Oxidation and Reduction
Mr Field
Using this slide show

The slide show is here to provide structure to the lessons, but not
to limit them….go off-piste when you need to!

Slide shows should be shared with students (preferable electronic to
save paper) and they should add their own notes as they go along.

A good tip for students to improve understanding of the
calculations is to get them to highlight numbers in the question and
through the maths in different colours so they can see where
numbers are coming from and going to.

The slide show is designed for my teaching style, and contains only
the bare minimum of explanation, which I will elaborate on as I
present it. Please adapt it to your teaching style, and add any notes
that you feel necessary.
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Menu:
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Lesson 1 – Oxidation and Reduction
Lesson 2 – Redox Equations
Lesson 3 – Reactivity Series
Lesson 4 – Voltaic Cells
Lesson 5-6 – Electrolytic Cells
Lesson 7 – HL – Standard Electrode Potentials
Lesson 8 – HL – Ecell and Non-Standard Conditions
Lesson 9 – HL – Advanced Electrolysis
Lesson 10 – HL – Quantitative Electrolysis
Lesson 11–12 – Internal Assessment
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Lesson 1
Oxidation and Reduction
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Overview

Copy this onto an A4 page. You should add to it as a
regular review throughout the unit.
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Assessment

This unit will be assessed by:

An internal assessment at the end of the topic

A joint test at the end of the Acids and Bases topic
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Lesson 1: Oxidation and Reduction

Objectives:

Reflect on prior knowledge of oxidation and reduction

Understand oxidation and reduction in terms of electron
transfer

Calculate oxidation numbers
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Reflecting on Redox

Write down everything you know on oxidation and
reduction

You have 60 seconds
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Defining Oxidation and Reduction
O
I
L
- oxidation
- is
- loss of electrons

R
I
G
- reduction
- is
- gain of electrons

Often (but far from always) in practice:


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Oxidation is gain of oxygen or loss of hydrogen


This results in the loss of electrons
Reduction is loss of oxygen or gain of hydrogen

This results in the gain of electrons
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Oxidation States / Oxidation Numbers

Oxidation state is the charge an atom would have if all it’s bonds were ionic


It is the number of electrons an atom has gained or lost by forming bonds
You even talk about oxidation state of covalent compounds!

It is important as the oxidation state of an atom has a significant impact on its
chemistry

Fe(II)
Fe(III)

Cr(III)
Cr(VI)

Mn(II)
Mn(VII)
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Calculating Oxidation States
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
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The oxidation state of an
element is zero

The oxidation states of a
neutral compound sum to
zero, and of an ion sum to
the charge on the ion
The more electronegative
atom in an ion assumes a
negative oxidation state, the
less electronegative one a
positive oxidation state

Some rules of thumb:
Element
Oxidation
State
Notes
Fluorine
-1
Always
Oxygen
-2
Except in peroxides
where -1, and F2O
where +2
Chlorine
-1
Except with O or F
where +1
Gp I Metal
+1
Always
Gp II Metal
+2
Always
Hyrdogen
+1
Except metal hydride
where -1
Start with these, work the
others out.
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For example

Determine the oxidation states of each atom in the following:

CO2

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H2SO4


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O, -2
H, +1
S, +6
BaO2 (barium peroxide)



O, -2
C, +4
Ba,
O,
CO32

+2
-1
O, -2
C, +4
-2 except with F or a peroxide
to balance out the 2 lots of ‘-2’
-2 except with F or peroxide
+1 except in metal hydrides
+6 since four lots of ‘-2’ and two of ‘-1’ sum to -6
+2 since Gp II metal
-1 since peroxide
-2 except with F or peroxide
+4 since 3 x -2 and +4 sums to the charge, -2
Note: oxidation states must be written with the sign in front: +2 NOT 2+
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Determine oxidation states for each atom in:

H2O

HClO
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S8

KMnO4

CH4

IO3-

H3PO4

Cr2O72-

CCl4

Cr(H2O)63+
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Oxidation States and Names

Oxidation states are used in the names of compounds

‘ate’ means an element is in a positive oxidation state

Usually because it is bonded with oxygen
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‘ide’ means an element is in a negative oxidation state

Oxidation state of transition metals is given in Roman numerals

FeCl2 – iron (II) chloride
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FeCl3 – iron (III) chloride


Iron (II) means Fe in the +2 ox. State
Chloride means the chlorine is in a negative oxidation state
Iron (III) mean Fe in the +3 ox. State
KClO3 – potassium chlorate

Chlorate tells you the chlorine is in a positive oxidation state
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Name the following:

