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 of electrons by a molecule, atom or ion (From +2 to +4)Reduction describes the gain of electrons 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 charge3 Ox No of F is always (-1), H is (+1) in most compounds, O(–2)

• Sum of Ox No’s in a neutral molecule must add up to zero or the charge on the ion• The more electronegative atom (strength to attract electrons) has the (-)Ox No, C in CO2 is (+4), in CH4 is (–4), because carbon more electronegative than hydrogen

SO42– overall oxidation = –2 Ox state of O = –2 total = –8 Ox state of S = +6

Fe in FeCl3 is +3 Cl in NaCl is -1 C in CO is +2 C in CCl4 is +4 O in O2– is –½

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– + 2Fe3+

2 half reactions: MnO4– + 8H+ + 5e– Mn2+ + 4H2O NO2

– + H2O NO3– + 2e– + 2H+

Balancing equations: (x2) 2MnO4– + 16H+ + 10e– 2Mn2+ + 8H2O (x5) 5NO2

– + 5H2O 5NO3– + 10e– + 10H+

Overall: 2MnO4– + 5NO2

– + 6H+ 2Mn2+ + 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 decrease3ClO-(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 =4 Total increase =1 Cl atom ox. no. +1 to ClO3- ion ox. no. +5 =4

understand the procedures and principles involved in the use ofpotassium manganate(VII) to estimate reducing agents and

potassium iodide and sodium thiosulphate to estimate oxidising agentsTitrations involving potassium manganate(VII) ions,  MnO4

-, are used to estimate concentrations of reducing agents like ethanedioate ions, C2O4

2- and iron II ions,Fe2+ The ionic equations are: 2MnO4

- + 16H+ + 5C2O42- ---> 2Mn2+ + 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 + 3H2O2MnO4

- + 10I- + 16H+ ---> 2Mn2+ + 8H2O + 5I2 The iodine in the flask is titrated with standardised aqueous sodium thiosulphate in the burette.  2S2O3

2- + I2 = 2I- + S4O6

2- 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 reactionsThe anti-clockwise rule: the more positive electrode potential on the bottom then anticlockwise arrows for direction of spontaneous change:     <---------- Eo

Cu2+(aq) +e-  ---> Cu(s) +0.34V Ag+ (aq) + e- ---> Ag(s)  +0.80V   (most +ve at bottom)     ----------> spontaneous reaction is Ag+(aq) + Cu(s) ----> Ag(s) + Cu2+(aq)

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 Eθ

cell is(+) change from left to right is favoured, thermodynamically feasible, in general

understand why these predictions may be wrong in practiceStandard 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 Eo values are close (or Ecell is small) then non-standard conditions can change Eo values.  In these cases  a reaction does not always happen as expected

understand disproportionation reactions in terms of standard electrode potentialsDisproportionation is the oxidation and reduction of the same element in the same reaction.              <-------------- Cu2+(aq) + e- -----> Cu+(aq) Eo = +0.15VCu+(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) => 2Fe2+(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 = - 0.36V Reaction at positive lead (IV) oxide plate is reduction (so it is the cathode) PbO2(s) + 4H+(aq) + SO4

2-(aq) + 2e- PbSO4(s) + 2H2O(l)  Eo = +1.69V When 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) + SO4

2-

(aq) + 2e-

The overall reaction for charging is: 2PbSO4(s) + 2H2O(l) Pb(s) + PbO2(s) + 4H+(aq) + 2SO4

2-(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 in –ate (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. This gives a greenish-blue solution.

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

Zn2+ + 2e– ⇌ Zn – 0.76Cu2+ + 2e– ⇌ Cu + 0.34NO3

– + 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 likelyOR NO3

– being the oxidised form of a redox couple with a more positive Eo than Eo H +/½ H2 is 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/VCu2+ + e- ⇌ Cu+ + 0.15I2 + 2e- ⇌ 2I- + 0.54

(i) From Eo values why would this reaction not occur 2Cu2+ (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 dm–3 sodium thiosulphate solution. The mean titre was 16.55 cm3

2Cu2+ (aq) + 4I - (aq) → 2CuI(s) + I2 (aq)2S2O3

2- (aq) + I2 (aq) → 2I - (aq) + S4O6 2- (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 molamount Cu2+ in 250cm3 = 1.655 × 10-2 molmass 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) Eo = +0.15 V(i) Combine the half-equations above to produce the redox equation for the disproportionation reaction.

