AS YEAR OUTLINE



A2 – Unit 5

CHEMISTRY

Revision

Booklet

Redox Equilibria

Transition Metals

Organic Chemistry Arenes,

Nitrogen compounds

and Synthesis

5.1 – Redox Chemistry

What is meant by the term redox?

Use the following half equations to write a balanced equation for the reaction between MnO4- and Fe2+.

MnO4- + 8H+ + 5e- [pic] Mn2+ + 4H2O

Fe2+ [pic] Fe3+ + e-

MnO4- + 8H+ + 5Fe2+ ( Mn2+ + 4H2O + 5Fe3+

Describe the titration used to find the concentration of Iodine in a solution.

Include a balanced equation.

2S2O32- + I2 ( 2I- + S4O62-

• The iodine solution is a yellow / brown colour.

• Titrate a known volume with a known concentration of sodium thiosulphate solution.

• Near the end point of the titration, when it becomes very pale, starch should be added turning the solution a dark blue/black colour.

• The end-point is the colourless solution produced by the titration of the blue/black solution. 

Using redox reactions

Mg2+(aq) + 2e- [pic] Mg(s) Eø = -2.37V

Zn2+(aq) + 2e- [pic] Zn(s) Eø = -0.76V

Cu2+(aq) + 2e- [pic] Cu(s) Eø = +0.34V

Fe2+(aq) + 2e- [pic] Fe(s) Eø = -0.44V

Sn2+(aq) + 2e- [pic] Sn(s) Eø = -0.14V

O2(g) + 2H2O(l) + 4e-[pic] 4 OH-(aq) Eø = +0.40V

Br2(aq) + 2e- [pic] 2Br-(aq) Eø = +1.09V

Cl2(aq) + 2e- [pic] 2Cl-(aq) Eø = +1.36V

MnO4-(aq) + 5e- + 8H+(aq) [pic] Mn2+(aq) + 4H2O(l) Eø = +1.51V

‘The standard electrode potential for Cu2+(aq)/Cu(s) is +0.34V’.

What do you understand by this?

This is the electric potential between a solution and ions (of a half – cell) measured relative to the Standard Hydrogen Electrode under standard conditions of 298K, 1atm and in a solution of concentration 1 mol dm-3.

When combined with a hydrogen electrode a copper electrode will give a potential difference of +0.34V

What is a standard electrode and why is one needed in order to measure electrode potentials?

The standard hydrogen electrode contains hydrogen gas at a pressure of 1 atm in equilibrium with a 1.0 mol/dm3 solution of hydrogen ions at the surface of a platinum electrode.

The standard electrode potential of a metal and its solution cannot be measured directly. The voltmeter requires two connections and a difference in potential needs to be measured.

Only potential differences between a metal and a standard electrode can be measured.

Calculate Eøcell for cells made from the following standard half cells:

i) Mg2+(aq)/Mg(s) with Zn2+(aq)/Zn(s)

Ecell = Eoright - Eoleft = Eobottom - Eotop = -0.76 - - 2.37 = + 1.61 V

ii) Zn2+(aq)/Zn(s) with Cu2+(aq)/Cu(s) = -0.76 – 0.34 = - 1.1 V

iii) Br2(aq)/2Br-(aq) with Cl2(aq)/2Cl-(aq) = 1.09 - 1.36 = - 0.27 V

The equation for the first stage of rusting is: 2Fe(s) + O2(g) + 2H2O(l) → 2Fe(OH)2(s). Calculate the E ø and show the reaction is feasible.

E = +0.84V The value is greater than zero so the reaction is feasible.

Use the E ø values to explain why zinc is used in preference to tin for preventing corrosion of steel car bodies.

Zn oxidises preferentially to Fe

If Sn used, Fe oxidises preferentially

Ecell for Zn being oxidised by O2 is more positive than for Fe being oxidised by O2.

The Eøcell value for Mg2+(aq)/Mg(s) with Ag+(aq)/Ag(s) is 3.18V (Mg electrode -ve).

What is Eø for Ag+(aq)/Ag(s)?

The negative electrode is the left electrode (gives a positive Ecell)

Ecell = Eoright - Eoleft

3.18 = Eoright - - 2.37

3.18 – 2.37 = Eoright = + 0.81 V

You can make Cl2 by the oxidation of Cl-. What oxidising agent, from the list above, is capable of doing this? Write an equation for the reaction.

