chemistry Unit 5.1

8/3/2019 chemistry Unit 5.1

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Topic 5.1: Redox equilibria

Unit 1.5 knowledge

Redox (reduction oxidation reaction) describes/all chemical reactions in which atoms have their oxidation number/state changedOxidation describes the loss ofelectrons by a molecule, atom or ion (From +2 to +4)

Reduction describes the gain ofelectrons by a molecule, atom or ion (From +4 to +2)

Oxidation number(state) Charge on an atom if the element/compound were ionic

1 Elements have Ox No zero(H2, Br2, Na, Be, K)

2 Monatomic ions Ox No same as charge

3 Ox No of F is always (-1), H is (+1) in most compounds, O(2)

Sum of Ox Nos in a neutral molecule must add up to zero or the charge on the ion

The more electronegativeatom (strength to attract electrons) has the (-)Ox No, C in CO2 is (+4), in CH4 is (4), because carbonmore electronegative than hydrogen

SO42

overall oxidation =2 Ox state of O =2 total =8 Ox state of S = +6Fe in FeCl3 is +3 Cl in NaCl is -1 C in CO is +2 C in CCl4 is +4 O in O2

is

Write ionic half equation for reduction of bromine to bromide ions Br2 + 2e 2Br

Write ionic half equation for oxidation of Fe2+ ions to Fe3+ ions Fe2+ Fe3+ + e

Hence write overall ionic equation for reaction of Fe2+

ions with bromine Br2 + 2Fe2+ 2Br

+ 2Fe

3+

2 half reactions: MnO4

+ 8H+

+ 5eMn

2++ 4H2O NO2

+ H2O NO3

+ 2e

+ 2H

+

Balancing equations: (x2) 2MnO4

+ 16H+

+ 10e

2Mn2+

+ 8H2O (x5) 5NO2

+ 5H2O 5NO3

+ 10e

+ 10H+

Overall: 2MnO4

+ 5NO2+ 6H+ 2Mn

2+ + 5NO3+ 3H2O

Separate into half reactions: 2FeCl2 + Cl2 2FeCl3 Cl2 + 2e 2Cl

Fe2+ Fe3++ e

Relate changes in oxidation number to reaction stoichiometry (relation between the quantities of substances that take part in a

reaction)

The total increase in oxidation number in a reaction = the total decrease

3ClO-(aq) ---> 2Cl-(aq) + ClO3-(aq) the oxidation numbers are

+1 -1 +5

Total decrease 2 Cl atoms at +1 to 2 Cl ions at -1 =4Total increase =1 Cl atom ox. no. +1 to ClO3- ion ox. no. +5 =4

understand the procedures and principles involved in the use of

potassium manganate(VII) to estimate reducing agents and

potassium iodide and sodium thiosulphate to estimate oxidising agents

Titrations involving potassium manganate(VII) ions, MnO4-, are used to estimate concentrations of reducing agents likeethanedioate ions, C2O4

2- and iron II ions,Fe2+

The ionic equations are: 2MnO4-+ 16H

++

5C2O4

2----> 2Mn

2++ 8H2O + 10CO2

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

The purple aqueous manganate VII is added from the burette and the end point is signalled by a permanent pink colour in the

flask. The reaction with ethanedioate needs a temperature of about 60oC. Both reactions requires excess dilute sulphuric acid.

Titrations involving iodine, I2 and thiosulphate ions, S2O32- are used to estimate concentrations of oxidising agents like manganate

VII ions, iodate V ions, IO3-or chlorine. In each case a known amount of the oxidising agent reacts with iodide ions to liberate

iodine in a conical flask.

IO3- + 5I- + 6H+---> 3I2 + 3H2O

2MnO4- + 10I- + 16H+ ---> 2Mn2+ + 8H2O + 5I2

The iodine in the flask is titrated with standardised aqueous sodium thiosulphate in the burette. 2S2O32-

+ I2 = 2I-+

S4O62-The iodine solution in the flask begins with a yellow /brown colour.

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

Titration of this and the end-point is a colourless solution

recall the definition of standard electrode/reduction potential and understand the need for a standard electrode cell diagrams are

not required

Standard electrode potential Eo The emf of a half-cell measured relative to the standard hydrogen electrode, all solutions at

1moldm-3

conc and gases at 1 atm pressure, 298KElectrode potential E Power/potential of an (aq)species to oxidise/reduce

predict the likely direction of spontaneous change of redox reactions

The anti-clockwise rule: the more positive electrode potential on the bottom then anticlockwise arrows for direction of

spontaneous change: Cu(s) +0.34V

Ag+ (aq) + e- ---> Ag(s) +0.80V (most +ve at bottom)