MnO2

CuO

Cu2O

KMnO4

K2Cr2O7
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Redox Reactions

Whenever an oxidation occurs, a reduction also occurs, hence REDOX

For example:
Zn(s) + Cu2+(aq)  Cu(s) + Zn2+(aq)
0
+2
0
+2

Zinc is oxidised – it loses two electrons


Copper is reduced – it gains two electrons
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Zinc is the reducing agent because it reduces copper
Cu2+ is the oxidising agent because it oxidises the zinc
The number of electrons gained by species is always equal to the number
of electrons lost by species.
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Disproportionation*

When some atoms of an element are oxidised and others are
reduced

For example:
2 H2 O2
+1 -1



2H2O +
+1 -2
O2
0
The oxygen ending in the H2O loses an electron and is oxidised
The oxygen ending in the O2 gains an electron and is reduced
*This is not on the syllabus but is useful and interesting
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Identify the species that are oxidised and reduced in each
reaction, stating the number of electrons gained or lost and
stating whether disproportionation takes place

Fe2O3 + 2 Al  Al2O3 + 2 Fe
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AgNO3 + NaCl  AgCl + NaNO3

3 Cl2 + 6 OH− → 5 Cl− + ClO3− + 3 H2O

H2SO4 + 2HBr  Br2 + SO2 + 2H2O

Cu + 4 HNO3  Cu(NO3)2 + 2 NO2 + 2 H2O
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Key Points

Oxidation state tells us the number of electorns an atom
has gained or lost

The most electronegative atom in a bond gains electrons,
to form a negative oxidation state and vice versa

Oxidation reactions are always accompanied by
reductions
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Lesson 2
Redox Equations
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Refresh

Fertilizers may cause health problems for babies because
nitrates can change into nitrites in water used for
drinking.
a)
Define oxidation in terms of oxidation numbers.
b)
Deduce the oxidation states of nitrogen in the nitrate, NO3–,
and nitrite, NO2–, ions.
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Lesson 2: Redox Equations

Objectives:

Deduce simple half-equations

Combine half-equations to form full equations
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Conduct a series of redox reactions
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Use H+ and H2O to balance redox equations
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Half-equations

Half equations show the changes to individual species in a
redox reaction.

Fe2O3 + 2 Al  2 Fe + Al2O3
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
Fe3+ + 3 e-  Fe ….this is the reduction

Al  Al3+ + 3 e- ….this is the oxidation
A wide variety of half equations can be found in the data
booklet
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Combining Half Equations and Balancing
Redox Reactions – Example 1
Step
Example 1: Reaction of iodine with
copper
Write out half-equations side by side and the number
of electrons lost or gained…can be found in the data
booklet
½ I2 + e-  I- … 1 e- gained
Cu  Cu2+ + 2 e- …2 e- lost
Combine them to form a single reaction, multiplying
each half-equation in order to balance the electrons
gained/lost
I2 + 2 e- + Cu  2 I- + Cu2+ + 2 e-
Ensure all atoms other than O and H balance
I2 + 2 e- + Cu  2 I- + Cu2+ + 2 e- no change needed
Cancel out any electrons that are duplicated on both
sides
I2 + Cu  2 I- + Cu2+
Balance oxygen atoms by adding H2O to the side that
needs extra
I2 + Cu  2 I- + Cu2+
Balance the hydrogens by adding H+ to the side that
needs extra
I2 + Cu  2 I- + Cu2+
- no change needed
Count the total charge on each side and add enough
electrons to the more positive side to balance it
I2 + Cu  2 I- + Cu2+
- no change needed
- the iodine ½-equation is doubled to make
. 2 e- gained, the other remains unchanged
- 2 e- appear on each side so are cancelled out
- no change needed
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Combining Half Equations and Balancing
Redox Reactions – Example 2
Step
Example 2: Reaction of nickel with manganate
ions
Write out half-equations side by side
and the number of electrons lost or
gained
MnO4- +5e-  Mn2+ …5 e- gained by Mn
Ni  Ni2+ + 2 e- … 2 e- lost by Ni
Combine them to form a single reaction
2MnO4- + 5Ni + 10e- 2Mn2+ + 5Ni2+ + 10e•Ni ½-equation multiplied by 5
•MnO4- ½-equation multiplied by 2
Ensure all atoms other than O and H
balance
2MnO4- + 5Ni + 10e- 2Mn2+ + 5Ni2+ + 10e•No change needed
Cancel out any electrons that are
duplicated on both sides
2MnO4- + 5Ni  2Mn2+ + 5Ni2+
•10 e- cancelled
Balance oxygen atoms by adding H2O to
the side that needs extra
2MnO4- + 5Ni  2Mn2+ + 5Ni2+ + 8H2O
•8H2O on the right balance the 8 O on the left
Balance the hydrogens by adding H+ to
the side that needs extra
2MnO4- + 5Ni + 16H+  2Mn2+ + 5Ni2+ + 8H2O
•16H+ added on left to balance 16 H on the right
Count the total charge on each side and
add electrons to the more positive side
to balance it
2MnO4- + 5Ni + 16H+  2Mn2+ + 5Ni2+ + 8H2O
•Total charge balances so no change needed’
•This step only needed on half-equations
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Microscale Redox Reactions