Only Acidified MnO4-(aq)

2MnO4- + 16H+ + 10Cl- ( 2Mn2+ + 8H2O + 5Cl2

How can Eø values be used to predict the likely spontaneity of redox reactions?

If Ecell is positive then the reaction is spontaneous

What two limitations are there to the predictions made using Eø values?

Kinetics: although the position of equilibrium can be predicted, the rate of reaction may be too slow for equilibrium to be attained.

Non-standard conditions. If the conditions (concentrations, pH, temperature) are not the standard ones, the electrode potentials can vary widely.

How is E øcell related to ΔStot?

ΔStot = -zFEcell / T Ecell ( ΔStotal

How is E øcell related to the equilibrium constant?

ΔStotal ( ln K. Ecell ( ln K

Briefly describe two useful applications of electrode potentials.

Fuel cells – A hydrogen electrode and an oxygen electrode.

Breatherlysers – Measure Ecell and indicate if a driver is over the blood alcohol limit.

Finding homogeneous catalysts – Ecell for the catalyst is between the Ecells of the two half electrodes involved in the reaction.

5.2 – Transition Metals

What is a transition metal?

Elements which form one or more stable ions with a partially-filled d-subshell.

Give three characteristics transition metals.

1. Variable oxidation states

2. Formation of complex ions

3. Coloured compounds

4. Catalytic properties

Why do transition metals share similar properties?

Outer electrons are in a d-subshell.

Draw out ‘electrons in boxes’ and use the s p d notation to represent the electron configuration of the following atoms and ions.

(a) Sc Shows [Ar]3d14s2

(b) Sc3+ Shows [Ar]

(c) Fe2+ Shows [Ar]3d6

(d) Fe3+ Shows [Ar]3d5

Suggest why Fe2+ ions are readily oxidised to Fe3+ ions but Mn2+ ions are not readily oxidised to Mn3+ ions.

Fe3+: [Ar]3d5 has less electron repulsion within the 3d subshell.

Fe2+: [Ar]3d6 has electron repulsion when a pair of electrons are in the same d orbital.

Mn2+ is already 3d5 which is more stable than 3d4.

Why is Zn2+ colourless?

[Ar]3d10 Has no unpaired electrons in the d-subshell.

Even if the orbitals split in energy, no electron excitation in the visible region is possible.

Describe the bonding a complex ion.

Lone pairs of electrons on the ligand form co-ordinate bonds or dative covalent bond to the central ion.

Ligands may contain covalent bonds.

Give an example of: a monodentate, a bidentate and a polydentate ligand.

Monodentate ligand is a molecule or negative ion which forms one co-ordinate bond to a metal ion; e.g. H2O, CN–.

Bidentate make two bonds. E.g. 2HN-CH2CH2-NH2

Polydentate make more than two bonds. E.g. EDTA4-

Draw, give the shape and name the following complexes:

(a) [Fe(H2O)6]3- Octahedral Hexaaqua iron(III) ion

(b) [Cu(NH3)4(H2O)2]2+ Octahedral Tetraammine diaqua copper(II) ion

(c) [CrCl4]- Tetrahedral Tetrachloro chromate(III) ion

(d) [CuCl2]- Linear Dichloro cuprate(I) ion

(e) [Pt(NH3)2Cl2] Square planar Diammine dichloro platinum(II)

Explain why the complexes above, apart from[CuCl2]- , are coloured.

When ligands pack around a metal ion, the d-orbitals no longer have exactly the same energies. If they are partially filled, it is possible for an electron to jump from a lower-energy d-orbital to an unoccupied higher-energy d-orbital.

These “d–d transitions” are of an energy corresponding to absorption in the visible region, and so the compound appears to be coloured. The colour is that of the light which is not absorbed.

Explain why the complexes above have different colours.

The “d–d splitting distance” varies. So different energy light which is absorbed.

Give an example of a ligand exchange reaction involving [CuCl4]2-.

What observations that would be made druing the reaction.

Cu(H2O)62+ + 4Cl–(aq) ( CuCl42– + 6H2O

The colour changes from blue through lime-green to yellow-green.

