Transition Metals: Some Chromium Chemistry

AQA, Edexcel, OCR A level Chemistry (2017)

Principles of transition metal chemistry.

Learning Objectives related to chromium chemistry.

Edexcel

15/22. understand, in terms of the relevant Eo values, that the dichromate(VI) ion, Cr2O72−:

•    i  can be reduced to Cr3+ and Cr2+ ions using zinc in acidic conditions. 


•    ii  can be produced by the oxidation of Cr3+ ions using hydrogen peroxide in alkaline conditions (followed by acidification). 


•     

15/23. To know that the dichromate(VI) ion, Cr2O72−, can be converted into chromate(VI) ions as a result of the equilibrium:

2CrO42−+ 2H+ ⇌ Cr2O72−+ H2O.

15/24. To be able to record observations and write suitable equations for the reactions of Cr3+(aq), Fe2+(aq), Fe3+(aq), Co2+(aq) and Cu2+(aq) with aqueous sodium hydroxide and aqueous ammonia, including in excess.

15/25. To be able to write ionic equations to show the difference between ligand exchange and amphoteric behaviour for the reactions in (24) above.

6/38 To understand the reactions of alcohols with:

iii  potassium dichromate(VI) in dilute sulfuric acid to oxidise primary alcohols to aldehydes (including a test for the aldehyde using Benedict’s/Fehling’s solution) and carboxylic acids, and secondary alcohols to ketones. 


In equations, the oxidising agent can be represented as [O]. 


AQA

Primary alcohols can be oxidised to aldehydes which can be further oxidised to carboxylic acids.

Secondary alcohols can be oxidised to ketones. Tertiary alcohols are not easily oxidised.

Acidified potassium dichromate(VI) is a suitable oxidising agent.

Students should be able to:

•   write equations for these oxidation reactions

•   equations showing [O] as oxidant are acceptable 


•   explain how the method used to oxidise a primary alcohol determines whether an

aldehyde or carboxylic acid is obtained.

OCR

Redox reactions


(k) redox reactions and accompanying colour changes for:

(ii) interconversions between Cr3+ and Cr₂ O₇2—

Cr3+ can be oxidised with H2O2/OH– and Cr2O72– reduced with Zn/H+.

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 Chromium Chemistry

Some chromium salts:

Potassium dichromate(VI) K₂Cr₂O₇. 

Chromium(III)oxide Cr2O3

Potassium chromate(VI)

In college and A level chemistry we first come across chromium chemistry in the oxidation of alcohols.

Oxidation of alcohols

A particular oxidising agent can be used to oxidise primary and secondary alcohols to aldehydes and ketones respectively.

This oxidant is potassium dichromate(VI) K₂Cr₂O₇.

Now I’ll tell you an interesting fact.  A student of mine once appeared at the end of class to show me a red rock he’d brought back he said from the Canary Islands where he’d been on holiday.  Washing this red rock under water produced a red solution.  It turned out to be potassium dichromate(VI) K₂Cr₂O₇.  But it wasn’t naturally occurring rather a synthetic product from the growing of crystals.  You can find instructions to grow these crystals here.

They are beautiful to look at.

Potassium dichromate(VI) acts as an oxidant in acidic solution. It is usual to use dilute sulphuric acid (H₂SO₄).

With primary alcohols:

With primary alcohols like ethanol CH₃CH₂OH the products are first the aldehyde ethanal (CH₃CHO) then if sufficient oxidant is available the aldehyde is oxidised to the carboxylic acid ethanoic acid (CH₃COOH).

i) CH₃CH₂OH   +   [O]      ⟶       CH₃CHO    +    H₂O 

The product is distilled over as it is formed in the reaction vessel using apparatus like this below:

ii) CH₃CHO     +   [O]      ⟶      CH₃COOH

To oxidise the alcohol further to the carboxylic acid the reaction mixture needs to be refluxed.

Potassium dichromate(VI) provides the oxygen [O].

With secondary alcohols:

With secondary alcohols like propan-2-ol (CH₃CH(OH)CH₃) the alcohol is oxidised only as far as the ketone (CH₃COCH₃).  Oxidation stops at the ketone because the ketone has no α hydrogen attached to the carbonyl group.

CH₃CH(OH)CH₃     +     [O]       ⟶         CH₃COCH₃    +    H₂O

Tertiary alcohols cannot be oxidised using potassium dichromate(VI) because they do not possess the α hydrogen.

Compare these structures and you can see the absence of the α hydrogen:

The alpha hydrogen is that hydrogen attached to the carbon atom attached to the hydroxyl group. 

The tertiary alcohol has no α hydrogen.

Let’s look now at the full redox equation for the oxidation of a primary alcohol to an aldehyde.

