Sciencemadness Discussion Board

Balanced equation makes no sense

CHRIS25 - 14-1-2015 at 00:53

I have been making copper nitrate using this stoichiometry many times and never questioned it:
Cu + 4HNO3 = Cu(HNO3)2 + 2H2O +2NO2

However I do not understand why the above is necessary since the equation is balanced: Cu + 2HNO3 = Cu(NO2)2 + H2

And the equation is balanced with these half reactions:
e- + HNO3 = NO2 + OH-
Cu = Cu2+ + 2e-

Multiply first half by 2:
2HNO3 + Cu = Cu2+ + 2NO2 + 2OH-

So my question is this, why do wikpedia and a many other papers insist on using 4 moles of Nitric acid and getting Nitrogen dioxide when the theory says you can use 2 moles and get just Hydrogen evolving?

(PS, I still can not use superscript due to web forum error)

chornedsnorkack - 14-1-2015 at 00:56

Just because you can balance a reaction does not mean that it is the one that actually happens. You may instead have a different reaction that you also can balance.

CHRIS25 - 14-1-2015 at 01:53

Quote: Originally posted by chornedsnorkack  
Just because you can balance a reaction does not mean that it is the one that actually happens. You may instead have a different reaction that you also can balance.

Give me an example then of a simple reaction that is balanced with both half equations and the standard normal balancing technique, but in reality you have to use different stiochiometric quantities AND different products are yielded. Maybe this will help me to understand. Because the nitric acid above yields Nitrogen dioxide in practise, but BOTH balancing techniques yield Hydrogen only and half the stoichiometric amounts are needed. I admit to a lack of understanding which is why I have posted this example.

Fulmen - 14-1-2015 at 04:11

Your half reactions are wrong, the reaction is a bit more complex. First HNO3 oxidizes the copper to copper oxide, this produces NO2. In dilute solutions this NO2 will react with water to form nitric acid and NO, but NO2 can also oxidize more copper or escape to air depending on the conditions. The copper oxide then reacts with more nitric acid to produce copper nitrate.

CHRIS25 - 14-1-2015 at 04:50

Quote: Originally posted by Fulmen  
Your half reactions are wrong, the reaction is a bit more complex. First HNO3 oxidizes the copper to copper oxide, this produces NO2. In dilute solutions this NO2 will react with water to form nitric acid and NO, but NO2 can also oxidize more copper or escape to air depending on the conditions. The copper oxide then reacts with more nitric acid to produce copper nitrate.


Cu loses 2 e- in Cu = Cu2+ + 2e- copper oxidised by HNO3 is a completely wrong concept in all the books? it is oxidised by NO and NO3 at the SAME time is the accurate version?

As for the half reactions here was my thinking:

HNO3 is reduced per:
HNO3 = NO2 + OH-
oxidation number of N in HNO3 = 5+ ( H+ + N + O32- ) ( -6+1 = +5 for the N) but the H is lost in solution to an OH, this leaves then NO3 .
NO3 is now N at 5+ and O3 at 2-; this is not balanced so therefore N has been reduced (by what?) to NO2 and one O has gone into solution. Here then we have N at 4+ and O2 at 4-

So this is where I got the final half reactions:
HNO3 has been reduced and gained one electron in
e- + HNO3 = NO2 + OH-

I am relatively new to this way of looking at reactions and am really trying to understand things, so what's wrong with what I am doing please?

And just to complicate things further I can not see why Ag only needs 2 moles of Nitric to yield EXACTLY the same products where Cu takes 4 moles of Nitric. Both Cu and Ag begin with a Zero net charge and the rest is EXACTLY the same. So this is why I started to question the logic of all this balancing. Hence I can not see any difference at all between the two reactions.