---------->

spontaneous reaction is Ag+(aq) + Cu(s) ----> Ag(s) + Cu

2+(aq)
http://en.wikipedia.org/wiki/Electronhttp://en.wikipedia.org/wiki/Moleculehttp://en.wikipedia.org/wiki/Atomhttp://en.wikipedia.org/wiki/Ionhttp://en.wikipedia.org/wiki/Electronhttp://en.wikipedia.org/wiki/Moleculehttp://en.wikipedia.org/wiki/Atomhttp://en.wikipedia.org/wiki/Ionhttp://en.wikipedia.org/wiki/Ionhttp://en.wikipedia.org/wiki/Atomhttp://en.wikipedia.org/wiki/Moleculehttp://en.wikipedia.org/wiki/Electronhttp://en.wikipedia.org/wiki/Ionhttp://en.wikipedia.org/wiki/Atomhttp://en.wikipedia.org/wiki/Moleculehttp://en.wikipedia.org/wiki/Electron


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For half equations written as above, Driving Force/Cell potential is diff between the 2E values Ecell = Eobottom - Eotop

Copper is a stronger reducing agent than silver as the Eo value of copper is more negative than that of Ag.If Ecell is(+) change from left to right is favoured, thermodynamically feasible, in general

understand why these predictions may be wrong in practice

Standard conditions of 1 molar solutions, 1 atmosphere pressure and 298K must exist for these rules to apply.

In practise the conditions in many reactions are not standard.

If the E

o

values are close (or Ecell is small) then non-standard conditions can change E

o

values.In these cases a reaction does not always happen as expected

understand disproportionation reactions in terms of standard electrode potentials

Disproportionation is the oxidation and reduction of the same element in the same reaction.

Cu+(aq) Eo = +0.15V

Cu+(aq) + e- ----> Cu(s) Eo = +0.52V

---------------->

understand the applications of electrode potentials in connection with corrosion and to the solution of problems caused by

corrosion

To prevent corrosion iron is coated (galvanised) with zinc. Zinc has a more negative electrode potential than iron, therefore zinc is

oxidised in preference to iron. i.e. zinc corrodes in place of iron therefore the iron is protected.Iron rusting : 4H+(aq) + O2(g) + 2Fe(s) => 2Fe

2+(aq) +2H2O(l) Zn(s) + Fe2+(aq) Zn2+(aq) + Fe(s)

understand the application of electrode potential to the construction of simple storage cellsA simple/primary cell has 2 electrodes (one metal more reactive than the other), each dipping into an electrolyte with the

electrolytes connected by a salt bridge.

Zn is very reactive (gives away its electrons) and goes into solution, making a -ve electrode. Cu becomes the +ve electrode.

A primary cell cannot have its reactants regenerated by charging.

A lead-acid battery (e.g. a car battery) is a secondary cell or storage cell which can be recharged. It consists of a lead plate, a

lead(IV)oxide in a lead grid plate and a sulphuric acid electrolyte.

DURING DISCHARGING

Reaction at negative lead plate is oxidation (so it is the anode) PbSO4(s) + 2e- Pb(s) + SO4

2-(aq)

Eo

= - 0.36VReaction at positive lead (IV) oxide plate is reduction (so it is the cathode) PbO2(s) + 4H

+(aq) + SO42-(aq) + 2e- PbSO4(s)

+ 2H2O(l) Eo

= +1.69VWhen both plates are connected, a current flows. When all the Pb and PbO2 have been converted to PbSO4 the reaction stops.

The spontaneous reaction (E = +2.05V) for discharging the cell is Pb(s) + PbO2(s) + 4H+(aq) + 2SO4

2-(aq)2PbSO4(s) +

2H2O(l)

The cell is recharged by the reverse reaction.

DURING CHARGING

the reaction at the positive lead plate is reduction (so it is the cathode): PbSO4(s) + 2e- Pb(s) + SO4

2-(aq)

the reaction at the negative lead(IV)oxide plate is oxidation (so it is the anode) PbSO4(s) + 2H2O(l) PbO2(s) + 4H+(aq) +

SO42-(aq) + 2e-

The overall reaction for charging is: 2PbSO4(s) + 2H2O(l) Pb(s) + PbO2(s) + 4H+(aq) +

2SO42-

(aq)

Hydration of metal ions In solution metal ions attract water molecules forming complex ions([M(H2O)6]x+)

The metal ion is joined to 6 water ligands by dative covalent bonds.

If complex is a cation, named by using prefix for ligands followed by name of central ion with its Ox

No(hexaaquairon(II)[Fe(H2O)6]2+

)

If complex is an anion, name of central ion is changed to ending inate (ferrate(Fe)cuprate(Cu))(hexacyanoferrate(II)[Fe(CN) 6]

4)

Roman numeral denotes(original) Ox No of central cation not charge on complex ion

Hexaaquacopper(II) hydrated copper(II) ions

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. Thisgives a greenish-blue solution.

