Transition metals: some copper chemistry

AQA, Edexcel, OCR A level Chemistry (2017)

Principles of transition metal chemistry

Learning Objectives related to copper chemistry.

AQA:

Students could carry out test-tube reactions of metal-aqua ions e.g. Cu²⁺ with NaOH, NH3 and Na2CO3 .

Edexcel:

15/26. understand that ligand exchange, and an accompanying colour change, occurs in the formation of:

i)              [Cu(ΝΗ3)4(Η2Ο)2]2+ from [Cu(Η2Ο)6]2+ via Cu(OH)2(Η2Ο)4

ii)            [CuCl4]2− from [Cu(Η2Ο)6]2+


15/27. understand that the substitution of small, uncharged ligands (such as H2O) by larger, charged ligands (such as Cl−) can lead to a change in coordination number

15/28. understand, in terms of the large positive increase in ΔSsystem, that the substitution of a monodentate ligand by a bidentate or multidentate ligand leads to a more stable complex ion.

OCR:

(k) redox reactions and accompanying colour changes for:

 (iii) reduction of Cu²⁺  to Cu⁺ and disproportionation of Cu+ to Cu2+ and Cu.

Cu2+ can be reduced with I–. In aqueous conditions, Cu+ readily disproportionates.

Learners will not be required to recall equations but may be required to construct and interpret redox equations using relevant half-equations and oxidation numbers.

Some copper chemistry

Several aspects of copper chemistry occur in UK A level specifications produced by the three main examination boards.

There are simple test tube reactions that you may (I hope you will) have carried out in the lab.  If not, you will find videos of them on You Tube.

With sodium hydroxide (NaOH):

The reaction with sodium hydroxide is a precipitation reaction.  Pale blue copper(II)hydroxide is an insoluble solid:

[Cu(H₂O)₆]²⁺(aq)  +   2OH^—(aq)   ⟶    [Cu(H₂O)₄(OH)₂](s)   +   2H₂O(l)  

Blue solution                                           pale blue solid

With ammonia (NH₃): 

The reaction of aqueous copper(II) ions with ammonia solution, dilute or concentrated, takes place in two stages. 

The first stage involves the formation of the pale blue copper(II)hydroxide precipitate as with sodium hydroxide solution. 

[Cu(H₂O)₆]²⁺(aq)  +   2OH^—(aq)   ⟶    [Cu(H₂O)₄(OH)₂](s)   +   2H₂O(l)  

Blue solution                                           pale blue solid

But the second stage involves ligand substitution.  The ammonia acts as a monodentate ligand and forms a new complex ion: tetra amminecopper(II). 

Four ammine ligands bond to the central copper(II) ion and displace two water molecules and two hydroxide ions.

An entropy increase drives the reaction forward.  Though there are five particles on both sides of the equation the particles on the right hand side are more disordered, there being three types, as opposed to two types on the left.  And the reaction moves from a solid and an aqueous solution to a solely aqueous solution.

[Cu(H₂O)₄(OH)₂](s)  +  4NH₃(aq)  ⟶  [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2H₂O +  2OH^—

Blue solution                                           deep blue solution

With sodium carbonate solution (Na₂CO₃(aq)):

This is another precipitation reaction that leads to the formation of green insoluble copper(II)carbonate. 

[Cu(H₂O)₆]²⁺(aq)  +   2CO₃^2—(aq)    ⟶       CuCO₃(s)   +   6H₂O(l)  

Blue solution                                           green solid

However there is another reaction that takes place simultaneously with the above.  This additional reaction occurs because the hexaaquacopper(II) ion is acidic and the carbonate ions react with the acidic hydrogens in the copper complex. 

This reaction results in the evolution of carbon dioxide gas and the formation of copper hydroxide. 

[Cu(H₂O)₆]²⁺(aq)  +  CO₃^2—(aq) ⟶ [Cu(OH)₂(H₂O)₄](s)  + CO₂(g) + H₂O(l)  

blue solution                                  blue solid

The copper(II)hydroxide combines with the carbonate to form what is commonly known as basic copper(II)carbonate.

CuCO_3.Cu(OH)₂(H₂O)₄

Ligand substitution

I’ve already discussed the reaction of copper(II) ions with ammonia solution as both a precipitation reaction and a ligand substitution reaction.