[Edited on 14-1-2015 by CHRIS25]

Fulmen - 14-1-2015 at 05:14

You have to start with the actual reactions, these can of course be broken down to quite a number of half reactions. Here's what I assume is happening (unbalanced equations):
1) HNO3 + Cu => CuO + HNO2
2) HNO2 => NO2 + NO + H2O
Depending on the conditions the NO2 can either escape from solution or react with water:
3) NO2 + H2O => HNO3 + HNO2
So the total of these reactions could be either:
4) HNO3 + Cu => CuO + NO + H2O
or
5) HNO3 + Cu => CuO + NO2 + NO + H2O

The final step is simpler:
6) HNO3 + CuO => Cu(NO3)2 + H2O

chornedsnorkack - 14-1-2015 at 05:58

Quote: Originally posted by Fulmen  
Your half reactions are wrong, the reaction is a bit more complex. First HNO3 oxidizes the copper to copper oxide, this produces NO2.


I don´t see why copper oxide needs to form as an intermediate. After all, copper could be oxidized on anode, simply
Cu -> Cu2+ + 2e-

HgDinis25 - 14-1-2015 at 06:17

You seem to have some misconceptions regarding Redox reactions. Just because you can balance an equation, it doesn't mean that reaction is going to work.

A Redox reaction can only happen if the sum of the potentials of the half-reactions gives a positive value.

http://chem.wisc.edu/deptfiles/genchem/sstutorial/Text14/Tx1...

Before any more explanation is given, please read that link. It will help you a lot.

CHRIS25 - 14-1-2015 at 06:23

I appreciate your response Fulmen but it does not help to answer my questions. Sorry. This really needs to be taken apart if I am to understand things:

The first thing that happens when Nitric acid is added to water is the following: HNO32- = H3O+ + NO32-
Now you add Cu and it is immediately attacked/oxidized by ??????? what exactly
If you have the nitrate molecule in solution which part of this is doing the oxidizing, the NO or the NO3 or the NO2
The N in HNO3 has that oxidation state of 5+ whereas in NO it must be 2+ and in NO3 it needs to be 4+ It can not be the the first one because that has dissociated, so that leaves NO and NO2. When I look at the stoichiometry I see that if you use 4 moles Nitric and 1 mole Cu you get NO2. If you use 4 moles of nitric and 3 moles of Cu you get NO only. Copper oxide is black right? When I perform this reaction as I have done many times, the solution goes extreme black before it turns to that nice blue. Maybe you are right? Then strictly speaking why the hell is this not included in any half reaction, it can not be seen with playing around with electrons can it?

But none of this answers any of my original questions, especially the case with silver.

HgDinis25 - 14-1-2015 at 07:36

@CHRIS25
Quote:

The first thing that happens when Nitric acid is added to water is the following: HNO32- = H3O+ + NO32-


What the hell is that nonsense? Are you saying HNO3 is an ion with charge 2-? And the Nitrate ion has a charge of 2-?
Nitric Acid ionizes as follows:
HNO3(aq) + H2O(l) --> H3O+(aq) + NO3-(aq)
Quote:

If you have the nitrate molecule in solution which part of this is doing the oxidizing, the NO or the NO3 or the NO2


Nitrate is an ion not a molecule. And you have NO3- in solution present, not NO or NO2.

Did you even read the link I provided?

woelen - 14-1-2015 at 08:23

Let's reset to the original question of CHRIS25 before people are accusing others of writing nonsense or not understanding things at all . . .

------------------------------------------------------------------------

First, acids can have different modes of oxidation. One mode of oxidation is that H(+) ions pick up electrons and change to hydrogen gas. The half reaction for that is:

2H(+) + 2e --> H2

This reaction, however, only occurs with metals which are sufficiently reactive (e.g. zinc, iron, aluminium). Copper, silver, gold, mercury do not react in this way. This reaction is the typical acid reaction. It is a redox reaction, but acids, which only show this mode of redox-reaction usually are not considered oxidizers. Examples are hydrochloric acid, acetic acid, hydroiodic acid.

