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A2 Chem 5.1: Redox equilibria & electrochemistry

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J Amuah-Fuster

on 8 July 2016

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Transcript of A2 Chem 5.1: Redox equilibria & electrochemistry

A2 Chemistry EDEXCEL 9080
Unit 5.1: Redox equilibria
Redox reactions
Standard Electrode Potential
Redox Equilibria
Assessment for Learning
Electrode potentials and the feasibility of reactions
Assessment for Learning
Applications of Electrochemistry
Half-reactions and Oxidation numbers
Electrochemical cells
Standard Electrode Potential
Batteries, protection and sacrificial nodes
Sacrificial protection
Sacrificial protection
aluminum sheets riveted using magnesium alloy rivets
The aluminium and magnesium alloys used in aircraft and ships will also corrode.
Corrosion is common when the two elements are in contact; the magnesium-aluminium emf = +0.71 V).
Such corrosion is prevented by regular maintenance and protection using water repellent materials and painting.
Corrosion of other metals
Where air is restricted (e.g., behind the rust under paint) the rust formed is Fe(OH)2 (s) – which is found in black rust.
Black rust
In presence of O2, Fe2+ ions are oxidised to Fe3+:
4Fe2+ (aq) + 2H2O (l) + O2 (g)  4Fe3+ (aq) + 4OH- (aq)
The Fe3+ ions react with the OH- ions to form brown iron(III) hydroxide:
Fe3+ (aq) + 3OH- (aq)  Fe(OH)3 (s)  Fe2O3.3H2O (s) [rust]
Corrosion of iron
Corrosion of iron
Lead-acid battery
The lead acid battery is used in cars. Each cell consists of two lead plates. The cathode is coated with solid lead(IV) oxide. The electrolyte is a fairly concentrated solution of sulfuric acid.
Lead-acid battery
Cells or batteries fall into two groups: rechargeable and disposable.
The rechargeable cells (e.g. lead-acid or nickel-cadmium) are recharged by passing the current in reverse direction.
Practical cells and batteries (storage cells)
Sodium thiosulphate is used to titrate iodine. The thiosulfate is oxidised to tetrathionate ions, S4O62-, while the iodine forms iodide ions:

2S2O32- (aq) + I2 (aq)  S4O62- (aq) + 2I- (aq)

Starch is added to act as an indicator near the endpoint – the blue black complex disappears at the endpoint.
The iodine is produced in another reaction, e.g.,

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

The liberated iodine can be titrated with sodium thiosulfate. The results of the titration can be used to work out an unknown in the other reaction.
Titrations – sodium thiosulfate titrations
The potassium manganate(VII) solution is run into an acidified standard solution of sodium ethanedioate at about 60 oC.
The reaction starts off slow until Mn2+ ions are produced that catalyses the reaction. [autocatalytic]
Electrode potentials shows that MnO4-/H+ is a very powerful oxidising agent and will oxidise almost any reducing agent which requires titration. [hence HCl cannot be used as the source of H+]
16H+ + 2MnO4− + 5C2O42−  2Mn2+ + 8H2O + 10CO2
Potassium manganate(VII) is a purple substance that acts as its own indicator.
It is placed in the burette so that at the end point the slight excess amount of manganate(VII) in the reaction mixture would produce a slight pink colouration.
The manganate(VII) solution has to be made fresh as it easily oxidises organic materials, including specks of dust.
Manganate(VII) solutions have to be standardised by titration.
Sodium ethanedioate Na2C2O4 is commonly used as a standard.
Titrations – potassium manganate(VII) titrations
Disproportionation reactions
If an element exists in three different oxidation states (which could include the zero state if the uncombined element), disproportionation becomes a possibility.
The feasibility of such a reaction can be predicted from standard electrode potentials.
Disproportionation reactions
the emf of the cell would be -0.39 V. This predicts that the reaction does not take place under standard conditions.
But as CuI (s) is precipitate the concentration of Cu+ is almost zero. This drives the eqm of the Cu2+/Cu+ reaction to the right, increasing its electrode potential and make the emf of the cell positive. The reaction is now feasible.
Consider the redox reaction:
Cu2+ (aq) + 2I- (aq)  CuI (s) + ½ I2 (s)

Explain why this reactions will take place, given that:

Cu2+ (aq) + e-  Cu+ (aq) +0.15 V
½ I2 (s) + e-  I- (aq) +0.54 V
The data provided are always standard electrode potentials. If the concentration of a reactant or product is not 1 mol dm-3, the emf of the cell will differ from that calculated if standard conditions.
This might result in a reaction taking place that is predicted to be unfeasible.
Non-standard conditions
All redox reactions reach equilibrium but the position of the equilibrium is so far to the left or right that this can be ignored.
However, if the electrode potentials are very close then there is a tendency for the reactions to go in unfavorable directions. If something escapes from the system (a gas or complex forms) then the equilibrium is disturbed and will shift in a particular direction.
The electrode potential values are very close and/or something escapes from the system
If we add copper to dilute acid then use the electrode potential for the Cu2+|Cu system we see that copper has a more positive system and its oxidised from, Cu2+, is the only reactant.

