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. 

Transition metals: Some Vanadium Chemistry

Edexcel A level Chemistry (2017)

Topic 15: Principles of transition metal chemistry

Learning Objectives related to vanadium chemistry

15/20. To know what the colours of the oxidation states of vanadium (+5, +4, +3 and +2) are in its compounds.

15/21. To understand the redox reactions for the inter-conversion of the oxidation states of vanadium (+5, +4, +3 and +2), in terms of the relevant Eo values.

Some Vanadium Chemistry

Vanadium has several stable oxidation states at room temperature.

Vanadium (II)    V²⁺          purple

Vanadium (III)   V³⁺          green

Vanadium (IV)   VO²⁺         blue

Vanadium (V)    VO₂⁺        yellow

Each of these states can exist in a redox equilibrium with the one below in acid solution.

Each redox equilibrium has its own redox potential.

You add these to an oxidation number chart or redox potential chart.

Here they are:

 V³⁺(aq)   +   e^—     ⟶      V²⁺(aq)    E^o  = —0.26v

 [VO²⁺(aq) + 2H⁺(aq)] + e^—  ⟶   V³⁺(aq) + H₂O(l)   E^o  = +0.34v

 [VO₂⁺(aq) + 2H⁺(aq)] + e^—  ⟶   [VO²⁺(aq) + H₂O(l)]   E^o  = +1.00v

Here is the oxidation number chart for the reduction of these vanadium ions:

Reduction of Vanadium(V) to Vanadium(II)

As the zinc/zinc(II) reduction potential is —0.76v then addition of zinc amalgam (zinc dissolved in mercury) or zinc powder to a solution of vanadium(V) ions will bring about a reduction of the yellow vanadium(V) eventually to purple vanadium(II).

3Zn(s)  +  2VO₂⁺(aq) + 8H⁺(aq) ⟶  2V²⁺(aq)  +  4H₂O(l) + 3Zn²⁺(aq)

yellow                            purple

Reduction of Vanadium(V) to Vanadium(III)

Adding tin powder should reduce the vanadium(V) to vanadium(III) since the tin/tin(II) reduction potential lies at —0.14v.

2Sn(s)  +  2VO₂⁺(aq) + 8H⁺(aq) ⟶  2V³⁺(aq)  +  4H₂O(l) + 2Sn²⁺(aq)

yellow                            green

Reduction of Vanadium(V) to Vanadium(IV)

Adding iodide ions to a solution of vanadium(V) will reduce the vanadium(V) down to vanadium(IV) since the iodine/iodide reduction potential lies at +0.54v.

I^—(aq)  +  VO₂⁺(aq) + 2H⁺(aq) ⟶  VO²⁺(aq) +  H₂O(l) + ½I₂(aq)

yellow                            blue                        brown

However the resultant brown colour of the product iodine will mask the blue of vanadium(IV). 

To remove this brown colour add thiosulphate ions.

   I₂(aq)      +   2S₂O₃^2—(aq) ⟶     2I^—(aq)   +     S₄O₆^2—(aq)      

brown                                       colourless

You can view the experiment to show these changes in vanadium chemistry here.

Transition metals: Heterogeneous catalysis

Edexcel A level Chemistry (2017)

Topic 15: Principles of transition metal chemistry

Learning Objectives related to heterogeneous catalysis

15/29. To know that transition metals and their compounds can act as heterogeneous and homogeneous catalysts

15/30. To know that a heterogeneous catalyst is in a different phase from the reactants and that the reaction occurs at the surface of the catalyst

15/31. To understand, in terms of oxidation number, how V2O5 acts as a catalyst in the contact process

15/32. To understand how a catalytic converter decreases carbon monoxide and nitrogen monoxide emissions from internal combustion engines by:

i adsorption of CO and NO molecules onto the surface of the catalyst 


ii weakening of bonds and chemical reaction 


iii desorption of CO2 and N2 product molecules from the surface of the catalyst

Transition metals as catalysts (2)

Catalysts are substances that change the rate of chemical reaction and remain unchanged at the end of that reaction. 

They act so as to reduce the activation energy of a chemical change.

The reduction of the activation energy of a reaction means that more molecules within that reaction possess the activation energy.

A greater percentage of collisions between these molecules can be effective if the activation energy is lower.  It is this increase in the number of effective collisions that leads to an increase in the reaction rate.  

With transition metals catalysts can be either homogenous or heterogeneous.