Give an example of a deprotonation reaction involving Cr(H2O)63+.

What observations that would be made druing the reaction.

Cr(H2O)63+ + 6OH- ( Cr(OH)63– + 6H2O

The Cr(H2O)63+ ion is purple solution.

Initially Cr(OH)3 is obtained as a green precipitate.

It is amphoteric and dissolves in excess sodium hydroxide to form a green solution of Cr(OH)63–.

Why is Cr(H2O)63+ more acidic than Cr(H2O)62+?

Cr3+ has a higher charge density than Cr2+.

It is able to polarise the water ligands and makes then more acidic.

Describe the reaction between Cu(H2O)62+ and NH3.

The solution starts blue because of the Cu(H2O)62+ ion.

It first forms a pale blue precipitate of Cu(OH)2, and then this dissolves to give a deep blue coloured solution, containing the Cu(NH3)4(H2O)22+ ion.

Use the following half equations to explain why copper(I) compounds are very unstable.

Cu2+ + e- [pic] Cu+ Eo = +0.15V

Cu+ + e- [pic] Cu Eo = +0.52V

Cu+(aq) ions disproportionate spontaneously to give Cu2+(aq) ions and Cu metal.

Ecell = 0.52 – 0.15 = 0.37 = Positive so spontaneous.

Explain what is meant by the term disproportionation.

ions disproportionate spontaneously to give Cu2+(aq) ions and Cu metal.

Ecell = 0.52 – 0.15 = 0.37 = Positive so spontaneous.

What colour are the following species?

Cr(OH)3(s) Cr(H2O)67+ Ni(OH)2(s)

Grey green Orange Lime green

Fe(OH)2(s) Fe(OH)2(s) Mn(OH)2(s)

Green Brown Buff

Zn(OH)2(s)

White

Describe a use for [Pt(NH3)2Cl2]

Anticancer drug – cis platin

1. Ammonium vanadate(V), NH4VO3 reacts with dilute sulphuric acid to form a solution containing yellow VO2+ ions.

a. Write an ionic equation for the reaction of the anion in NH4VO3 with dilute sulphuric acid.

VO3- + 2H+ => VO2+ + H2O

b. Is the reaction in (a) a redox reaction? Justify your answer.

No because the oxidation number of V is unchanged

c. Addition of zinc to the solution containing VO2+ ions causes the colour to change from yellow to green then to blue, followed by green again and finally violet. State the formulae of the ions responsible for each of these colour changes.

First green colour: VO2+ and VO2+

Second green colour: V(H2O)63+

Violet colour: V(H2O)62+

2. A few drops of NH3 is added to [Ni(H2O)6]2+ and a green precipitate forms. More NH3 is added, the precipitate redissolves and a blue solution forms.

a. What type of bonds are present in the [Ni(H2O)6]2+ ion?

Dative covalent and covalent

b. Write an equation to show the formation of the green precipitate.

[Ni(H2O)6]2+ + 2NH3 => [Ni(OH)2(H2O)4] + 2NH4+

c. Explain why the first reaction is deprotonation.

H+ is removed from H2O

d. Name the type of reaction occurring in the second step.

Ligand exchange

e. Give an equation for the reaction in the second step.

[Ni(OH)2(H2O)4] + 6NH3

f. Explain why to [Ni(H2O)6]2+ is coloured.

D subshell split in energy by ligands

Light in the visible region is absorbed

Electron is promoted to a higher energy level

3. Brass is a widely used alloy that contains copper and zinc. There are many varieties of brass with different compositions. In the volumetric analysis of the composition of brass, the first step is to react a weighed sample of the alloy with nitric acid. This gives a greenish-blue solution. Use the following standard electrode potentials for the question:

Zn2+ + 2e- ( Zn Eθ = -0.76V

Cu2+ + 2e- ( Cu Eθ = +0.34V

NO3- + 2H+ + e- ( NO2 + H2O Eθ = +0.81V

a. Use the information above to calculate the standard electrode potential for the reaction between zinc and nitric acid derive the equation.

E = + 1.57V

Zn + 2NO3- + 4H+ → Zn2+ + 2NO2 + 2H2O

b. Suggest why zinc does not produce hydrogen with nitric acid.

