Download solutions can`t be made accurately by weighing the solid, must be

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 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 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–
2+
3+
Write ionic half equation for oxidation of Fe ions to Fe ions
Fe2+  Fe3+ + e–
2+
Hence write overall ionic equation for reaction of Fe ions with bromine
Br2 + 2Fe2+  2Br– + 2Fe3+
–
+
–
2+
2 half reactions:
MnO4 + 8H + 5e  Mn + 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 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 60 oC. 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- +
S4O62The 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
o
Standard electrode potential E 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 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:
<---------Eo
2+
Cu (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 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)
o
+ 2H2O(l)
E = +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) + 2eThe overall reaction for charging is:
2PbSO4(s) + 2H2O(l)  Pb(s) + PbO2(s) + 4H+(aq) +
22SO4 (aq)
Hydration of metal ions In solution metal ions attract water molecules forming complex ions([M(H 2O)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:
o
2+
E /V
– 0.76
–
Zn + 2e ⇌ Zn
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 a more positive E than E H +/½ H2 is a stronger oxidising
agent than H +
o
o
(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:
o
E /V
+ 0.15
+ 0.54
Cu2+ + e- ⇌ Cu+
I2 + 2e- ⇌ 2Io
(i) From E values why would this reaction not occur
2Cu2+ (aq) + 4I - (aq) → 2CuI(s) + I2 (aq)
o
E 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 NH 3 (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) + 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 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)
o
Cu+ (aq) + e- → Cu(s) E = +0.52 V
o
Cu2+ (aq) + e- → Cu+(aq) E = +0.15 V
(i) Combine the half-equations above to produce the redox equation for the disproportionation reaction.
2Cu+ (aq) → Cu(s) + Cu2+ (aq)
o
(ii) Use the E 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(NH 3)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
o
E /V
3+
2+
Fe (aq) + e
Fe (aq)
+0.77
Cl2 (aq) + 2e 2Cl - (aq)
+1.36
+
2+
MnO4 (aq) + 8H (aq) + 5e
Mn (aq) + 4H2O(l)
+1.51
(a) (i) Use the data to explain why dil HCl acid is not used to acidify solutions of KMnO4
o
o
Δ E = +0.15V (MnO4-/ Mn2+) more positive or greater than E (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 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 MnO 4 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 MnO4Moles 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
o
Pb2+ (aq) + 2ePb(s)
PbO2 (s) + 4H+ (aq) + 2e-
Pb2+ (aq) + 2H2O(l)
E /V
–0.13
+1.46
o
(i) Calculate the standard e.m.f. of the cell.
E = +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
2+
-
2Fe(s) + O (g) + 2H O(l)  2Fe (aq) + 4OH (aq)
2
2
Calculate Eo for the reaction and show that it’s feasible
θ
ΔE
react
= +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
θ
2+
E Zn /Zn more negative than for Fe
θ
2+
OR E Zn/Zn more positive than for Fe
OR E
θ
cell
for Zn being oxidised by O is more positive than for Fe being oxidised by O
2
θ
2
OR similar E arguments related to preferential oxidation with Sn
2. (a) (i) Give the electronic configuration of Fe and Fe 2+
6
2
2+
Fe
6
[Ar] 3d 4s
Fe [Ar] 3d
(ii) Draw the structure of the hexaaquairon(II) ion, [Fe(H 2O)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.
