Topic 5.1: Redox equilibria
Unit 1.5 knowledge
Redox (reductionoxidation reaction) describes / all chemical reactions in which atoms have their oxidation number/state changed
Oxidation 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 charge
3 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 electronegativeatom (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 = –2Ox state of O = –2 total = –8 Ox state of S = +6Fe in FeCl3 is +3Cl in NaCl is -1C in CO is +2C in CCl4 is +4O in O2– is –½
Write ionic half equation for reduction of bromine to bromide ionsBr2 + 2e– 2Br–
Write ionic half equation for oxidation of Fe2+ ions to Fe3+ ionsFe2+ Fe3+ + e–
Hence write overall ionic equation for reaction of Fe2+ ions with bromineBr2 + 2Fe2+ 2Br– + 2Fe3+
2 half reactions:MnO4– + 8H+ + 5e– Mn2+ + 4H2ONO2– + 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 2FeCl3Cl2 + 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 =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 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 like ethanedioate ions, C2O42- 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 + 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 electrodecell diagrams are not required
Standard electrode potentialEo 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, 298K
Electrode potentialE 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:
<------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 valuesEcell = Eobottom - Eotop
Copper is a stronger reducing agent than silver as the Eo value of copper is more negative than that of Ag.
If Eθcell 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 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 potentials
Disproportionation is the oxidation and reduction of the same element in the same reaction.
<------
Cu2+(aq) + e------> 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) => 2Fe2+(aq) +2H2O(l) Zn(s) + Fe2+(aq) <=> Zn2+(aq) + Fe(s)
understand the application of electrode potential to the construction of simple storage cells
A 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) + SO42-(aq) Eo = - 0.36V
Reaction 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 = +1.69V
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) + 2SO42-(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) + SO42-(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 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 – 0.76
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+ 2H2OE = +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 Eo than EoH +/½ 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 occur2Cu2+ (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)
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 ×10-3mol
amount Cu2+in 250cm3= 1.655 ×10-2mol
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) 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 Eovalues 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 CuCl42-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 separation
Eo/V
Fe3+ (aq) + e- Fe2+ (aq) +0.77
Cl2 (aq) + 2e - 2Cl - (aq) +1.36
MnO4- (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) 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 200cm3in 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 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 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-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 up
between 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
Write an overall equation for the 1st stage in the rusting of iron
2Fe(s) + O2(g) + 2H2O(l) 2Fe2+(aq) + 4OH-(aq)
Calculate Eo for the reaction and show that it’s feasible
ΔEθreact = +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 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 possibleHigh 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+ + 4H2O
SO42– overall oxidation = –2Ox 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
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 versa
Not 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 another
Connection 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 electrode
Half equationH+(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, 298K
If eqtn is halved or doubled Eθ mustn’t be altered but if eqtn reversed sign of Eθ 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) Eθ = -0.76VCu2+(aq) + 2e- Cu(s) Eθ = +0.34V
1sthalf 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)Eθ = +1.7VFe3+(aq) + e- Fe2+(aq)Eθ = +0.77V
2nd 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 wrong
2 EACT of the reaction is too high
Adding copper to dil acidCu2+(aq) + 2e- Cu(s) Eθ = +0.34V2H+(aq) + 2e- H2(g) Eθ = 0V
Hydrogen 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 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 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