Reaction of copper and nitric acid?
Hello dear chemists,
I've been studying redox reactions recently (10th grade). I don't find it particularly difficult to set up redox reactions. But understanding the chemical background is more difficult.
As an example, I have the reaction of copper and nitric acid to form copper ions and nitrogen dioxide.
I am particularly interested in the reduction, which as we know looks like this:
HNO3 + e- + H+ —> NO2 + H2O
As the oxidation number of nitrogen changes from V to IV, it absorbs an electron, which must be balanced with an H ion.
But I find that pretty hard to imagine. After all, the nitrogen atom doesn't gain an electron. Instead, one of the electrons from the electron pair bond with the "lost" oxygen atom is no longer attracted to it more strongly. If nitrogen had "really" gained an electron, the actual charge would also have to change, right?
I have the same problem with the positively charged H ion: Together with the "lost" H atom and O atom, it forms a water molecule. Shouldn't this water molecule also be positively charged?
For me, the partial reaction would make more sense if it looked like this:
HNO3 + H —> NO2 + H20
Maybe someone can explain to me where my mistake lies.
LG
(2 votes)
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Hi.
First of all, you have the right that nitrogen does not absorb electrons in the reaction of HNO3 with copper, but instead reduces its oxidation number from +5 to +4. This is done by removing an oxygen atom from nitrogen to form nitrogen dioxide (NO2).
The equation you have shown is a reduction half reaction that represents the electron transfer. In this case, the copper atom emits an electron which is absorbed by the nitric acid in order to form the NO2 molecule. The H+ ion acts as a catalyst and helps to initiate the reaction.
Regarding the H+ ion which forms a water molecule with the lost H atom and O atom, it should be noted that the sum of the charges in a redox reaction is always maintained. In this case, the H+ ion carries a positive charge while the water molecule is neutral. If the H+ ion takes up an electron to form H2O, the H+ ion becomes H, which is now also neutral.
Thank you. The “free” electrons are therefore not absorbed by the nitrogen but rather by the hydrogen, are they? This is tricky;-)
Correct! In the reduction half equation, the H+ ion absorbs the electron, not the nitrogen. The electron originates from the copper atom which emits an electron in the reaction to become copper ion. The H+ ion is merely a catalyst and helps to start the reaction and transmit the electrons.
Yeah, that’s right! The proton transfer from H+ to OH leads to the formation of H2 O, and there is no change in the oxidation number, so this is not a redox reaction. The electron of copper provided is then taken up by the nitrogen atom, which leads to a reduction of the nitrogen atom and thus causes a change in the oxidation number and the actual charge.
😀 I have my own fun.
Thank you. I must have said something misunderstandable. I actually meant that the detergent proton H+ after OH (which has dissolved off from the NH3O) splits with the OH- an electron, which of course is drawn more strongly to the O atom.
So OH-+H+ —> H2O (no change in oxidation numbers, no redox reaction)
The electron provided by the copper is then taken up by the nitrogen atom, which then also causes a change in the oxidation number (and also the actual charge). (reduction)
I hope that’s right.
By the way, I think it’s very nice that you have responded so patiently to my partly senseless assumptions.
No, in redox reactions, electrons are exchanged between the particles involved (atoms, ions or molecules) without a catalyst having to be involved. However, a catalyst can increase the reaction rate by allowing the reaction path with a lower activation energy barrier, which increases the probability that particles react with one another and exchange electrons. If a catalyst is involved, however, it normally does not absorb the electrons, but rather facilitates the electron transfer between the particles by changing the reaction path.
Thanks, so in redox reactions, is it always the catalyst (only if there is one) that takes up the electrons?
Yeah, that looks right! The oxygen molecule actually takes along the entire binding electrons when it leaves the bond. The nitrogen is thereby charged twice positively. The nitrogen then absorbs an electron and is simply positively charged. The resulting molecule is then completely neutral.
Thank you for answering me again! I just looked at the structure of nitric acid and a nitrate ion.
The nitric acid emits an H-proton, which makes it simply negatively charged. Now I’m not sure how it goes on with the electron:
Would it look like that?
The oxygen molecule leaving the binding takes the entire binding electrons “with” so that the nitrogen would be twice positively charged. The nitrogen then absorbs the electron, is simply positively charged and the molecule as a whole is neutral.
I’m sorry if I made a mistake or misunderstood. The electron is not absorbed by the nitric acid itself, but by the nitrate ion (NO3-), which is part of the nitric acid (HNO3). The nitrate ion is reduced in the reaction by absorbing the electron emitted by the copper atom. The reduced nitrate ion is then converted to nitrogen dioxide (NO2). I hope this will explain your confusion!
Thank you. I’m still a little uncertain. In your answer, you wrote, the electron is taken up by nitric acid. Where would it go? Or did I get you wrong?
It is possible to approach the task with various possible solutions. An example would be:
nitric acid is present in a more or less dilute solution. Oxonium ions and nitrations are then present.
Can such an ion mixture react with copper?
The Redox series is looking for suitable redox systems and finds the following with the corresponding standard potentials:
NO2 + 3 H2O ⇌ NO3– + 2 H3O+ + e–+0.80 V
Cu2++ 2 e–+0.34 V
You can see that the copper redox system has the smaller standard potential from the two, i.e. is unedlerable. This has the two partial equations. The less noble copper system is oxidized (emitting electrons) and the redox system with the more positive standard potential is reduced (takes electrons). This equation must then be written in reverse.
In order to summarize the two equations for the total redox equation, the electron numbers in both equations must be equal. The coefficients in the upper equation must therefore be doubled. The overall equation is then written without electrons.
Cu + 2 NO3– + 4 H3O+ ⇌ Cu2++ 2 NO2 + 6 H2O
Thank you. I always think it’s great if something is explained so vividly. The division of NO3 to NO2 was very helpful!
So it’s really different when I first thought: the nitrogen atom really picks up an electron and it’s not only “less taken away”. When the O is released from the NO3 bond, it takes along the entire divided electrons, so that the N atom is now twice positively charged. On the same train, however, it takes up the electron provided by the copper.
This also explains the changed oxidation number. The binding electron with the oxygen, which has already been drawn more strongly to this one, is lost and replaced by a new electron, which is now “all the N atom belongs”
Do I see that right?