(a) The following standard electrode potentials are needed for this question:


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Eo/V

Zn2+

+ 2e

Zn 0.76

Cu2+

+ 2e

Cu + 0.34

NO3

+ 2H+

+ e

NO2+ H2O + 0.81

(i) Calculate the standard electrode potential for the reaction between zinc and nitric acid and derive the equation.

Zn + 2NO3

+ 4H+

Zn2+

+ 2NO2+ 2H2O E = +1.57V

(ii) Suggest why zinc does not produce hydrogen with nitric acid.

E reaction for the production of hydrogen is + 0.76V smaller than reaction in (i) so is less likely OR NO3

being the oxidised form of a redox couple with amore positive E

othan E

oH

+/ H2is a stronger oxidising

agent than H+

(iii) If the greenish-blue solution is diluted with water it turns light blue and contains hydrated copper(II) ions.

Name the light blue complex ion and draw its structure so as to show its shape.

Hexaaquacopper(II)

(iv) If conc HCl 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.

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

(b) The light blue solution from (a)(iii) is then neutralised, and reacted with an excess of potassium iodide solution.

The following standard electrode potentials are needed:

Eo/V

Cu2+

+ e-Cu

++ 0.15

I2+ 2e-2I

- + 0.54

(i) From Eovalues why would thisreaction not occur 2Cu

2+(aq) + 4I

-(aq) 2CuI(s) + I2 (aq)

Eo

for the reaction is -0.39V(so not feasible)

(ii) Explain why, in practice, the reaction in (i) does occur and iodine is liberated.

CuI is a solid, equilibrium moves to favour RHS

(iii) When the ppt formed in reaction (i) is filtered off and then dissolved in conc NH3 (aq), a colourless solution is produced.

Suggest the formula of the cation in this solution.

[Cu(NH3)4]+

(iv) If the colourless solution from (iii) 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+

(c) In a determination of the composition of a sample of brass, 1.5g 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 mol

dm3 sodium thiosulphate solution. The mean titre was 16.55 cm3

2Cu2+

(aq) + 4I-(aq) 2CuI(s) + I2 (aq)

2S2O32-

(aq) + I2 (aq) 2I-

(aq) + S4O62-

(aq)

(i) State which indicator you would use for the titration and the colour change seen at the end point.Starch, blue-black to colourless

(ii) Explain why the indicator is not added until the reaction is nearly complete.

If added too early, insoluble complex, not all the iodine is titrated

(iii) Calculate the percentage of copper by mass in this brass.

Amount thiosulphate = 0.01655 dm3

0.1 mol dm-3

= amount Cu2+

in 25cm3

= 1.655 10-3

mol

amount Cu2+

in 250cm3

= 1.655 10-2

mol

mass of Cu (in sample) = 1.655 10-2

63.5 = 1.051 g

% Cu in brass = 1.051 100/1.5 = 70 %

2. (a) Identify the substances and conditions used in the standard chlorine half-cell.

Pt electrode, chlorine gas at 1 atm, chloride ions at 1.0 moldm-3

(b) Cu+

(aq) + e- Cu(s) Eo = +0.52 V

Cu2+

(aq) + e- Cu+(aq) E

o= +0.15 V

(i) Combine the half-equations above to produce the redox equation for the disproportionation reaction.

2Cu+(aq) Cu(s) + Cu

2+(aq)


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(ii) Use the Eo

values to explain why this reaction is feasible.

Ecell = +0.37 V is positive

(iii) Explain why the above reaction can be classified as a disproportionation reaction.

The Cu+

is oxidised to Cu2+

and Cu+

also reduced to Cu

(c) Complete the electronic configuration for a copper(I) ion, Cu+.

(1s2) 2s

22p

63s

23p

63d

10

(d) When excess conc NH3 is added to a solution of hydrated copper(II) ions, the complex [Cu(NH3)4 (H2O)2]2+

is formed.

(i) State the type of reaction occurring and give the colour of the complex.

ligand substitution, deep/dark blue(ii) Explain why this complex has a colour.

d-orbitals split (in energy) by ligands absorbs energy(light in visible region) electron moves to a higher energy

level

(iii) Explain why copper(I) complex ions are not coloured.

full d subshell Therefore d-d transitions impossible

(e) When conc HCl acid is added to an aqueous solution of Cu2+

ions, the complex CuCl42-

is formed.

