[Polleke]

Hallo all,

I was wondering if any of the chemical experts here could tell me what I am doing wrong or ok with the next test:

I have been asked to guide some students true a lab test, the synthesis of CuCl.

Now I have the following information:

I start with 5 grams of Cu , I then add 4ml of HNO3
This causes the following reactions: Cu + 4 HNO3 ==> Cu(NO3)2 +2 NO2 +2 H2O.

And I tell them the HNO3 is not acting as an acid but as a oxidizing agent.
But what if they ask why this is? What should I then best answer?



After this step I have been told to add 50ml of H2O , and then there is the formation of a complex with Cu and water. But why and how can I "draw" this complex? What structure?

After this step I simple add some Na2CO3 to neutralize the excessive amount of HNO3 ( Na2CO3 + HNO3 = NaNO3 + CO2 + H2O)

after neutralizing this HNO3 , it states that I will get a CuCO3 complex , but why does the Cu that is first bound to the NO3 (in the water complex?) bound with the CO3 ?  Is this simply because the NO3 will forms bounds with the Na ? and thus meaning that I not only get the rid of the excessive part of HNO3 but all the HNO3 ?


Then I have to use HCl to disolve the CuCO3 when the CuCO3 is dissolved I have to add again some Cu , now how can I best describe the action of the HCl ? What would you say I anyone asked you what happens when adding the HCl ?
And then I have to boil this untill it reaches the boiling point and I need to wait till the temperature changes to a darker color.

Why does this happend and why do I need to boil it?



I know that this are a lot of questions but it has been a while since I did some chemistry and I have to fill in for a sick teacher and have not been able to get more information from him, so I hope someone here can help me so I can guide the students true their lab experiment.

[Astrokel]

hey polleke

And I tell them the HNO3 is not acting as an acid but as a oxidizing agent.
But what if they ask why this is? What should I then best answer?

It's a redox reaction because of reduced and oxidized species observed, however it is not observed in an acid base reaction.

But why and how can I "draw" this complex? What structure?

hexaaquacopper(II) ion, it's an octahedral complex.

After this step I simple add some Na2CO3 to neutralize the excessive amount of HNO3 ( Na2CO3 + HNO3 = NaNO3 + CO2 + H2O)

you need to precipitate the insoluble copper carbonate first then forming the copper chloride complex below.

now how can I best describe the action of the HCl ? What would you say I anyone asked you what happens when adding the HCl ?

to form CuCl₂(green) then the complex [CuCl₄]^2- which is a brown solution

Why does this happend and why do I need to boil it?

because you need to drive the equilibrium towards the formation of CuCl₃^-

[Borek]

And I tell them the HNO3 is not acting as an acid but as a oxidizing agent.
But what if they ask why this is? What should I then best answer?

Activity series. Copper doesn't dissolve in acids.

after neutralizing this HNO3 , it states that I will get a CuCO3 complex

Complex? To some extent for sure, but I would expect copper carbonate preipitate.

but why does the Cu that is first bound to the NO3 (in the water complex?)

Copper is not bound to NO₃^-, this is solution, ions are freely swimming around. If anything, copper is bound to six molecules of water (complex that you have asked about earlier).

bound with the CO3 ?  Is this simply because the NO3 will forms bounds with the Na ? and thus meaning that I not only get the rid of the excessive part of HNO3 but all the HNO3 ?

Honestly, I have no idea what you mean so I can't answer. Still: this is solution, most of things you are listing are unbound ions.

Then I have to use HCl to disolve the CuCO3 when the CuCO3 is dissolved I have to add again some Cu , now how can I best describe the action of the HCl ? What would you say I anyone asked you what happens when adding the HCl ?

Is it all the time the same solution, or have you in the meantime filtered out the precipitate? Because if you are doing it in the same solution, carbonate step seems superficial to me.

And then I have to boil this untill it reaches the boiling point and I need to wait till the temperature changes to a darker color.

Why does this happend and why do I need to boil it?

Boiling - most likely just to speed things up. But could be there is more to it, no idea.

Partially that's what Astrokel already wrote, but I have started to answer long ago and then got distracted - and he posted in the meantime

[Polleke]

now how can I best describe the action of the HCl ? What would you say I anyone asked you what happens when adding the HCl ?

to form CuCl₂(green) then the complex [CuCl₄]^2- which is a brown solution

Why does this happend and why do I need to boil it?

because you need to drive the equilibrium towards the formation of CuCl₃^-

[/quote]

Ok, the following thing is a bit confusing:

to form CuCl₂(green) then the complex [CuCl₄]^2- which is a brown solution

you state that there will be CuCl₂(green) and then [CuCl₄]^2- which is a brown solution

but then you write:
Why does this happend and why do I need to boil it?[/quote]
because you need to drive the equilibrium towards the formation of CuCl₃^-


you ment CuCl₄^-2 ? in stead of CuCl₃^-

Or did you mean CuCl₂ wich would be even more confusing because then you would try to go back from the brown solution towards the green one.