First the half equation for the reduction of potassium dichromate(VI) in acidic solution:

Cr₂O₇^2–   +    14H⁺    +    6e–       ⟶        2Cr³⁺     +    7H₂O 

Second the half equation for the oxidation of a primary alcohol to an aldehyde:

CH₃CH₂CH₂OH       ⟶      CH₃CH₂CHO   +    2H⁺     +    2e–

So to combine the two half equations means that the oxidation half equation needs to by multiplied up by three to: 

3CH₃CH₂CH₂OH       ⟶      3CH₃CH₂CHO   +    6H⁺     +    6e–

and added to the reduction half equation:

3CH₃CH₂CH₂OH       ⟶      3CH₃CH₂CHO   +    6H⁺     +    6e–

Cr₂O₇^2–   +    14H⁺    +    6e–       ⟶        2Cr³⁺     +    7H₂O 

Giving this as the final equation:

Cr₂O₇^2–   +  8H⁺   +  3CH₃CH₂CH₂OH    ⟶   2Cr³⁺  +   3CH₃CH₂CHO  +    7H₂O

orange                                                 green

Redox Reactions of Chromium(III)

i) reduction of dichromate(VI) in acid.

Let’s look now at the action of zinc in acid solution to reduce chromium(VI) to chromium(III) and chromium(II).

Here are the relevant electrode potentials:

Zn²⁺(aq)     +   2e–    ⇋   Zn(s)               –0.76v

Cr³⁺(aq)     +   e–    ⇋     Cr²⁺(aq)            –0.41v

Cr₂O₇^2– + 14H⁺ + 6e–  ⇋  2Cr³⁺ + 7H₂O       +1.33v

These electrode potentials suggest that zinc (top right) will reduce orange chromium(VI) (bottom left) to dark green chromium(III) and then to blue chromium(II).

If the blue solution once formed is left in the air it rapidly oxidises to the chromium(III) state. 

Equations:

Cr₂O₇^2– + 14H⁺ +  3Zn(s) ⇋   3Zn²⁺(aq)  +  2Cr³⁺ + 7H₂O

then

2Cr³⁺(aq)  +   Zn(s)   ⇋      2Cr²⁺(aq)    +   Zn²⁺(aq)    

green                             blue

(ii) Oxidation of chromium(III) in alkaline solution to dichromate(VI).

Here are the relevant electrode potentials

CrO₄^2–(aq)  +  4H₂O + 3e– ⇋   Cr(OH)₃(s)  +  5OH^–      –0.13v

H₂O₂(aq)  + 2H⁺(aq) + 2e– ⇋  2H₂O(l)               +1.77v

These equations suggest that hydrogen peroxide in alkaline conditions will oxidise green chromium(III) up to yellow chromate(VI).  Addition of acid will produce the required orange dichromate(VI)

Equations

2Cr(OH)₃(s) +  10OH^– ⇋   2CrO₄^2–(aq)  +  8H₂O + 6e–

and

  +  3H₂O₂(aq)  + 6H⁺(aq) + 6e– ⇋  6H₂O(l)

giving the overall redox equation to be:

2Cr(OH)₃ + 4OH^–  + 3H₂O₂(aq)   ⇋  2CrO₄^2–(aq)  +  8H₂O

Acidifying the solution turns the chromate(VI) into the dichromate(VI) this is not a redox reaction merely an effect of decreasing the pH of the solution.

2CrO₄^2–(aq)  +2H⁺ (aq)       ⇋     Cr₂O₇^2– (aq)  +   H₂O(l)

yellow                                   orange

Precipitation and ligand exchange reactions of chromium(III) Cr³⁺(aq)

Precipitation:

Hydrated chromium(III) ions are amphoteric meaning that they combine with both acidic and alkaline solutions.

So addition of sodium hydroxide solution to chromium(III) ions produces the chromium(III) hydroxide precipitate Cr(OH)₃ which is mauve.  The image below shows this reaction.

However, addition of further sodium hydroxide solution causes the precipitate to dissolve forming the green solution of Cr(OH)₆^3–(aq) so demonstrating the amphoteric properties of this precipitate.

Overall we can write:

Cr(H₂O)₆³⁺(aq) +  3OH^–(aq)  ⇌ Cr(H₂O)₃(OH)₃(s) +  3H₂O(l)

green                                    grey-green

Cr(H₂O)₃(OH)₃(s)  +  3OH^–(aq) ⇌  Cr(OH)₆(aq)   +  3H₂O(l)

grey-green                                    green

The reaction of the hydrated chromium(III) ion with ammonia is initially an acid—base reaction forming the Cr(OH)₃ precipitate.

Cr(H₂O)₆³⁺(aq) +  3OH^–(aq)  ⇌ Cr(H₂O)₃(OH)₃(s) +  3H₂O(l)

As above with sodium hydroxide solution, the hydroxide ions in the ammonia solution abstract a proton from a water ligand on the chromium ion, three times to form the precipitate.  See the diagram below:

Ligand substitution:

Only when the precipitate is treated with further ammonia solution do the ammonia ligands replace the water and hydroxide ligands to form the hexaamminechromium(III) complex and the solution turns purple.