Heterogeneous transition metal catalysts

Heterogeneous catalysts are not in the same state as the reagents in the chemical change.

More often than not heterogeneous catalysts are solids over which reactants in the gaseous or liquid state pass. 

Since the catalyst is in a different phase to the reactants the reaction occurs on the surface of the catalyst. 

There are several examples of this type of catalysis. 

First off we will examine the Contact process for the manufacture of sulphuric acid.

 

You probably learned about this process for the manufacture of sulphuric acid in a previous chemistry course. If you did, you will remember that in a key part of the process a catalyst of vanadium(V)oxide is used.

The Contact process

Stage 1:  Combustion

In Stage 1 sulphur is burned in oxygen

S_{8 (l)}   +     8O_2(g)       ⟶    8SO_2(g)

Stage 2: Oxidation

In Stage 2 sulphur dioxide is oxidised to sulphur trioxide.  Oxidation takes place in the presence of vanadium(V)oxide (V₂O₅) catalyst.

2SO_2(g)    +    O_2(g)    ⟶    2SO_3(g)

Stage 3: Dissolution

In Stage 3, sulphur trioxide is added to concentrated sulphuric acid to form oleum.  Sulphur trioxide is better dissolved in sulphuric acid than in water since the dissolution of the gas in water is so exothermic a fog is formed which is very difficult to condense.

  2SO_3(g)     +     H₂SO_4(l)     ⟶   H₂S₂O_7(l)    

Stage 4: Dilution

In Stage 4, oleum is diluted with water to form concentrated sulphuric acid doubling the number of moles of the acid we started with. 

   H₂O_(l)      +      H₂S₂O_7(l)    ⟶       2H₂SO_4(l)                    

The role of the catalyst vanadium(V)oxide.

Sulfur dioxide and oxygen react like this:

2SO_2(g) + O_2(g) ⇌ 2SO_3(g) : ΔH = -197 kJ·mol⁻¹

Since this reaction is exothermic, a lower temperature would shift the chemical equilibrium to the right.  That would increase the percentage yield.

But too low a temperature lowers the rate of formation to an uneconomical level. So to increase the rate, high temperatures (450 °C), medium pressures (1-2 atm), and  vanadium(V)oxide (V₂O₅) are used.  These conditions give a 95% conversion.

All the catalyst does is increase the rate of reaction.   

It does not change the position of equilibrium.

The catalyst acts in two steps:

Step 1:  Oxidation of SO₂ into SO₃ by V⁵⁺:

2SO₂ + 4V⁵⁺ + 2O²⁻ → 2SO₃ + 4V⁴⁺

Step 2: Oxidation of V⁴⁺ back into V⁵⁺ by oxygen to regenerate the catalyst

4V⁴⁺ + O₂ → 4V⁵⁺ + 2O²⁻

This summarises the action of the V₂O₅ catalyst.

How a catalytic converter works

Between the car engine and the exhaust box and silencer there is another box: the catalytic converter.

It is usually fitted close to the exhaust manifold as it is heated using the hot exhaust gases.

Some need to be at about 700^oC to operate at maximum efficiency.

 

The diagram shows us that inside the steel box there is a ceramic structure, a honeycomb that has a very high surface area roughly equivalent to a couple of soccer pitches!!

On the ceramic surface are particles of noble metal platinum and rhodium.  It is on the surface of these metals that the conversion of carbon monoxide to carbon dioxide and nitrogen oxides to nitrogen will occur.  

You can see that in the diagram above two changes take place in the converter.

First, a reduction involving nitrogen oxides and carbon monoxide that can be summarised in the following equation:

NO(g)    +     CO(g)      ⟶        CO₂(g)     +    ½N₂(g) 

Second, an oxidation involving residual oxygen, unburnt hydrocarbons and carbon monoxide.

CO   +    CH₄   +   2½O₂     ⟶       2CO₂     +     2H₂O

These two reactions take place on the surface of the catalyst.

We can see below how this process works when oxygen and carbon monoxide are involved.

A: Adsorption of the gas to the platinum surface

B: Formation of weak bonds between the gas and the surface platinum atoms.

C: Coming together of the two gases on the platinum surface close enough to collide with a very lower E_a activation energy.

D: Collision and conversion of carbon monoxide to carbon dioxide

E: Desorption of the carbon dioxide product from the surface as the bonds with the Platinum are now too weak to hold it there.