Suggest the shape of this complex, state the bond angle and suggest why it has this shape.

tetrahedral range 109110o 4 (bonding) pairs of electrons repel to a position of maximum separation

Eo/V

Fe3+

(aq) + e- Fe

2+(aq) +0.77

Cl2 (aq) + 2e

-

2Cl

-

(aq) +1.36MnO4

-(aq) + 8H

+(aq) + 5e

- Mn

2+(aq) + 4H2O(l) +1.51

(a) (i) Use the data to explain why dil HCl acid is not used to acidify solutions of KMnO4

Eo

= +0.15V (MnO4-/Mn

2+) more positive or greater than E

o(Cl2/Cl

-)

(so) MnO4reacts with Cl- OR Cl

-ions form Cl2 OR KMnO4 reacts with HCl

(ii) Explain why titrations involving KMnO4 solution do not require the addition of an indicator.

Colour change of colourless to pink

(b) (i) The ionic equation for the oxidation of iron(II) ions by manganate(VII) ions in acidic solution is

MnO4-

(aq) + 5Fe2+

(aq) + 8H+

(aq) Mn2+

(aq) + 4H2O(l) + 5Fe3+

(aq)

Explain, in terms of the half equations listed above, why the ratio of MnO4 ions to Fe2+

ions is 1 : 5 in this reaction.

Have to multiply iron half equation by 5 to cancel out/balance electrons

(ii) Tablets containing FeSO4.7H2O mass 6g, were dissolved in distilled water and made up to 200cm3

in a volumetric flask.

25cm3portions of this solution were titrated against a 0.02mol dm

-3solution of acidified KMnO4

The mean titre was 20.10 cm3. Calculate % of FeSO4.7H2O in the tablets. [Molar mass FeSO4.7H2O = 278 g mol1]

Moles MnO4-= (0.02x20.1) /1000 = 0.000402 mol MnO4

-

Moles Fe2+

per 25 cm3 = 5 x 0.000402= 0.00201 mol Fe2+

Moles Fe2+

per 200 cm3 = 0.00201 x 200/25 mol Fe2+ = 0.01608 mol Fe2+

Mass of FeSO4.7H2O = 0.01608 x 278 = 4.47g or via concentrations

Percentage purity = 4.47/6 x 100% = 74.5%

(c) An important application of redox reactions is in car batteries. The electrolyte is aqueous sulphuric acid

Eo/V

Pb2+

(aq) + 2e- Pb(s) 0.13

PbO2 (s) + 4H+

(aq) + 2e- Pb

2+(aq) + 2H2O(l) +1.46

(i) Calculate the standard e.m.f. of the cell. Eo

= +1.46( -0.13) = +1.59V

(ii) A single cell in a car battery has an e.m.f. of 2V Suggest why this value is different from the answer calculated in (i).

PbSO4 ppted OR [H+(aq)] not 1moldm

-3 OR [Pb

2+(aq)] not 1moldm

-3

OR the conditions (in the car battery) are not standard

(b) (i) When a metal is placed in a solution of its ions, the electrical potential set upbetween the metal and the solution cannot be measured without using a reference

electrode. Explain why this is so.

(ii) Label the diagram of the standard hydrogen electrode


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Write an overall equation for the 1st

stage in the rusting of iron

2Fe(s) + O2(g) + 2H

2O(l) 2Fe

2+

(aq) + 4OH-

(aq)

Calculate Eofor the reaction and show that its feasible

E

react= +0.84V Greater than zero therefore feasible

Use the Eovalues above to explain why zinc is used in preference to tin for preventing corrosion of steel car bodies.Zn oxidises preferentially to Fe/Zinc acts as sacrificial (anode)

If Sn used (and damaged), Fe oxidises preferentiallyE

Zn2+

/Zn more negative than for Fe

OR E

Zn/Zn2+

more positive than for Fe

OR E

cellfor Zn being oxidised by O

2is more positive than for Fe being oxidised by O

2

OR similar E

arguments related to preferential oxidation with Sn

2. (a) (i) Give the electronic configuration of Fe and Fe2+ Fe

[Ar] 3d6

4s2

Fe2+

[Ar] 3d6

(ii) Draw the structure of the hexaaquairon(II) ion, [Fe(H2O)6]2+

so as to clearly show its shape.

(iii) Give the equation for the complete reaction of sodium hydroxide solution with a solution of hexaaquairon(II) ions.

OR

(iv) State what you would see if the product mixture in (iii) is left to stand in air.Green ppt/solidred

(v) Give the equation for a reaction in which iron metal is used as a catalyst.


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(b) Consider the half reaction(i) Define the term standard electrode potential with reference to this electrode.