@Borek



Copper doesn't dissolve in acids.

==> so I will still see the copper in his original form? 




Is it all the time the same solution, or have you in the meantime filtered out the precipitate? Because if you are doing it in the same solution, carbonate step seems superficial to me.


==> ah yes, you are right!
I forget an important step:

After adding the Na2CO3 I have to filter the precipate (=the CuCO3)

[Astrokel]

you ment CuCl4-2 ? in stead of CuCl3-

no, i meant CuCl₃^-

After adding HCl you will get CuCl₂ followed by forming the complex [CuCl₄]^2- and lastly with the adding of copper, it is reduced to CuCl₃^-.

[Polleke]

you ment CuCl4-2 ? in stead of CuCl3-

no, i meant CuCl₃^-

After adding HCl you will get CuCl₂ followed by forming the complex [CuCl₄]^2- and lastly with the adding of copper, it is reduced to CuCl₃^-.

Oh I see

Ok thanks a lot.

I'am gonna use your information and boreks to try and get the total picture.

[Borek]

Copper doesn't dissolve in acids.

==> so I will still see the copper in his original form?

No, that why you use an OXIDIZING acid - to get copper dissolved. Later you will convert it to copper chloride. In the case of iron, or zinc, you could simply put the metal into hydrochloric acid - and it will dissolve. Copper won't dissolve in hydrochloric acid, but it will dissolve in concentrated nitric.

[Astrokel]

After adding HCl you will get CuCl2 followed by forming the complex [CuCl4]2- and lastly with the adding of copper, it is reduced to CuCl3-.

im sorry i meant CuCl₃^2- so is my first post mistake

[Polleke]

im sorry i meant CuCl32- so is my first post mistake

ok I am getting confused now

correct me if I am wrong here:

first you will get CuCl2 followed by forming the complex [CuCl4]^-2 and at last with the adding of copper, it is reduced to CuCl₃^-2

and to shift the reaction towards CuCl₃^-2 you boil it?


one general question then:

first you get CuCl2 but this changes towards [CuCl4]^-2  , why?

and why does the extra copper makes it even possible to get CuCl₃^-2 ? is this because of the reducing capacities of Cu?

[Borek]

first you get CuCl2 but this changes towards [CuCl4]^-2  , why?

Because you have lots of Cl^- in teh solution and CuCl₄^2- complex is quite stable.

and why does the extra copper makes it even possible to get CuCl₃^-2 ? is this because of the reducing capacities of Cu?

Yes, metallic copper reduces Cu²⁺ to Cu⁺ (and gets oxidized to the same Cu⁺).

[Astrokel]

Hi, i had my A level paper 2 today and guess what, 21/60 marks was assigned to inorganic and the question happened to be the topic of this thread! In the question, it stated, there are two ways of synthesising CuCl, starting from CuCl₂.

First way: CuCl₂ to X then reduced it with SO₂ to CuCl (two steps)
Second way: CuCl₂ with Cu to Y then add large amount of water to CuCl (two steps)

So in first step of first and second way, it requires exactly two moles of HCl to react with one mole of CuCl₂ forming the intermediate X and Y as the only single compound formed. Write a balanced equation for it and X and Y are intermediate complex formed with X being yellowish green solution and Y colourless.

1) CuCl₂ + 2HCl ---> H₂CuCl₄
2) CuCl₂ + 2HCl  ---> H₂CuCl₃ + Cl^-

My problem is in 1) should i write it in dissociate formed or in this compound formed as they suggested one single compound. I have no idea how to write for 2 and i don't think i have identified Y wrongly.

[Polleke]

Hi, i had my A level paper 2 today and guess what, 21/60 marks was assigned to inorganic and the question happened to be the topic of this thread! In the question, it stated, there are two ways of synthesising CuCl, starting from CuCl2.

First way: CuCl2 to X then reduced it with SO2 to CuCl (two steps)
Second way: CuCl2 with Cu to Y then add large amount of water to CuCl (two steps)

So in first step of first and second way, it requires exactly two moles of HCl to react with one mole of CuCl2 forming the intermediate X and Y as the only single compound formed. Write a balanced equation for it and X and Y are intermediate complex formed with X being yellowish green solution and Y colourless.

1) CuCl2 + 2HCl ---> H2CuCl4
2) CuCl2 + 2HCl  ---> H2CuCl3 + Cl-

My problem is in 1) should i write it in dissociate formed or in this compound formed as they suggested one single compound. I have no idea how to write for 2 and i don't think i have identified Y wrongly.

you found the answer allready Astrokel?

[Astrokel]

I have no idea but i think the ions was X [CuCl₄]^2- and Y [CuCl₃]^2-, but i do not understand what the question means by one single compound only. And if Y is the ion i suggested, it only needs one mole of HCl. I really don't know.