Most oxo-acids, however, also can exhibit another mode of oxidation. In this mode of oxidation, the central atom changes oxidation state and not the hydrogen ions. This mode of oxidation does not occur with the ionic dissociated acids (e.g. HNO3 split into H(+) and NO3(-)), but with the undissociated covalent acid molecules. Some examples of such redox reaction's half reactions:

2HNO3 + e --> H2O + NO2 + NO3(-) (one nitrogen atom goes from +5 to +4 oxidation state)
2H2SO4 + 2e --> 2H2O + SO2 + SO4(2-) (one sulphur atom goes from +6 to +4 oxidation state)
4HNO3 + 3e --> 2H2O + NO + 3NO3(-) (one nitrogen atom goes from +5 to +2 oxidation state)

In these situations, these acids do not really act as acids, they are what classically is called "strong oxidizers". The oxidizing power now does not lay in the acidic properties of the compound, but in the high oxidation state of the central atom (in these examples N and S). One acid can even have different modes, as shown by the example of HNO3, depending on concentration, the reducing agent and temperature. In practice, if an acid has different modes, then these reactions even can occur at the same time, leading to mixes of reaction products (in this case NO and NO2).

More examples of such oxo-acids reactions can be given, maybe you can work them out yourself: HClO4, HClO3, HBrO3, H2SeO4

These redox reactions only occur with acids at high concentration, when at least part of the acid is undissociated, or when the reductor is very strong (then the reaction may be a combination of formation of H2 and the other reaction, e.g. when you add zinc to dilute H2SO4 you get mostly H2, but also some SO2 and even a small amount of H2S).
Non-oxoacids, such as HCl and HBr cannot show the type of redox reaction as described immediately above, they can only act as oxidizer through H(+) ions.



[Edited on 14-1-15 by woelen]

CHRIS25 - 14-1-2015 at 08:45

@HgDinis We must have cross-posted because I have only just seen your response. And I am very appreciative of that link, it helps to understand. Also I know that NO3 is an ion in water and that HNO3 is not an ion, perhaps I was too quick in assuming that others would understand that by writing HNO3 = NO3 + H3O would have been obvious that I was referring to dissociation in water. The only mistake I made was by adding 2- to the No3 that was stupid, but then I am untrained, uneducated and learning in a sinusoidal waveform with stepladders attached and gaping holes :) as opposed to being in a college/university linear situation, in which case I would not even have needed to post this question, and many others I ask. Appreciate the link greatly though.

@Woelen This helps to clear things up a bit , thankyou. Need to digest it though. But thanks.

chornedsnorkack - 14-1-2015 at 08:57

Quote: Originally posted by woelen  


Most oxo-acids, however, also can exhibit another mode of oxidation. In this mode of oxidation, the central atom changes oxidation state and not the hydrogen ions. This mode of oxidation does not occur with the ionic dissociated acids (e.g. HNO3 split into H(+) and NO3(-)), but with the undissociated covalent acid molecules.

Then dilute HNO3 should be reduced to H2, not NO by metals, because being a strong acid it is completely dissociated in solution.
Quote: Originally posted by woelen  

More examples of such oxo-acids reactions can be given, maybe you can work them out yourself: HClO4,

I cannot. HClO4 is only oxidant at high concentrations or temperatures, where it also is explosive.
What are the predominant products of metal reduction of concentrated HClO4?
HClO3? ClO2? Cl2? HCl?
Quote: Originally posted by woelen  
HClO3, HBrO3, H2SeO4

These redox reactions only occur with acids at high concentration, when at least part of the acid is undissociated,

Aren´t HClO3, HBrO3 and HMnO4 strong oxidants even in dilute solutions, or as salts in basic conditions?
Quote: Originally posted by woelen  

Non-oxoacids, such as HCl and HBr cannot show the type of redox reaction as described immediately above, they can only act as oxidizer through H(+) ions.

There are few non-oxo oxidizing acids because, e. g. HI3 is commonly not counted as a compound.

DraconicAcid - 14-1-2015 at 09:14

Quote: Originally posted by CHRIS25  

So my question is this, why do wikpedia and a many other papers insist on using 4 moles of Nitric acid and getting Nitrogen dioxide when the theory says you can use 2 moles and get just Hydrogen evolving?


Why not ask your piece of copper why it doesn't simply react with the acid to give off hydrogen, instead of insisting on giving off nitrogen dioxide?