Emf of cell / potential difference of cell containing Fe2+

and Feand standard hydrogen electrode/half cell OR hydrogen electrode and 1 mol

dm3

H+

and 1 atm H2

1 mol dm3

Fe2+

(ii) Explain, with the aid of an equation, why the value ofEsuggests that iron will react with an aqueous solution of an acid to

give Fe2+ ions and hydrogen gas

Emf of hydrogen electrode is zerostated or implied(e.g. if calculate Ecell

= +0.44 (V))

Fe + 2H+

Fe2+

+ H2

Potential for the reaction is positive so reaction is feasible

OR H+

and ()H2has a more +ve electrode potential than Fe

2+

and Fe (1)

H+ will oxidise Fe / H+ is an oxidising agent / Fe is a reducing agent for H+/ other correct redox statement (1)

Fe + 2H+

Fe2+

+ H2

(iii) State why Evalues cannot predict that a reaction will occur, only that it is possible High Ea so slowreaction / reactants are kinetically stable

(c) Use the following standard electrode potentials to explain why iron(III) iodide does not exist in aqueous solution

2Fe3+

+ 2I

2Fe2+

+ I2or words E

0

= (+) 0.23 V

So I-

would reduce Fe3+

/ Fe3+

would oxidise I-

/ E0

positive so reaction LR

OR Fe3+

and Fe2+

has a more positive electrode potential than I2and I

-

(1)

I-

will reduce Fe3+

/ Fe3+

will oxidise I-

Ox Nos Oxidation of iron(II) by manganate (VII) Stoichiometric ratio(relative numbers in the eqtn) of MnO4- to Fe2+ 1:5

MnO4- + 5Fe2+Mn2+ + 5Fe3+ 4O atoms requires 8H+ MnO4- + 5Fe2+ + 8H+Mn2+ + 5Fe3+ + 4H2OSO4

2overall oxidation =2 Ox state of O =2 total =8 Ox state of S = +6

For every reducing agent there is an oxidised form which is a potential oxidising agent. The scale is based on the tendency of the

oxidised form of a redox pair to take up electrons forming the reduced form, the more positive the value the more likely this is to

happen

Half equation linking reducing agent and its oxidised as reversible reductions

Fe3+(aq) + e- Fe2+(aq) Electrical conductor (inert platinum plate)

Oxidised form Reduced form Measuring pressure of electrons using magic voltmeter

Standard electrode arrangement, standardise measurements

(oxidised and reduced forms present at molar concs in reduction half equation) Gases at 105Pa/1atm, 298K


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If Fe

2+(aq) was better at reducing than Fe

3+(aq) was at oxidising then Fe

2+(aq) would deposit electrons on the plate as it changed

into Fe3+(aq) giving a negative reading on the magic voltmeter & vice versa

Not the solution to the problem Cant devise a magic voltmeter as electron pressure isnt like water pressure at the end of a pipe.

An electric circuit is needed if we are to use a voltmeter. As soon as we complete the circuit by putting a 2nd

Pt plate in the

mixture we shall get another equal and opposite electron pressure on the voltmeter & the instrument will register 0.00VOnly way to measure this pressure is to play one electrode off againstanother

Connection between 2 solutions mustnt be of metals, a salt bridge is

used(a conducting solution of KCl in jelly(or a wet filter paper bridge))

This simple cell arrangement pumps electrons from the more (-

)electrode to the less(-) one. Voltmeter will not measure either E buttheir difference.

To arrive at a value for different electrodes, one of them must be given

an arbitrary value, eg standard hydrogen electrode

Half equation H+(aq) + e- H2(g) E

=0.00V

Standard electrode potential E

of an electrode is its potential(emf) measured relative to the standard hydrogen electrode: conc

of all ions is 1moldm-3

pressure of all gases is 100kPa, 298KIf eqtn is halved or doubled E

mustnt be altered but if eqtn reversed sign ofE

must be changed

Feasiblity of redox reactions in (aq)solution is assessed by comparing E (+3 is more positive than +1)

bigger/larger than(smaller/less than is incorrect) more negative/positive always including signs of numbers

Will zinc metal reduce copper(II)ions or will copper metal reduce zinc ions?