Cu2+ (aq) + 2e-  Cu (s) +0.34 V
2H+ (aq) + 2e-  H2 (g) 0.00 V

Copper should not react with dilute acids to give hydrogen.
However, we would expect to form copper when hydrogen gas is bubble into a solution of Cu2+ ions.
Unfortunately, you will be disappointed. The covalent H-H bond would have to be broken first and this requires a large activation energy, and hydrogen is almost completely insoluble in water as well.
The activation energy of the reaction is too high
MnO4- (aq) + 8H+ (aq) + 5Fe2+ (aq)  Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)
MnO4- (aq) + 4H+ (aq) + 3Fe2+ (aq)  MnO2 (s) + 2H2O (l) + 3Fe3+ (aq)
The reaction usually occurs when the sulfuric acid is not added to the titration mixture producing a ugly brown precipitate.
When the acid is added the titration results in a clear and almost colourless solution because the following reaction is preferred:
Suppose we want to know whether manganate(VII) ions will oxidise iron(II). The electrode potentials are:

MnO4- (aq) + 4H+ (aq) + 3e-  MnO2 (s) + 2H2O (l) +1.70 V
Fe3+ (aq) + e-  Fe2+ +0.77 V

The overall reaction would be:
MnO4- (aq) + 4H+ (aq) + 3Fe2+ (aq)  MnO2 (s) + 2H2O (l) + 3Fe3+ (aq)
The reaction takes a more favoured course than the one we thought of
The reaction takes a more favoured course than the one we thought of.
The activation energy of the reaction is too high.
The electrode potential values are very close and/or something escapes from the system.
Non-standard conditions.
When does a feasible reaction not happen?
+1.05 V
Sn2+ (aq) + 2e- Sn (s) = -0.14 V
Mn (s) Mn2+ (aq) + 2e- = +1.19 V
Mn2+ (aq) + 2e- Mn (s) -1.19 V

Sn2+ (aq) + 2e- Sn (s) -0.14 V

The electrode potential for the Mn half reaction is more negative so it is the electron provider. The equation is reversed.
Add the electrode potential values gives a positive value; hence the reaction is feasible.
The half equations for standard electrode potentials are all reduction reactions.
Therefore, one of the half equations in a redox reaction will have to be reversed.
The half-equation with the more negative reduction potential will reveal the reducing agent (the half-equation goes backward and the sign of the electrode potential changes)
The half-equation with the more positive reduction potential will reveal the oxidising agent. The half-equation will go forwards.
NOTE: if you double or halve an equation you must not alter E⊖.
In an electrochemical cell the voltmeter measures the difference in the electrode potentials of the redox reactions.
The standard electrode potentials are measured using the hydrogen electrode as the reference electrode.
The standard hydrogen electrode, SHE, has a potential of zero.
Standard conditions: 1.0 mol dm-3 solutions; gases at a pressure of 1.0 atm and system at 298 K.
If the substance is a gas, the electrode consists of a platinum plate dipping into a 1.0 mol dm-3 solution of ions of the element with the gaseous element, at 1.0 atom pressure, bubbling over the surface of the platinum.
Measuring standard electrode potentials, E⊖
Standard electrode potentials
Zinc metal and iron (II) salts are good reducing agents.
Aqueous chlorine and potassium manganate(VII) are good oxidising agents.

But how good?

Chemists have a scale which measures the power or potential of an aqueous species to oxidise or reduce.
It is called the
standard electrode potential
redox potential
reduction potential scale
The more positive the value (on the scale) the more likely the reduction is to happen.
Electrode potentials
oxidation is always on the left
Half equations are equilibriums as the reactions are reversible.
The solutions are connected by a salt bridge – a concentrated solution of an inert electrolyte, such as Na or K chloride or nitrate. The anions move towards the anode and the cations towards the cathode.
The current is carried in the wire by the flow of electrons from the zinc rod to the copper rod.
Electrochemical cells
oxidation is always on the left
Half equations are equilibriums as the reactions are
Oxidation occurs at the zinc rod (the anode) and reduction takes place at the copper rod (cathode).
The redox reaction produces a potential difference (emf) between the two metal rods. The emf is measured using a high resistance voltmeter.
Electrochemical cells
Electrochemical cells
When zinc is placed in a solution of copper(II) nitrate, a reaction takes place.
This reaction involves the movement of electrons from the zinc atoms to the copper ions.
This reaction can also take place without copper ions and the zinc atoms coming into contact in an electrochemical cells.
Electrochemical cells
Half equations show the reduction and oxidation reactions separately:
Identify the change in oxidation number – this determines the number of electrons involved.
Oxidation half equation: electrons in products
Reduction half equation: electrons in reactants
Add H+/H2O and balance.
Writing half equations
Since the carbon goes from +3 to +4, it is oxidized.
Since the manganese goes from +7 to +2, it is reduced.
MnO4− + C2O42-  Mn2+ + CO2
First, we assign oxidation numbers.
…worked example
Consider the reaction between MnO4− and C2O42− :