CHRIS25 - 14-1-2015 at 10:01

Quote: Originally posted by DraconicAcid  
Quote: Originally posted by CHRIS25  

So my question is this, why do wikpedia and a many other papers insist on using 4 moles of Nitric acid and getting Nitrogen dioxide when the theory says you can use 2 moles and get just Hydrogen evolving?


Why not ask your piece of copper why it doesn't simply react with the acid to give off hydrogen, instead of insisting on giving off nitrogen dioxide?

He's not a chemist...


@HgDinis, Actually in response to this"
QUOTE:
If you have the nitrate molecule in solution which part of this is doing the oxidizing, the NO or the NO3 or the NO2

And you have NO3- in solution present, not NO or NO2.
END QUOTE
The Nitrogen gains an electron in: e- + HNO3 = NO2 + OH-
therefore since the HNO3 has dissociated leaving the NO3 and this NO3 further splits into NO2 and O, this is why I asked, what is doing the oxidizing? Oxidation number of N in HNO3 is 5+, but the Oxidation number of N in NO3 is 4+. A little confused here actually because nitrogen has gained an electron and is reduced therefore copper is being reduced by Nitrogen Dioxide? Strictly speaking then it's not the HNO3 that is doing the oxidizing but the NO2, is this not the case?




[Edited on 14-1-2015 by CHRIS25]

HgDinis25 - 14-1-2015 at 11:05

Your post is very hard to read. Please be more careful in the future.

Copper oxidizes as follows:
Cu(s) --> Cu2+(aq) + 2e-

Nitrate reduction goes as follow:
NO3-(aq) + e- + 2H+(aq) --> NO2(g) + H2O(l) (x2)

If you were to write the full equation you would get:
Cu(s) + 2NO3-(aq) + 4H+(aq) --> Cu2+(aq) + 2NO2(g) + 2H2O(l)

However, you have four "H+" but only two "NO3-". They are supplied in pairs by the HNO3, so if there are 4 H+ what happens to the other 2 Nitrate ions? They keep the solution's electroneutrality. But, it takes two Nitrate ions to compensate for the Copper ion. Therefore, the final equation is written as:

Cu(s) + 4HNO3(aq) --> Cu(NO3)2(aq) + 2H2O(l) + 2NO3(g)

CHRIS25 - 14-1-2015 at 11:36

Quote: Originally posted by HgDinis25  
Your post is very hard to read. Please be more careful in the future.

Thankyou for your explanation and the electroneutrality is not something I am familiar with, and I do not see it's relevance in balancing equations since I have never had problems with balancing equations before, electroneutrality is not featured in the lecture videos when coming to working out reactions stoichiometrically. Even in the half equations I have never seen any tutor mention this. So why then does Silver not need 4 moles of nitric acid, yet it still produces NO2? The reduction of NO3 to NO2 is without question the same thing that happens with Silver, yet only 2 moles are required.

Also you said: If you were to write the full equation you would get:
Cu(s) + 2NO3-(aq) + 4H+(aq) --> Cu2+(aq) + 2NO2(g) + 2H2O(l)

In the previous half equation you have 2 H+ so why have you just added an extra 2 here from nowhere? and written 4?

You know what just leave it, the whole damn thing is too confusing, it's all over the blasted place, though this is the first time I have had difficulty, maybe it's the nitric acid.

The above comment, however, is unwelcome since you have provided no explanation. My original question at the beginning was very clear indeed, concise and to the point, if there had been a direct answer by you like the one you just provided then there would have been no further confusion on my part. Instead of nit-picking half way down maybe you could have just answered before I put my confused thoughts into writing.