Half equations written as reductions: Zn2+

(aq) + 2e-

Zn(s) E

= -0.76V Cu2+

(aq) + 2e-

Cu(s) E

= +0.34V1sthalf eqtn represents a more(-)E (Zn2+|Zn electrode potential more(-)so 1st eqtn provides electrons by going backwards)

Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq) Reaction feasible but will it happen?Reasons preventing a feasible reaction from occurring

1 Reaction takes a more favoured course than the one thought of

Do manganate(VII) ions oxidise iron(II) consulting reduction tables finding:

MnO4-(aq) + 4H+(aq) + 3e-MnO2(s) + 2H2O(l) E

= +1.7V Fe3+(aq) + e- Fe2+(aq) E = +0.77V

2nd

eqtn E is more(-) so, MnO4-(aq) + 4H

+(aq) + 3Fe

2+MnO2(s) + 2H2O(l) + 3Fe

3+(aq)

However 1mol of MnO4-(aq) oxidising 5mol of Fe2+(aq) gives a clear solution(above equation giving a brown ppt)

MnO4-(aq) + 8H+(aq) + 5Fe2+MnO2(s) + 4H2O(l) + 5Fe

3+(aq) Yes oxidation occurred but choice of reduction product was

wrong

2 EACT of the reaction is too high

Adding copper to dil acid Cu2+

(aq) + 2e-

Cu(s) E

= +0.34V 2H+(aq) + 2e

-H2(g) E

= 0V

Hydrogen half eqtn represents a more(-)E. Hydrogen should reduce a solution of copper sulphate to give copper, but bubble H2(g)

through & youll be disappointed. Covalent H-H would have to be broken which requires large EACT and hydrogen is almostcompletely insoluble in water

3 Values of E are very close and/or something escapes from the system

All redox reactions reach equil but POE is usually so far to left or right that its ignored. This doesnt happen when E is very

close together unless something escapes (forms a gas/complex) from the system and disturbs the equil. It accounts for the

tendency of some reactions to go in unfavourable directions


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Warming a little (red)copper(I)oxide with dil H2SO4 and you will obtain a brown ppt of copper and a blue solution of

copper(II)sulphate, copper(I)ions disproportionate

E = +0.34V 2Cu+(aq) Cu(s) + Cu2+(aq)

Ox No 2(+1) 0 +2

Cu

+

(aq) + e

-

Cu(s) E

= +0.53V Cu

2+

(aq) + e

-

Cu

+

(aq) E

= +0.15VDriving Force/Cell potential is diff between the 2E values here: 0.53 -0.15 = 0.38V, 2

Cu+(aq) Cu(s) + Cu

2+(aq) E

cell = +0.38V

Findings support dispropotionation, in general if Ecell is(+) change from left to right is favoured

Does an iron(II)salt disproportionate into iron and an iron(III)salt? 3Fe2+(aq) 2Fe3+(aq) + Fe(s)

Fe2+(aq) + 2e- Fe(s) E

= -0.44V Fe3+(aq) + e- Fe2+(aq) E

= +0.77V

2Fe3+(aq) + Fe(s) 3Fe2+(aq) Ecell = +1.21V Equation opposite of what were looking for,

Non-standard conditions Temp & conc changes are likely to affect rate of reaction rather than its direction

Ionic reactions in solution are very fast, so seldom noticeable. Most significant effect of conc is usually with acids. Small change of conc

but changes of conc of H3O+

with pH are enormous

Cells A device that by chemical reaction produces an electric current

In principle a cell can be made by opposing any 2 electrode systems of different electrode potential

The electrode with more negative electrode potential provides electrons(reduced form changes to oxidised form)thus the negative pole of

versa In the cell a redox reaction occurs. Unlike a simple reductionoxidation in a mixed solution, electrons used are not passed uselessly frthe oxidant

Reduction of the oxidising agent and oxidation of the reducing agent occur in separate half-cells and electrons pass(current) through an e

the pressure difference(voltage) caused by the difference in the electrode potentials of the 2 half-cells


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Large size, low voltage, inconvenient liquid content makes it obsolete

E and associated half equations(for half cells):

Cu2+(aq) + 2e- Cu(s) E

= +0.34V Zn2+(aq) + 2e- Zn(s) E

= -0.76V

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) E = +1.1V

Zinc electrode electron provider so zinc is(-) pole, copper is(+) pole of the cell

Will cobalt reduce gallium(III)salts in solution to gallium? 3Co(s) + 2Ga3+(aq) 3Co2+(aq) + 2Ga(s)

2 reduction half equations Co2+(aq) + 2e- Co(s) E = -0.28V Ga3+(aq) + 3e- Ga(s) E

= -0.56V

3Co(s) + 2Ga3+(aq) 3Co2+(aq) + 2Ga(s) Ecell = -0.28V E

cell for reaction is(-) reverse reaction favoured no

If emf (overall/reaction potential) is positive the reaction in the direction written is thermodynamically feasible

Mn2+

(aq) + 2e-

Mn(s) E

= -1.19V Sn2+

(aq) + 2e-

Sn(s) E

= -0.14V Reversing anyone of these and adding,

Sn(s) + Mn2+(aq)Mn(s) + Sn2+(aq) E

= -1.05V Reaction in direction written isnt thermodynamically feasible

Max work the cell could perform = n x E x F Free G = -n x E x F (-)sign indicating fall in free energy if work nEF is done(E

emf of the cell max external voltage the cell could provide)(F is Faradays constant, charge on a mole of electrons)