2Cu+ (aq) → Cu(s) + Cu2+ (aq) (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) 2s22p63s23p63d10 (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 CuCl4

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

tetrahedral range 109 – 110o 4 (bonding) pairs of electrons repel to a position of maximum separationEo /V

Fe3+ (aq) + e- Fe2+ (aq) +0.77Cl2 (aq) + 2e - 2Cl - (aq) +1.36MnO4

- (aq) + 8H + (aq) + 5e- Mn2+ (aq) + 4H2O(l) +1.51(a) (i) Use the data to explain why dil HCl acid is not used to acidify solutions of KMnO4

Δ Eo = +0.15V (MnO4

-/ Mn2+) more positive or greater than Eo (Cl2 / Cl - )

(so) MnO4 reacts 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 isMnO4

- (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. 25cm3 portions of this solution were titrated against a 0.02mol dm-3 solution of acidified KMnO4

The mean titre was 20.10 cm3. Calculate % of FeSO4.7H2O in the tablets. [Molar mass FeSO4.7H2O = 278 g mol–1]

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 concentrationsPercentage 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) –0.13PbO2 (s) + 4H+ (aq) + 2e- Pb2+ (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 [Pb2+(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 referenceelectrode. Explain why this is so.

(ii) Label the diagram of the standard hydrogen electrode

Write an overall equation for the 1st stage in the rusting of iron2Fe(s) + O2(g) + 2H2O(l) 2Fe2+(aq) + 4OH-(aq)

Calculate Eo for the reaction and show that it’s feasibleΔEθreact = +0.84V Greater than zero therefore feasible

Use the Eo values 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 preferentially Eθ Zn2+/Zn more negative than for Fe OR Eθ Zn/Zn2+ more positive than for Fe OR Eθcell for Zn being oxidised by O2 is more positive than for Fe being oxidised by O2 OR similar Eθ arguments related to preferential oxidation with Sn

2. (a) (i) Give the electronic configuration of Fe and Fe2+ Fe [Ar] 3d64s2 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/solid red

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

(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 Fe and standard hydrogen electrode/half cell OR hydrogen electrode and 1 mol

dm – 3 H+ and 1 atm H2

1 mol dm – 3 Fe2+

(ii) Explain, with the aid of an equation, why the value of E suggests that iron will react with an aqueous solution of an acid to give Fe2+ ions and hydrogen gas

Emf of hydrogen electrode is zero – stated 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 (½)H2 has a more +ve electrode potential than Fe2+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 E values cannot predict that a reaction will occur, only that it is possible High Ea so slow reaction / 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+ + I2 or words E0 = (+) 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 I2 and I- (1) I- will reduce Fe3+ / Fe3+ will oxidise I-

Ox No’s 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

2– overall 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 happenHalf equation linking reducing agent and its oxidised as reversible reductionsFe3+(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

If Fe2+(aq) was better at reducing than Fe3+(aq) was at oxidising then Fe2+(aq) would deposit electrons on the plate as it changed into Fe3+(aq) giving a negative reading on the magic voltmeter & vice versaNot the solution to the problem Can’t devise a magic voltmeter as electron pressure isn’t 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 0.00V

Only way to measure this pressure is to play one electrode off against anotherConnection between 2 solutions mustn’t 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’ but their difference. • To arrive at a value for different electrodes, one of them must be given an arbitrary value, eg standard hydrogen electrodeHalf equation H+(aq) + e- ½H2(g) Eθ =0.00V

Standard electrode potential Eθ 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 Eθ mustn’t be altered but if eqtn reversed sign of Eθ must be changedFeasiblity 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 numbersWill zinc metal reduce copper(II)ions or will copper metal reduce zinc ions?Half equations written as reductions: Zn2+(aq) + 2e- Zn(s) Eθ = -0.76V Cu2+(aq) + 2e- Cu(s) Eθ = +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 occurring1 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) Eθ = +1.7V Fe3+(aq) + e- Fe2+(aq) Eθ = +0.77V2nd eqtn E is more(-) so, MnO4

-(aq) + 4H+(aq) + 3Fe2+ MnO2(s) + 2H2O(l) + 3Fe3+(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) + 5Fe3+(aq) Yes oxidation occurred but choice of reduction product was wrong2 EACT of the reaction is too high Adding copper to dil acid Cu2+(aq) + 2e- Cu(s) Eθ = +0.34V 2H+(aq) + 2e- H2(g) Eθ = 0VHydrogen half eqtn represents a more(-)E. Hydrogen should reduce a solution of copper sulphate to give copper, but bubble H2(g) through & you’ll be disappointed. Covalent H-H would have to be broken which requires large EACT and hydrogen is almost completely insoluble in water3 Values of E are very close and/or something escapes from the systemAll redox reactions reach equil but POE is usually so far to left or right that it’s ignored. This doesn’t 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

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

Eθ = +0.34V 2Cu+(aq) Cu(s) + Cu2+(aq) Ox No 2(+1) 0 +2

Cu+(aq) + e- Cu(s) Eθ = +0.53V Cu2+(aq) + e- Cu+(aq) Eθ = +0.15V‘Driving Force’/Cell potential is diff between the 2E values here: 0.53-0.15 = 0.38V, 2Cu+(aq) Cu(s) + Cu2+(aq) Eθ

cell = +0.38V Findings support dispropotionation, in general if Eθ

cell is(+) change from left to right is favouredDoes an iron(II)salt disproportionate into iron and an iron(III)salt? 3Fe2+(aq) 2Fe3+(aq) + Fe(s)Fe2+(aq) + 2e- Fe(s) Eθ = -0.44V Fe3+(aq) + e- Fe2+(aq) Eθ = +0.77V2Fe3+(aq) + Fe(s) 3Fe2+(aq) Eθ

cell = +1.21V Equation opposite of what we’re 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 hardly affects E but changes of conc of H3O+ with pH are enormousCells 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 the cell & vice versa• In the cell a redox reaction occurs. Unlike a simple reduction – oxidation in a mixed solution, electrons used are not passed ‘uselessly’ from the reactant to the oxidant• Reduction of the oxidising agent and oxidation of the reducing agent occur in separate half-cells and electrons pass(current) through an external wire, under the pressure difference(voltage) caused by the difference in the electrode potentials of the 2 half-cells