Cr(H₂O)₃(OH)₃(s)  +   6NH₃(aq)    ⇌    Cr(NH₃)₆(aq)   +  3OH^—(aq) +  3H₂O(l)

Transition Metals: Some Manganese Chemistry

AQA, Edexcel, OCR A level Chemistry (2017)

Principles of transition metal chemistry

Learning Objectives related to manganese chemistry.

AQA

The redox titrations of Fe²⁺ and C₂O₄^2— with MnO₄^—

Students should be able to perform calculations for these titrations and similar redox reactions.

A homogeneous catalyst is in the same phase as the reactants.

When catalysts and reactants are in the same phase, the reaction proceeds through an intermediate species.

To explain, with the aid of equations, how Mn²⁺ ions autocatalyse the reaction between C₂O₄^2— and MnO₄^—

Edexcel

15/35. To know the role of Mn²⁺ ions in autocatalysing the reaction between MnO₄^— and C₂O₄^2— ions

14/18. To be able to carry out both structured and non-structured titration calculations including Fe²⁺/ MnO₄^—.

Some Manganese Chemistry

Manganese chemistry at A level and college level revolves around the use of potassium manganate(VII) (MnO₄^—) in redox titrations and the use of manganese(II) ions (Mn²⁺) as a homogenous catalyst.

A) Manganate(VII) redox titrations

These are usually styled back titrations.

Let’s look at an example of one typically used to measure the percentage purity of an ethandioate (oxalate) salt.

You can check out the actual practical here from the RSC.

A student weighed out a 2.29g sample of impure hydrated potassium iron(III) ethandioate, K₃[Fe(C₂O₄)₃]3H₂O and dissolved it in water.  This solution was added to a 250ml volumetric flask and made up to 250ml with distilled water.

A 25ml aliquot was pipetted into a conical flask and an excess of sulphuric acid was added.  The mixture was warmed to 65^oC and titrated with 0.02M KMnO₄ solution.  26.40ml of the manganate(VII) solution was needed for complete reaction.

Calculation steps

1: Find the number of moles of the manganate(VII)

Use n=cV, so moles MnO₄^—  =   

=   5.28 × 10^—4 moles

2: Construct the manganate(VII)—ethandioate equation.

2MnO₄^-(aq) + 16H⁺(aq) + 5C₂O₄^2-(aq)  ⟶  2Mn²⁺(aq) + 8H₂O(l) + 10CO₂(g)

So if 2 moles of manganate(VII) react with 5 moles of ethandioate ions

then 5.28 × 10^—4 moles of manganate(VII) react with 5/2 × 5.28 × 10^—4 moles of ethandioate ions.

3. The total moles of ethandioate is 10 times the answer to 2 because a 25ml aliquot is a tenth of the 250ml volumetric flask.

Total moles ethandioate =  10 × 5/2 × 5.28 × 10^—4  =  132 × 10^—4 

4. Therefore total moles of the compound potassium iron(III)ethandioate is

total moles = 

44  ×  10^—4 

since one mole of the compound produces 3 moles of ethandioate ions.

5. Next calculate the molar mass of the hydrated potassium iron(III)ethandioate

It turns out to be 491g.mol^—1

6. From the M_r of the potassium iron (III) ethandioate we can find the mass of pure material in the 2.29g.

Use n= m/M_r  and rearrange to give: 

mass (m)  = amount (n)  ×  molar mass (M_r)

therefore

mass (m)   =   44  ×  10^—4   ×  491    =  2.16g

Lastly, percentage purity of the potassium salt is given by

% purity =  mass of pure salt/mass of impure salt   ×  100

therefore

% purity =  2.16g/2.29   ×  100  =  94.3%

B) Manganese(II) ions (Mn²⁺) as an homogenous catalyst, an example of autocatalysis.

2MnO₄^-(aq) + 16H⁺(aq) + 5C₂O₄^2-(aq)  ⟶  2Mn²⁺(aq) + 8H₂O(l) + 10CO₂(g)

In this reaction the Mn²⁺ ions catalyse the reaction once they form.

Initially, the reaction is very slow because the reaction is between two negatively charged ions.  These two ions will more likely repel each other and there will, therefore, be a very high activation energy for this reaction.

The action of Mn²⁺ as a catalyst is two fold.

First, the Mn²⁺ reacts with the manganate(VII) ion producing unstable Mn³⁺ ions.

4Mn²⁺   +  MnO₄^—   +  8H⁺    =   5Mn³⁺    +   4H₂O

Second, these Mn³⁺ ions then oxidise the ethandioate ions to form carbon dioxide and regenerate the Mn²⁺ ions. 

2Mn³⁺    +   C₂O₄^2—     =   2CO₂    +   2Mn²⁺

A plot of concentration of reactants vs time looks something like this:

The reaction rate is given by the gradient to the curve at each concentration.

As the catalyst Mn²⁺ concentration increases, the reaction begins to speed up.

But eventually the concentration of the reactants falls and the rate slows down to the point where the reaction stops.