2+
Emf of cell / potential difference of cell containing Fe and Fe
and standard hydrogen electrode/half cell OR hydrogen electrode and 1 mol
dm
–3
+
H and 1 atm H
2
1 mol dm
–3
Fe
2+
(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 E
Fe + 2H
+
cell
= +0.44 (V))
2+
 Fe + H
2
Potential for the reaction is positive so reaction is feasible
+
OR
2+
H and (½)H has a more +ve electrode potential than Fe and Fe (1)
2
+
+
+
H will oxidise Fe / H is an oxidising agent / Fe is a reducing agent for H / other correct redox statement (1)
+
2+
Fe + 2H → Fe + H
2
(iii) State why E values cannot predict that a reaction will occur, only that it is possible
reaction / reactants are kinetically stable
High Ea so slow
(c) Use the following standard electrode potentials to explain why iron(III) iodide does not exist in aqueous solution
3+
−
2+
0
2Fe + 2I → 2Fe + I or words E = (+) 0.23 V
2
3+
-
3+
-
0
So I would reduce Fe / Fe would oxidise I / E positive so reaction LR
OR
3+
2+
-
Fe and Fe has a more positive electrode potential than I and I (1)
2
-
3+
3+
I will reduce Fe / Fe will oxidise I
-
Ox No’s Oxidation of iron(II) by manganate (VII) Stoichiometric ratio(relative numbers in the eqtn) of MnO 4- to Fe2+ 1:5
MnO4- + 5Fe2+  Mn2+ + 5Fe3+
4O atoms requires 8H+
MnO4- + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O
2–
SO4 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
happen
Half equation linking reducing agent and its oxidised as reversible reductions
Fe3+(aq) + eFe2+(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 2 nd 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 equation
H+(aq) + e½H2(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) + 2eZn(s) Eθ = -0.76V
Cu2+(aq) + 2eCu(s) Eθ = +0.34V
st
2+
st
1 half eqtn represents a more(-)E (Zn |Zn electrode potential more(-)so 1 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 = +1.7V
Fe3+(aq) + e-  Fe2+(aq) Eθ = +0.77V
nd
+
2+
2 eqtn E is more(-) so, MnO4 (aq) + 4H (aq) + 3Fe  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 acid
Cu2+(aq) + 2eCu(s) Eθ = +0.34V
2H+(aq) + 2eH2(g) Eθ = 0V
Hydrogen half eqtn represents a more(-)E. Hydrogen should reduce a solution of copper sulphate to give copper, but bubble H 2(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 H 2SO4 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,
2
Cu+(aq)  Cu(s) + Cu2+(aq) Eθcell = +0.38V
Findings support dispropotionation, in general if Eθcell 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)
2+
θ
3+
2+
θ
Fe (aq) + 2e
Fe(s) E = -0.44V
Fe (aq) + e
Fe (aq) E = +0.77V
2Fe3+(aq) + Fe(s)
3Fe2+(aq) Eθcell = +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 h
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 reduction – oxidation in a mixed solution, electrons used are not passed ‘uselessly’ fro
the oxidant
• Reduction of the oxidising agent and oxidation of the reducing agent occur in separate half-cells and electrons pass(current) through an ex
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 obsolete
E and associated half equations(for half cells):
Cu2+(aq) + 2eCu(s) Eθ = +0.34V
Zn2+(aq) + 2eZn(s) Eθ = -0.76V
2+
2+
θ
Zn(s) + Cu (aq)  Zn (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+
2 reduction half equations
Co (aq) + 2e  Co(s) Eθ = -0.28V
Ga3+(aq) + 3e-  Ga(s) Eθ = -0.56V
3+
2+
θ
θ
3Co(s) + 2Ga (aq)  3Co (aq) + 2Ga(s) E cell = -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) + 2eMn(s) Eθ = -1.19V
Sn2+(aq) + 2eSn(s) Eθ = -0.14V
Reversing anyone of these and adding,
2+
2+
θ
Sn(s) + Mn (aq)  Mn(s) + Sn (aq)
E = -1.05V Reaction in direction written isn’t thermodynamically feasible
Max 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
θ
(E 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 F
If Eθ is positive, G will be negative and reaction is accompanied by a fall in free energy and reaction is thermodynamically feasible
If Eθ 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 & disposable
Rechargeable cells(lead-acid or nickel-cadmium)can have the redox processes which gave rise to the electricity reversed by ‘electrolysis’ – p
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 whi
manganese(IV)oxide on standing
solutions can’t be made accurately by weighing the solid, must be standardized against a solution which can be so made(primarily
commonly used Na2C2O4 Sodium ethanedioate
Reaction for oxidation of ethanedioate ions: C2O42-(aq)  2CO2(g) + 2eHence 2MnO4-(aq) + 5C2O42-(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
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 age
titration.
Sodium thiosulpate titrations Sodium thiosulphate used to titrate iodine, thiosulphate oxidised to tetrathionate ions, S 4O622S2O32-(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 H 2O 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 substance
be difficult to see, so starch solution often added near endpoint where solution turns blue-black due to formation of starch/iodine complex, w
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 w
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 mus
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) + 2eReleased electrons pass through
Acidic gases in the water(CO2)
the iron to some point where
reaction by removal of OH- ions
water and oxygen are present
OH-(aq) + CO2(g) HCO3-(aq)
(e-)s reduce mixture to OH ions:
2H2O(l)+O2(g)+4e- 4OH-(aq)
Iron(II)ions oxidised to iron(III) when they come into contact with O 2
(often at exposed metal where water entered under the paint):
4Fe2+(aq) + 2H2O(l) + O2(g)
4Fe3+(aq) + 4OH-(aq)
3+
In oxygenated region: Fe (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, the
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(Eθ =
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 w
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 da
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(repla
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 dama
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 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) in
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 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 CO 2 & 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 aren’t transition metals since Sc only forms Sc 3+ 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’s
Complex ion A metal ion associated with a No of anions or neutral molecules(ligands) (H 2O 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)
4–
6] )
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(SO 42–, 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)
4–
6] )
• Roman numeral denotes(original) Ox No of central cation not charge on complex ion
• Complexes owe their stability(relative)