E = + 0.76V for production of hydrogen

Smaller than value in (a) so is less likely

c. If the greenish-blue solution is diluted with water it turns light blue and contains hydrated copper (II) ions. Name the lights blue complex ion and name its shape.

Hexaaquacopper (II)

Octahedral

d. If concentrated hydrochloric acid is added to a portion of the light blue solution it turns green. State the type of reaction that occurs and give an equation for the reaction type.

Ligand exchange

[Cu(H2O)6]2+ + 4Cl- ( CuCl42- + 6H2O

e. The light blue solution is then neutralized and reacted with potassium iodide solution. Use the electrode potentials below to explain why you would not expect a reaction to occur.

Cu2+ + 2e- ( Cu+ Eθ = +0.15V

I2 + 2e- ( 2I- Eθ = +0.54V

E = -0.39V – is negative so is not feasible

f. Explain why, in practice, the reaction does occur and iodine is liberated.

CuI is a solid so conditions are not standard

Equilibrium is pulled over to favour the right hand side

g. When the precipitate formed in the reaction in (e) is filtered off and then dissolved in concentrated aqueous ammonia, a colourless solution is formed. Suggest the formula for the cation in this solution.

[Cu(NH3)4(H2O)2]+

h. If the colourless solution is left to stand in air for some time, it turns blue. State why this is so, naming the reactant responsible for the change.

Atmospheric oxygen

Oxidises Cu+ to Cu2+

i. In determination of the composition of a sample of brass, 1.50 g of the alloy was treated to give 250 cm3 of a neutral solution of copper (II) nitrate and zinc nitrate. Excess potassium iodide solution was added to 25.0 cm3 portions of this solution and the liberated iodine titrated with 0.100 moldm-3 sodium thiosulphate solution. The mean titration was 16.55 cm3.

a. State which indicator you would use for the titration and the colour change seen at the end point.

Starch

Blue-Black to colourless

b. Explain why the indicator is not added until the reaction is nearly complete.

Insoluble black solid complex formed, removes iodine from solution so not all the iodine is titrated. The titre is too low.

c. Calculate the percentage of copper by mass in the brass.

70%

4. a. Complete the electronic configuration of: Cr and Cr3+.

Cr: [Ar]3d54s1 Cr3+: [Ar]3d3

b. State and explain the shape of [Cr(H2O)6]3+.

Octahedral

6 electron pairs around the Cr ion

These repel each other to a position of minimum repulsion

d. State what you would see when dilute sodium hydroxide is added to a solution containing [Cr(H2O)6]3+ until it is in excess.

Green ppt

Dissolves to give a green solution

e. Give equations for the reactions taking place in (c). Identify the role of the hydroxide ion in the reactions.

[Cr(H2O)6]3+ + 3OH- → Cr(OH)3(H2O)3 + 3H2O

Cr(OH)3(H2O)3 + 3OH- → [Cr(OH)6]3- + 3 H2O

Base – removes H+ ions from water ligands

f. Give the structural formula of an organic compound that can be oxidized by potassium dichromate (VI) in dilute sulphuric acid and of an organic product of the reaction.

Any primary alcohol → aldehyde

Any primary alcohol → acid

Any secondary alcohol → ketone

Any aldehyde → acid

f. Both dichromate (VI) ions and manganate (VII) ions need hydrogen ions in order to act as oxidizing agents in titration experiments. Explain using electrode potentials whether hydrochloric acid could be used to provide the H+ ions for these oxidations.

Cr2O72- + 14H+ + 6e- ( 2Cr3+ + 7H2O Eθ = +1.33V

Cl2 + 2e- ( 2Cl- Eθ = +1.36V

MnO4- + 8H+ + 5e- ( Mn2+ + 4H2O Eθ = +1.51V

Ecell = +0.15V for MnO4- reacting with Cl-

Ecell = -0.03V for Cr2O72- reacting with Cl-

Manganate will oxidise chloride as positive Ecell so HCl can not be used

Dichromate can not oxidise chloride as negative Ecell so HCl can be used

g. When aqueous alkali is added to an aqueous solution containing dichromate (VI) ions, the following change takes place:

Cr2O72- + 2OH- ( 2CrO42- + H2O

Explain in terms of oxidation numbers why this is not a redox reaction.