I’ll copy again here what I said earlier:

The reaction of aqueous copper(II) ions with ammonia solution, dilute or concentrated, takes place in two stages. 

The first stage involves the formation of the pale blue copper(II)hydroxide precipitate as with sodium hydroxide solution. 

[Cu(H₂O)₆]²⁺(aq)  +   2OH^—(aq)   ⟶    [Cu(H₂O)₄(OH)₂](s)   +   2H₂O(l)  

Blue solution                                           pale blue solid

But the second stage involves ligand substitution.  The ammonia acts as a monodentate ligand and forms a new complex ion: tetra amminecopper(II). 

Four ammine ligands bond to the central copper(II) ion and displace two water molecules and two hydroxide ions.

An entropy increase drives the reaction forward.  Though there are five particles on both sides of the equation the particles on the right hand side are more disordered, there being three types, as opposed to two types on the left.  And the reaction moves from a solid and an aqueous solution to a solely aqueous solution.

[Cu(H₂O)₄(OH)₂](s)  +  4NH₃(aq)  ⟶  [Cu(NH₃)₄(H₂O)₂]²⁺(aq) + 2H₂O +  2OH^—

Blue solution                                           deep blue solution

A different ligand substitution reaction occurs with chloride ions in the form of concentrated hydrochloric acid (HCl(aq)) solution. 

In this reaction, the chloride ion is larger than the water molecules already bonded to the copper(II) ion. 

Consequently, whereas six water molecules could fit around the copper(II) ion now only four of the larger chloride ions can do so.

The shape of the complex ion changes from an octahedral complex to a tetrahedral complex. 

The complex ion coordination number has changed from 6 to 4.

There is also a marked colour change from blue to bright yellow green.

[Cu(H₂O)₆]²⁺(aq)  +   4Cl^—(aq)    ⟶     [CuCl₄]^2—(aq)    +   6H₂O(l) 

Blue solution                                  bright yellow green solution

Redox reactions

We can of course use redox potentials to predict the kind of redox reactions that aqueous copper(II) ions will undergo. 

Here are the relevant electrode potentials

Half equation                                                     E^o/v

(1) Cu²⁺ + e^— ⇌  Cu⁺                               +0.159v

(2) Cu²⁺ + 2e^— ⇌ Cu                                      +0.34v

(3) Cu⁺  + e^—  ⇌  Cu                                       +0.52v

(4) I₂    +  2e^—  ⇌  2I^—                                   +0.54v

(5) NO₃^–  + 4H⁺  + 3e^—     ⇌     NO   +   2H₂O     + 0.958v

We can easily predict from these half equations that copper(I) ions will disproportionate.

Compare equation (1) and equation (3) we see that the copper(I) ion reacts with itself to form both copper and copper(II) ions.

(1) Cu²⁺ + e^— ⇌  Cu⁺                              +0.159v

(3) Cu⁺  + e^—  ⇌  Cu                                     +0.52v

overall the disproportionation equation is:

2Cu⁺ (aq)   ⟶     Cu(s)     +    Cu²⁺ (aq)   E_cell  =  +0.361v

Oxidation of copper (Cu) to copper(II) (Cu²⁺) occurs using concentrated nitric acid (HNO₃).

If we compare equation(2) and equation(5)

(2) Cu²⁺ + 2e^— ⇌ Cu                                      +0.34v

(5) NO₃^–  + 4H⁺  + 3e^—     ⇌     NO   +   2H₂O     + 0.958v

then copper is oxidised to copper(II) ions and nitrogen(I)oxide (NO) is formed which instantaneously oxidises to nitrogen(IV)oxide (NO₂) on exposure to  oxygen in the air. 

Overall equation:

3Cu +  2NO₃^–  + 8H⁺     ⇌     3Cu²⁺ +  2NO   +   4H₂O

then      NO    +     ½O₂     ⟶     NO₂

a brown gas (NO₂) is observed coming off the reaction as the mixture turns blue green due to the presence of copper(II) ions (Cu²⁺). 

The increase in oxidation number of the copper is from 0 to + 6 in total and the decrease in oxidation number of the nitrogen is from +10 to +4, a decrease of —6.