[Edited on 14-1-2015 by CHRIS25]

woelen - 14-1-2015 at 12:54

Quote:
Then dilute HNO3 should be reduced to H2, not NO by metals, because being a strong acid it is completely dissociated in solution.
And indeed, this is what you observe. Add some metal powder to 1 M HNO3 and you will get mainly H2 (try it with e.g. Mg, Zn, Fe). With some metals like zinc or magnesium, you also will get some NO and that is because Zn (or Mg) is quite a strong reductor and is capable of reducing NO3(-) as well, besides reducing H(+). Ions like NO3(-) are fairly inert in aqueous solution, but this does not mean that they cannot be reduced at all. The same is true with dilute H2SO4, I mentioned that example already in my previous post.
Quote:
I cannot. HClO4 is only oxidant at high concentrations or temperatures, where it also is explosive.
What are the predominant products of metal reduction of concentrated HClO4?
HClO3? ClO2? Cl2? HCl?
Pure reduction will lead to Cl(-) ions. The reaction is extremely exothermic with anhydrous HClO4. If you mix HClO4 with combustible matter (including metals), then you get chloride, oxide and water. In practice I expect that quite some perchlorate will remain unreacted, not as HClO4, but as ion ClO4(-), which is MUCH more stable than HClO4.
Quote:
Aren´t HClO3, HBrO3 and HMnO4 strong oxidants even in dilute solutions, or as salts in basic conditions?
Yes, they are and they certainly can act as oxidizer, even in dilute solution. But this does not contradict what I said, it just adds to what I said. I explained the phenomenon of HNO3 reacting to NO2 and/or NO to CHRIS25, but life never is really simple. If looked up in more detail, then other reactions can (and will) occur as well. The nitrate ion also can react as oxidizer in solution, but with much more difficulty than the ones you mention. Sulfate is even less reactive. Perchlorate is the most inert one, it is very very difficult to reduce that ion in dilute aqueous solution. It is the most inert of oxoanions in dilute aqueous solution.
Quote:
There are few non-oxo oxidizing acids because, e. g. HI3 is commonly not counted as a compound.
Sure, I can mention more of them: HSbF6, HICl4, H2PtF6. These, however, are compounds, where another highly electronegative element takes the role of oxygen. These are more general examples of the theory, I explained in the previous post. I did not want to make things more difficult for CHRIS25 than they already were, but your question justifies the mention of these acids in this context. An acid like HSbF6 will react with metals, almost certainly not with production of H2, but with formation of the metal fluoride, HF and a lower fluoride of Sb.

HgDinis25 - 14-1-2015 at 13:07

@CHRIS25
Quote:

Thankyou for your explanation and the electroneutrality is not something I am familiar with, and I do not see it's relevance in balancing equations since I have never had problems with balancing equations before, electroneutrality is not featured in the lecture videos when coming to working out reactions stoichiometrically. Even in the half equations I have never seen any tutor mention this.


You don't need to call it electoneutrality. HNO3 provides H+ and NO3- in pairs. If 4 H+ are used and only 2 NO3- are used, there will be two more NO3- floating around the solution. When writing the equation you combine those two Nitrate ions with the Copper ion, Cu(NO3)2.
Quote:

So why then does Silver not need 4 moles of nitric acid, yet it still produces NO2? The reduction of NO3 to NO2 is without question the same thing that happens with Silver, yet only 2 moles are required.


Silver only loses one electron in its oxidation. Therefore you don't multiply the Nitrate reduction by a factor of two.
Quote:

In the previous half equation you have 2 H+ so why have you just added an extra 2 here from nowhere? and written 4?


When Copper was oxidized it lost 2 electrons. When Nitrate got reduced it consumed 2H+ but only one electron. So, you multiply this half equation by a factor of two. It now consumes 4H+ and uses the two electrons liberated by the oxidation of Copper.
Quote:

You know what just leave it, the whole damn thing is too confusing, it's all over the blasted place, though this is the first time I have had difficulty, maybe it's the nitric acid.


Really, giving up on the first obstacle? Let's pretend that was just a moment of weakness and let's move on.
Quote:

The above comment, however, is unwelcome since you have provided no explanation. My original question at the beginning was very clear indeed, concise and to the point, if there had been a direct answer by you like the one you just provided then there would have been no further confusion on my part. Instead of nit-picking half way down maybe you could have just answered before I put my confused thoughts into writing.



Your last comment was, indeed, hard to read. Read it yourself and you'll se. I do apologize if my first comment seemed too harsh.

j_sum1 - 14-1-2015 at 17:20

Quote: Originally posted by woelen  
Let's reset to the original question of CHRIS25 before people are accusing others of writing nonsense or not understanding things at all . . .