Oxidation number of Cr remains 6+

5. Storage cells and fuel cells are types of electrochemical cells used as sources of energy. Information about five redox systems that could be used in electrochemical cells is shown below.

Redox System 1: Fe2+ + 2e- ( Fe(s) Eθ = -0.44V

Redox System 2: 2H2O(l) + 2e- ( 2OH-(aq) + H2(g) Eθ = -0.83V

Redox System 3: 2H+ + 2e- ( H2(g) Eθ = 0.00V

Redox System 4: O2(g) + 2H2O(l) + 4e- ( 4OH-(aq) Eθ = +0.40V

Redox System 5: O2(g) + 4H+(aq) + 4e- ( 2H2O(l) Eθ = +1.23V

a. The standard electrode potential of redox system 1 can be measured by constructing an electrochemical cell. Draw a diagram to show how the standard electrode potential could be measured for redox system 1. State the conditions needed to measure this standard electrode potential.

Conditions: 1 moldm-3 for solutions, 298K and 1 atms

Ensure: complete circuit drawn with salt bridge and voltmeter; Pt electrode with the H2/H+ electrode, Fe electrode with the Fe2+ solution.

b. When an alkaline hydrogen-oxygen fuel cell is being used to produce electrical energy, chemical changes take place within the cell. Write half equations for the changes that take place at each electrode, write the overall equation for the cell reaction and calculate the standard cell potential of this fuel cell.

Oxygen electrode: O2(g) + 2H2O(l) + 4e- → 4OH-(aq)

Hydrogen electrode: H2(g) + 2OH-(aq) → 2H2O(l) + 2e-

Overall: 2H2(g) + O2(g) → 2H2O(l)

1.23V

c. State one important difference between a fuel cell and a conventional storage cell.

A fuel cell reacts a fuel with oxygen to produce a voltage.

d. People often assume that hydrogen-oxygen fuel cells are a source of energy that is carbon neutral. State one reason why this assumption may not be correct.

Fossil fuels are required to make the fuel cell

e. A student constructs a cell as follows:

• A half cell is made from a strip of chromium metal and a solution of chromium (III) sulphate.

• A second half cell is made from a strip of metal X and a solution of XSO4(aq).

• The two half cells are connected together and a current is allowed to pass for a length of time.

The chromium electrode gains 1.456 g in mass.

The electrode made of metal X loses 1.021 g in mass.

Determine the identity of metal X.

Mol of Cr = 0.028

3 mol X reacts with 2 mol Cr3+

Mol of X = 1.5 x 0.028 = 0.042 g

Molar mass of X = 1.021/0.042 = 24.3

X = Mg

6. a. Construct a balanced equation for the reaction between acidified dichromate (VI) ions and methanoic acid, HCOOH.

Cr2O72- + 3HCOOH + 8H+ → 2Cr3+ + 3CO2 + 7H2O

b. A student added some chromium metals to an acidified solution containing copper (II) ions. A reaction took place. The student confirmed that “chromium is more reactive than copper”.

i) Explain in terms of their electrode potentials why chromium is more reactive than copper.

Cr is a stronger reducing agent.

ii) When the experiment is carried out, the student observed some bubbles of gas. Suggest an explanation for this observation.

Cr reacts with H+ to produce H2

c. Methanoic acid, HCOOH, has the common name of “formic acid”. Direct-Formic acid fuel cells are being developed for use in small, portable electronics such as phones and laptops. Suggest two advantages of using methanoic acid as the fuel rather than hydrogen.

Methanoic acid is a liquid and easier to transport

Hydrogen is explosive

Organic Chemistry – Arenes

Describe the bonding in benzene.

The σ-bonds are built up in a similar way to ethene, leaving an unused 2p orbital on each of the six carbons. These orbitals (in the same orientation) can overlap sideways in both directions, resulting in a delocalised π-electron cloud, containing six π-electrons and stretching over all six atoms, in a “sandwich” which lies above and below the plane of the ring:

[pic]

Describe the evidence that supports your description above.

Electron density maps show - The C–C bonds in benzene are all equal, each of length between a single and a double bond.

Enthalpy studies show - The delocalised structure is much more stable (lower in enthalpy content) than one with three double bonds - probably by more than 150 kJ/mol -1

Chemical reactions show - Benzene is much less reactive than ethene, since reaction involves loss of at least part of this extra stabilisation.