(n is No of electrons transferred)

Max amount of work the cell could perform = (charge it gives) x (voltage that moves the charge through), amount of charge = n x F

If E

is positive, G will be negative and reaction is accompanied by a fall in free energy and reaction is thermodynamically feasibleIf E is (nearly) zero then reaction might be expected to reach an equil

Method of finding G only applicable to ionic reactions in solution

Cells/batteries(a collection of cells), 2 types: rechargeable & disposableRechargeable cells(lead-acid or nickel-cadmium)can have the redox processes which gave rise to the electricity reversed by electrolysis

reverse direction.

KMnO4 titrations Potassium manganate(VII)(purple) (slight excess at end point seen as a pink colouration)half-reaction for its reduction in acidic solution, usual conditions of use is: MnO4

-(aq) + 8H+(aq) + 5e-Mn2+(aq) + 4H2O(l)

KMnO4 is a strong oxidising agent & cant be obtained in a high state of purity, as it easily oxidises organic materials, eg specks of dust wh

manganese(IV)oxide on standing

solutions cant be made accurately by weighing the solid, must be standardized against a solution which can be so made(primarilcommonly used Na2C2O4 Sodium ethanedioate

Reaction for oxidation of ethanedioate ions: C2O42-

(aq) 2CO2(g) + 2e-

Hence 2MnO4-(aq) + 5C2O4

2-(aq) + 16H+(aq) 2Mn2+(aq) + 10CO2(g) + 8H2O(l)

Autocatalyticreaction where products catalyse the reaction KMnO4 solution run into a standard solution of sodium ethanedioate at about

Initially reaction is quite slow, but as soon as some manganese(II)ion is produced this catalyses the reaction which becomes much faster.

Ref to tables of electrode/redox potentials shows that MnO4-/H

+is a very powerful oxidising agent and will oxidise almost any reducing ag

titration.Sodium thiosulpate titrations Sodium thiosulphate used to titrate iodine, thiosulphate oxidised to tetrathionate ions, S4O6

2-

2S2O32-(aq) + I2(aq) S4O62-(aq) + 2I-(aq)- Substance to be analysed usually produces iodine in another reaction by oxidation by oxidation of iodide ions

- Copper(II)oxidises iodide ions: 2Cu2+(aq) + 4I-(aq) 2CuI(s) + I2(aq) liberated iodine can be titrated with sodium thiosulphate solution

- Titration can be used to determine amount of copper in brass, No of molecules of H2O of crystallisation in hydrated copper(II)sulphate

- As titrations of iodine with thiosulphate ions proceed solution becomes pale yellow, at endpoint colourless(unless other coloured substancbe difficult to see, so starch solution often added near endpoint where solution turns blue-black due to formation of starch/iodine complex,

at endpoint

- Sodium thiosulphate oxidised to sulphate ions by other halogens but not used for their analysis

Corrosion The conversion of a metal in its normal working environment (mostly iron to its ions) therefore oxidation and many circumstanc

corrosion occurs involve electrochemical cells

Rust is a hydrated iron oxide Fe2O3.xH2O rusting requires oxygen and a film of liquid water on the iron, water vapour is not enough, since

the electrolyte in which corrosion occurs

- Attack on iron occurs where there are impurities or points of strain; these cause minute variations in electrode potential of iron. Water mu

oxygen not necessary at point of attack- Pitting occurs on car bodies under loose paintwork where water can make its way in

- Iron dissolves to give a limited conc of iron(II)ions: Fe(s) Fe2+(aq) + 2e-

Released electrons pass through

the iron to some point where

water and oxygen are present

(e-)s reduce mixture to OH ions:

2H2O(l)+O2(g)+4e- 4OH-(aq)

Acidic gases in the water(CO2)

reaction by removal of OH- ion

OH-(aq) + CO2(g) HCO3-(a

Iron(II)ions oxidised to iron(III) when they come into contact with O2(often at exposed metal where water entered under the paint):

4Fe2+

(aq) + 2H2O(l) + O2(g) 4Fe3+

(aq) + 4OH-(aq)

In oxygenated region: Fe3+(aq) + 3OH-(aq) Fe(OH)3(s)(brown)which ages to give rust(a mixture of hydrates of iron(III)oxide) 2Fe(OH)3(s) Fe2O3.3H2O(s)

Rust is porous allowing further entry of water, where air is restricted(eg behind rust under the paint):