------------------------------------------------------------------------

First, acids can have different modes of oxidation. One mode of oxidation is that H(+) ions pick up electrons and change to hydrogen gas. The half reaction for that is:

2H(+) + 2e --> H2

This reaction, however, only occurs with metals which are sufficiently reactive (e.g. zinc, iron, aluminium). Copper, silver, gold, mercury do not react in this way. This reaction is the typical acid reaction. It is a redox reaction, but acids, which only show this mode of redox-reaction usually are not considered oxidizers. Examples are hydrochloric acid, acetic acid, hydroiodic acid.

Most oxo-acids, however, also can exhibit another mode of oxidation. In this mode of oxidation, the central atom changes oxidation state and not the hydrogen ions. This mode of oxidation does not occur with the ionic dissociated acids (e.g. HNO3 split into H(+) and NO3(-)), but with the undissociated covalent acid molecules. Some examples of such redox reaction's half reactions:

2HNO3 + e --> H2O + NO2 + NO3(-) (one nitrogen atom goes from +5 to +4 oxidation state)
2H2SO4 + 2e --> 2H2O + SO2 + SO4(2-) (one sulphur atom goes from +6 to +4 oxidation state)
4HNO3 + 3e --> 2H2O + NO + 3NO3(-) (one nitrogen atom goes from +5 to +2 oxidation state)

In these situations, these acids do not really act as acids, they are what classically is called "strong oxidizers". The oxidizing power now does not lay in the acidic properties of the compound, but in the high oxidation state of the central atom (in these examples N and S). One acid can even have different modes, as shown by the example of HNO3, depending on concentration, the reducing agent and temperature. In practice, if an acid has different modes, then these reactions even can occur at the same time, leading to mixes of reaction products (in this case NO and NO2).

More examples of such oxo-acids reactions can be given, maybe you can work them out yourself: HClO4, HClO3, HBrO3, H2SeO4

These redox reactions only occur with acids at high concentration, when at least part of the acid is undissociated, or when the reductor is very strong (then the reaction may be a combination of formation of H2 and the other reaction, e.g. when you add zinc to dilute H2SO4 you get mostly H2, but also some SO2 and even a small amount of H2S).
Non-oxoacids, such as HCl and HBr cannot show the type of redox reaction as described immediately above, they can only act as oxidizer through H(+) ions.



[Edited on 14-1-15 by woelen]

This is the clearest explanation of this that i have ever read.
Thank you.

CHRIS25 - 15-1-2015 at 02:00

Quote: Originally posted by HgDinis25  
@CHRIS25

Quote:

You know what just leave it, the whole damn thing is too confusing, it's all over the blasted place, though this is the first time I have had difficulty, maybe it's the nitric acid.


Really, giving up on the first obstacle? Let's pretend that was just a moment of weakness and let's move on.


I was referring to the fact that it felt like your tone was: - as if I should have known this by now... and ....I should be writing in a certain manner....(It is sometimes so that people answer questions with their knowledge rather than answering the persons question) It was to this that I was really giving up on, not the problem, not the chemistry. However having said this, your further detailed explanation clinches it, thankyou for the way you answered my questions, coupled with Woelens explanation I can at least understand what I need to further read up on now. So while this is still an obstacle, at least I have the tools now to climb.

this educational site then is misleading at this point especially when with an 8:3 ratio you get NO instead of NO2:
http://www.chem.uiuc.edu/webfunchem/Redox/HNO3Cu9.htm

Just realized this half equation theory leaves a gaping hole:
Cu(s) + 2NO3-(aq) + 4H+(aq) --> Cu2+(aq) + 2NO2(g) + 2H2O(l)
This I now understand having gone through every step to get here, I see this as a balanced equation with 2+ charge on the left and right sides. But from the net ionic equation alone how can you tell that you are now getting Cu(NO3)2? The 2 NO3's are supplying one oxygen and two NO2's, the 4 Hydrogen are supplying water molecules. So from this alone how on earth can one deduce that Cu(NO3)2 is being formed?

[Edited on 15-1-2015 by CHRIS25]