Large size, low voltage, inconvenient liquid content makes it obsoleteE and associated half equations(for half cells):Cu2+(aq) + 2e- Cu(s) Eθ = +0.34V Zn2+(aq) + 2e- Zn(s) Eθ = -0.76VZn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Eθ = +1.1VZinc 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) Eθ = -0.28V Ga3+(aq) + 3e- Ga(s) Eθ = -0.56V3Co(s) + 2Ga3+(aq) 3Co2+(aq) + 2Ga(s) Eθ

cell = -0.28V Eθcell 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) Eθ = -1.19V Sn2+(aq) + 2e- Sn(s) Eθ = -0.14V Reversing anyone of these and adding, Sn(s) + Mn2+(aq) Mn(s) + Sn2+(aq) Eθ = -1.05V Reaction in direction written isn’t thermodynamically feasibleMax work the cell could perform = n x Eθ x F Free G = -n x Eθ x F (-)sign indicating fall in free energy if work nEθF is done(Eθ emf of the cell max external voltage the cell could provide)(F is Faraday’s 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 FIf Eθ is positive, G will be negative and reaction is accompanied by a fall in free energy and reaction is thermodynamically feasibleIf Eθ is (nearly) zero then reaction might be expected to reach an equilMethod of finding G only applicable to ionic reactions in solutionCells/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’ – passing current in 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 & can’t be obtained in a high state of purity, as it easily oxidises organic materials, eg specks of dust which slowly deposit manganese(IV)oxide on standingsolutions can’t be made accurately by weighing the solid, must be standardized against a solution which can be so made(primarily standard) eg commonly used Na2C2O4 Sodium ethanedioate Reaction for oxidation of ethanedioate ions: C2O4

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

Hence 2MnO4-(aq) + 5C2O4

2-(aq) + 16H+(aq) 2Mn2+(aq) + 10CO2(g) + 8H2O(l)Autocatalytic – reaction where products catalyse the reaction KMnO4 solution run into a standard solution of sodium ethanedioate at about 60 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 agent which requires titration.Sodium thiosulpate titrations Sodium thiosulphate used to titrate iodine, thiosulphate oxidised to tetrathionate ions, S4O6

2-

2S2O32-(aq) + I2(aq) S4O6

2-(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 substances are present)can be difficult to see, so starch solution often added near endpoint where solution turns blue-black due to formation of starch/iodine complex, which disappears at endpoint- Sodium thiosulphate oxidised to sulphate ions by other halogens but not used for their analysisCorrosion The conversion of a metal in its normal working environment (mostly iron to its ions) therefore oxidation and many circumstances where corrosion occurs involve electrochemical cellsRust 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 water film forms 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 must be present but 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) assist this reaction by removal of OH- ionsOH-(aq) + CO2(g) HCO3

-(aq)

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, then almost black

before turning brown, alternatively(quick results)add NaOH(aq) to mixture of iron(II) & iton(III) salts in solutionRusting 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(Eθ = +0.71Vbolts with alloy componentsCorrosion kept on by maintenance & protection using water-repellent materials frequent cleaning, paintingSacrificial protection • The more reactive metal is sacrificed to preserve another(steel) electrochemistry used in prevention of rusting as well as being the cause 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 damaged, a cell is 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(replaced when necessary)Tinning Large amounts of zinc is toxic so steel coats tin instead, tin less easily oxidised than iron so doesn’t act as a sacrificed coat if damaged, tin just 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 to the terminals of the cell, what are the changes?(a) Emf of cell = 1.1V, Zn is(-), emf has sign of RH electrode Cu E = 1.1VIf 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) instead of 1moldmsolutions(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 cellLower conc of Zn2+ ions would make the (-)E of the Zn half cell even more (-)Diff between E’s 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 regular supply of water(and electrolytes) at high tide and air (with CO2) at low tide• Scandium, zinc aren’t transition metals since Sc only forms Sc3+ ion with no d electrons and Zn only Zn2+(3d10) with a full d subshell• Because the orbitals being filled are inner ones, change in chemistry across the transition series is less marked• Increase between successive IE’s compensated for by increased bond strengths or increased ΔHhyd of the ions, enabling the transition metal to have several Ox No’sComplex 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–, O2–) 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(SO4

2–, 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 in –ate (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)