MnO −(aq) + C O (aq) Mn (aq) + CO2(aq)
Worked example
The stoichiometry must be such that the
numbers of electrons cancel
Redox equations are balanced using the
half equation method

1. Assign oxidation numbers to determine what is oxidized and what is
2. Write the oxidation and reduction half-reactions.
3. Balance each half-reaction.
Balance elements other than H and O.
Balance O by adding H2O.
Balance H by adding H+.
Balance charge by adding electrons.
4. Multiply the half-reactions by integers so that the electrons gained
and lost are the same.
5. Add the half-reactions, subtracting things that appear on both sides.
6. Make sure the equation is balanced according to mass.
7. Make sure the equation is balanced according to charge.
Writing overall redox equations
The half-reactions for
Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq)

Sn2+(aq) Sn4+(aq) +2e-
2Fe3+(aq) + 2e- 2Fe2+(aq)

Oxidation: electrons are products.
Reduction: electrons are reagents.
Half equations
Redox reactions
What are the oxidation number of iron in Fe O ?
Assigning oxidation numbers (O.N.)- an
Electron transfer reactions are oxidation-reduction or redox reactions.
—loss of electron(s) by a species; increase in oxidation number; increase in oxygen.
—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen.
—electron acceptor; species is reduced.
—electron donor; species is oxidized.
Redox reactions
The following aspects will be examined:

Use changes in oxidation number to write balanced redox equations.
Understand the procedures and principles involved in the use of potassium manganate(VII) to estimate reducing agents and potassium iodide and sodium thiosulfate to estimate oxidising agents.
Define the term standard electrode potential (standard reduction potential) and understand the need for a standard electrode [cell diagrams are not required].
Predict the direction of a redox reaction using standard electrode potential data and understand why these predictions may not be borne out in practice [i.e. redox changes do not always happen under standard conditions].
Understand disproportionation reactions in terms of standard electrode potentials.
Learning objectives
Examination questions set in this topic will require the application of knowledge covered in unit 1.5;
rules for assigning oxidation numbers;
writing half equations
balanced overall redox equations.

Unit 5 is also a
Questions set on this topic will require application of knowledge covered in topics 5.2 – transition metal chemistry.
Notes from the specification
Zinc cannot be used to protect steel used to make food cans.
Large amounts of zinc are toxic.
Steel cans are coated with tin.
Tin is less easily oxidised than iron so it does not act as a sacrificial coat if it is damaged.
The tin coating acts to protect the iron with a non-toxic, tough coating.
Corrosion is the conversion of a metal (e.g., Fe) into its ions. [oxidation]
E⊖ = 0 V
Standard hydrogen electrode
16H+ + 2MnO4− + 5C2O42−  2Mn2+ + 8H2O + 10CO2
Determine the oxidation numbers for each atom.
Balance the increase and decrease in oxidation numbers.
Then balance the H and O atoms.
x 2
x 10
down +5
up +1
½ C2O42−  CO2

MnO4−  Mn2+
…worked example
The oxidation number of oxygen is usually –2, except in
(e.g. hydrogen peroxide [H2O2] where it is
) or combined with
, where it is
In monatomic ions, the oxidation number is equal to the charge on the ion.
The oxidation number of an element in its standard state is zero.
Assigning oxidation numbers - the rules
Redox equilibria [application]
C grade
E grade
A grade
What grade am I working at?
C grade
E grade
A grade
What grade am I working at?
The sum of the oxidation numbers in a neutral formula is zero.
The oxidation number of hydrogen is +1, except in metal hydrides where it is -1.
The sum of the oxidation numbers in an ion adds up to the charge on that ion.
3 4
Rule 6 states the O.N. of each O atom is -2; therefore three O atoms are 3 x -2 = -6.
The three iron atoms are +8 in total (rule 3). Thus, the average oxidation state of each iron atom is 8 / 3 = 2.667
The answer is not a whole number because two of the iron atoms are in the +3 state and one is in the +2 state
Try these
a) Mn in MnO b) O in H O

c) Cr in Cr O d) Cr in CrO

e) V in VO f) V in VO
a) +7 b) -1

c) +6 d) +7

e) +5 f) +4

(voltage) of electrochemical cell =
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