Explain why arenes undergo electrophilic substitution reactions rather than addition reactions.

Once an electrophile has added on to the ring, the subsequent step is likely to be loss of a proton, leading to electrophilic substitution rather than addition, since the stable delocalised π-system is regained.

Why are phenols often described as ‘activated arenes’?

Phenols are compounds in which the -OH group is directly attached to the benzene ring.

The lone pair of electrons on the oxygen atom delocalise into the arene ring.

Extra electron density makes the arene a stronger nucleophile.

Give an example reaction that supports this description.

Bromination – Doesn’t require a catalyst.

Give the reagents and conditions for the reactions below.

Draw an example mechanism showing Electrophilic Substitution

e.g. Nitration

HNO3 + 2H2SO4 [pic] NO2+ + H3O+ + 2HSO4–

The nitronium ion is a powerful electrophile, and this pulls out an electron pair from the π–system,

Nitrogen Compounds

Why are amines classed as bases?

The lone pair on the nitrogen can accept a proton (H+) ion

Why are short chained amines soluble in water?

The lone pair on the nitrogen hydrogen bond with water.

What class of compounds are formed when amines react with acyl chlorides?

Amides

Describe how Nylon (6,6) is made.

HOCO(CH2)4COOH + NH2(CH2)6NH2 (

–CO(CH2)4CO–NH(CH2)6NH–CO(CH2)4CO–NH(CH2)6NH–CO(CH2)4CO–NH(CH2)6NH–

Describe how a mixture of amino acids can be separated and identified.

Use thin layer chromatography. Polar amino acids move furterh in a polar solvent.

Identification – Use Rf value, visualise spots by spraying with ninhydrin.

What are the products from the following reactions;

R-C ≡N  + LiAlH4  ( Amine

CH3-CN + HCl + 2H2O ( Ethanoic aicd

CH3CH2CH(NH2)COOH (aq) + H+ (aq) ( CH3CH2CH(+NH3) COOH

Positively charged amino acid

CH3CH2CH(NH2)COOH (aq) + OH- (aq) ( CH3CH2CH(+NH3) COO-

Negatively charged amino acid

CH3CH2CH(NH2)COOH(aq) + CH2(NH2)COOH(aq) ( Dimer of amino acids ( Protein

Synthesis

Design a synthetic scheme for the following conversion.

Include the reagents and conditions needed in each step.

[pic]

Design a synthetic scheme for the following conversion.

Include the reagents and conditions needed in each step.

[pic]

Synthesis in action

Explain why organic reactions are often heated under reflux.

Either the reactant has a low boiling temperature or the reaction is slow at room temperature. Enables reactions to be heated at their maximum temperature without the loss of any volatile reagents or products.

Draw the apparatus for heating under reflux.

Condenser vertical, no bung, heated, water in at bottom.

Why are organic solvents added to organic reaction mixtures during purification?

To dissolve the organic molecules. Makes layers easier to see and separate later.

Why are organic reaction mixtures washed with water during purification?

To dissolve any water soluble impurities.

Why are organic reaction mixtures washed with sodium carbonate solution during purification?

To neutralize any unreacted acids present.

Name a drying agent that can be used during the purification of organic reaction mixtures.

Anhydrous calcium chloride

Describe the key points involved in the purification of an organic solid.

Choose a suitable solvent. 

Dissolve the impure sample in the minimum volume of hot solvent.

Filter the solution hot under reduced pressure and collect the filtrate.  This removes solid impurities which were insoluble in the solvent.

Allow the filtrate to cool so that crystals of the product form.

Again filter the mixture under reduced pressure.  Soluble impurities are now removed.

Wash the residue with a little cold solvent.

Describe the key points involved in the purification of an organic liquid.

Use distillation, collect product at official boiling point.

Correct thermometer position is vital.

Describe how sensitive organic compounds can be extracted from plant materials.

Use steam distillation.

Do not heat the sample directly – risk of thermal decomposition.

What is the major advantage of using combinatorial chemistry in drug research?

Many possible active compounds can be made at once. Quicker and cheaper.

-----------------------

Br

+ 3HBr

Br

OH

+ 3Br2 (

OH

Br

2,4,6-tribromophenol

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