possibly some NH3 for the case of magnesium metal even in dilute acid solutions.

[Edited on 8-2-2018 by AJKOER]

[Edited on 8-2-2018 by AJKOER]

AJKOER - 12-2-2018 at 13:20

Can across an interesting study examining the kinetics between metallic silver and nitric acid reaction, "Kinetic Investigation of Reaction between Metallic Silver and Nitric Acid Solutions in the Range 7.22−14.44 M", by Cengiz Özmetin, et al., in Ind. Eng. Chem. Res., 1998, 37 (12), pp 4641–4645, DOI: 10.1021/ie980121w .Link: https://pubs.acs.org/doi/abs/10.1021/ie980121w .

To quote from the abstract:

"In this study, the reaction kinetics between metallic silver and nitric acid solutions was investigated by taking into consideration the parameters of temperature, solid-to-liquid ratio, particle size, stirring speed, nitric acid concentration, and addition of sodium nitrite. It was determined that the dissolution rate of the process increased with decreasing particle size, solid-to-liquid ratio, and acid concentration, and increasing reaction temperature and stirring speed. It was found that the amount of sodium nitrite in the solution has no significant effect on the dissolution rate. "

Points I find interesting are the solid-to-liquid ratio and stirring speed, both of which suggests to me aspects of some surface chemistry. Note my prior reference, in the opening thread, to the implied need for another molecule or surface via the radical reaction:

.OH + .NO2 + M --> HNO3 + M

[Edited on 12-2-2018 by AJKOER]

a general warning re. conc. nitric acid plus metals

Sulaiman - 13-2-2018 at 01:15

Quote: Originally posted by AJKOER  
Interestingly, at some point with many metals in addition to copper, the amount of NxOy (see https://books.google.com/books?id=SPxJAQAAMAAJ&pg=PA1403... ) and the very vigor of the action of nitric acid on the metal can actually decline with increasing HNO3 concentration.

This may be due to the formation of an insoluble layer of the nitrate (passivation, see https://www.researchgate.net/publication/215895254_REACTION_... ), or I do believe I recall an experiment arguing for the required presence of some HNO2. So a uniform increase in reaction rate with concentration of HNO3 is not always observed.


With both copper and silver metal I have tried 69% HNO3 ... be super careful ...
cold conc. HNO3 does indeed seem to form a fairly effective passivation layer, but;
. adding water results in a NO2 volcano
. warming results in a sudden violent volcano
. doing nothing results in a NO2 volcano, after an indeterminate delay.
So if you are impatient for the action of dilute acid,
add the acid in small portions ensuring that most of the acid has reacted before adding more.

More dilute acid will dissolve more NO2 due to volume, boiling removes most of it, and more boiling will be required for concentration/crystallization.
Silver nitrate is so soluble (216 g/100ml @ 20oC) that you need to boil off virtually all liquid.
Copper nitrate is simialr at 125 g/100ml @20oC

P.S. since 69% nitric acid seemed to passivate silver metal fairly well, but subject to thermal runnaway,
I reasoned that red fuming nitric acid should passivate copper powder
IT MOST CERTAINLY DOES NOT :P

[Edited on 13-2-2018 by Sulaiman]