(not yet oxidised)Fe2+(aq)+2OH-(aq) Fe(OH)2(s) Results in mixture of iron oxides in diff Ox states(cause of black rust)ppt a little iron(II)hydroxide by adding excess NaOH(aq) to dil iron(II)sulphate(aq) and continuously shake with air, ppt will go greener, th


10/10

before turning brown, alternatively(quick results)add NaOH(aq) to mixture of iron(II) & iton(III) salts in solution

Rusting quantitatively most significant form of corrosion, Al and Mg alloys in aircraft, ships will also corrode,

Corrosion found where 2 diff metals are in contact eg Al alloys riveted with Mg alloy rivets, which forms a magnesium aluminium cell(E

bolts with alloy components

Corrosion kept on by maintenance & protection using water-repellent materials frequent cleaning, painting

Sacrificial protection The more reactive metal is sacrificed to preserve another(steel) electrochemistry used in prevention of rusting as wcause of it

A more reactive metal in contact with iron will corrode preferentially, eg in galvanising where object is coated in zinc. If zinc coating s d

formed and zinc will oxidise rather than iron

Iron(steel)ships, underground pipes, protected by blocks of Zn or Mg attached at intervals to the steel which corrode instead of steel(replanecessary)

Tinning Large amounts of zinc is toxic so steel coats tin instead, tin less easily oxidised than iron so doesnt act as a sacrificed coat if dam

protects the iron with a tough malleable non-toxic coating

(q) If Daniell cell Zn(a)|Zn2+(aq) : Cu2+(aq)|Cu(s) is set up under standard conditions & an opposing potential greater than 1.1V is applied t

the cell, what are the changes?

(a) Emf of cell = 1.1V, Zn is(-), emf has sign of RH electrode Cu E = 1.1V

If a more (-)potential is applied to Zn, electrons will reduce Zn2+ to Zn from the Cu electrode where Cu will be oxidised to Cu2+

Overall change: Cu(s) + Zn2+

(aq) Cu2+

(aq) + Zn(s)

(q) How would you expect emf of Daniell cell to alter (qualitatively) if 0.1moldm-3

ZnSO4 and 2moldm-3

CuSO4 were used (in same cell) i

solutions

(a) Increasing conc of (+)ion/oxidised form in each electrode/half cell makes emf more(+)

Greater conc of copper(II) ions would increase the already (+)E of the copper half cell

Lower conc of Zn2+ ions would make the (-)E of the Zn half cell even more (-)Diff between Es greater an magnitude of cell emf would increase

(q) Why is corrosion of iron pier supports worst in region between high & low tide levels?

(a) Additional to iron, air, water, rusting accelerated by CO2 & electrolytes. Region between high & low tides is the only one which gets a r

water(and electrolytes) at high tide and air (with CO2) at low tide

Scandium, zinc arent transition metals since Sc only forms Sc3+

ion with no d electrons and Zn only Zn2+

(3d10

) with a full dsubshell

Because the orbitals being filled are inner ones, change in chemistry across the transition series is less marked

Increase between successive IEs compensated forby increased bond strengths or increased Hhyd of the ions, enabling the

transition metal to have several Ox Nos

Complex ion A metal ion associated with a No of anions or neutral molecules(ligands) (H2O NH3 Cl-CN

-)

Ligand Anions/molecules firmly bonded to the central cation. Each ligand contains at least one atom with a lone pair of

electrons. These can be donated to the central cation forming a co-ordinate (dative)bond. The ligand is said to be co-ordinated to

the central ion.Hydration of metal ions In solution metal ions attract water molecules forming complex ions([M(H2O)6]

x+)

The metal ion is joined to 6 water ligands by dative covalent bonds.

The water molecules donate a lone pair of electrons to empty orbitals on the metal ion and an octahedral complex forms

( [Cu(H2O)6]2+(pale blue) Ions formed by transition metals are usually coloured)

Simple anions Non metals (Cl, O

2) Complex anions Where groups around the central metal ion are negative([Fe(CN)

6]4)

Polyatomic cations Several atoms bonded covalently, the whole structure having a positive charge (NH4+)

Polyatomic anions Several atoms bonded covalently, the whole structure having a negative charge(SO42

, NO3, CHCOO

,

MnO4) derived from acids by the loss of one or more hydrogen ions

Overall charge on complex ion is sum of charges on central cation & ligands

If complex is a cation, named by using prefix for ligands followed by name of central ion with its Ox

No(hexaaquairon(II)[Fe(H2O)6]2+)

If complex is an anion, name of central ion is changed to ending inate (ferrate(Fe) cuprate(Cu))(hexacyanoferrate(II)[Fe(CN)

6]4)

Roman numeral denotes(original) Ox No of central cation not charge on complex ion

Complexes owe their stability(relative)

chemistry Unit 5.1

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Transcript of chemistry Unit 5.1