Complete the experiment here, on microscale redox
reactions
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

Rather than doing it on a plastic sheet, use a dropping tile
For each change you observe, you should write a balanced
redox equation to describe it
Once you finish, you should practice balancing the redox
equations on the following slide
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Produce balanced redox equations for the
reactions of:

Bromine with sodium
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Copper (II) oxide with hydrogen
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Cu2+ + 2e-  Cu
½ H 2  H + + e-
Aluminium reacting with chromate ions
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½ Br2 + e-  BrNa Na+ + e-
Al  Al3+ + 3eCr2O72- + 6e- 2Cr3+
Iron (II) chloride reacting with manganate ions


Fe2+  Fe3+ + eMnO4- + 5 e- Mn2+
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Key Points
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Half-equations show the changes to each species in a
redox reaction

To combine half-equations into a full equation




Multiply each such that the electron-transfers balance
Add H2O to balance O
Add H+ to balance H
Add e- to balance charge
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Lesson 3
Reactivity Series
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Refresh

Nitric acid reacts with silver in a redox reaction:
__ Ag(s) + __ NO3–(aq) + ____ → __ Ag+(aq) + __ NO(g) + ____

Using oxidation numbers, deduce the complete balanced
equation for the reaction showing all the reactants and
products.
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Lesson 3: Reactivity Series

Objectives:

Deduce reactivity series from chemical observations

Use reactivity series to predict the feasibility of reactions
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Redox and Reactivity
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Redox behaviour is closely linked to reactivity
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The most reactive metals are the best reducing agents
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The most reactive non-metals are the best oxidising
agents
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The least reactive elements are neither good oxidising or
reducing agents
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Reactivity and Displacement Reactions

Reactive metals are better reducing agents than unreactive metals.

As such, a reactive metal can displace less reactive metals
from their compounds.
Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

As a good reducing agent, the zinc reduces the Cu2+,
causing it to gain two electrons.
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Constructing a Reactivity Series

Complete the experiment here in which you have to
construct a reactivity series.

Analysis

Use your reactivity series to predict the feasibility of the
reactions of:
 I


with Fe3+
Zn with Sn2+
Fe2+ with Cl
Compare your series with the order found on Table 14 of the
data booklet.
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Key Points

More reactive metals are better reducing agents

More reactive non-metals with oxidising agents

A more reactive metal will reduce (displace) ions of a less
reactive metal
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Lesson 4
Voltaic Cells
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Refresh

Consider the following three redox reactions.
Cd(s) + Ni2+(aq) → Cd2+(aq) + Ni(s)
Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s)
Zn(s) + Cd2+(aq) → Zn2+(aq) + Cd(s)
a)
Deduce the order of reactivity of the four metals, cadmium,
nickel, silver and zinc and list in order of decreasing
reactivity.
b)
Identify the best oxidizing agent and the best reducing agent.
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Lesson 4: Voltaic Cells

Objectives:

Explain in simple terms how voltaic cells use redox reactions
to produce electricity

Understand that oxidation occurs at the anode and reduction
at the cathode

Make a series of voltaic cells in order to better understand the
how they work
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Voltaic Cells

The reaction of Mg with Cu2+ ions:

Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)

This reaction involves two electrons being transferred from the Mg to the Cu:





The Mg reduces the copper ions as it is more reactive
This is an exothermic reaction, and the energy is normally released as heat
A voltaic cell forces each half of the reaction to take place in a separate
container, with the electrons moving through a circuit to get from one side
to the next


Mg  Mg2+ + 2eCu2+ + 2e-  Cu
This is an exothermic reaction, where the energy is released as electrical rather
than thermal energy
The reactions in Voltaic cells usually involve only metals but do not have to.
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Voltaic Cells Continued
-
+
Anode:
Cathode:
Where
oxidation
happens
Where
reduction
happens
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Key Parts of a Voltaic Cell

Anode

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
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Cathode


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


Electrode or ‘half-cell’ where reduction happens
Contains the less reactive metal
The positive electrode: accepts electrons



Contains a neutral salt such as potassium nitrate
Made of a tube of jelly or a filter paper soaked in salt solution
Ions diffuse in and out to balance charge and complete circuit
Voltmeter
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Measures the difference in potential between half-cells
Could be replaced with other circuitry to do useful work
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Anode-Oxidation
CaRe
Salt Bridge

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Electrode or ‘half-cell’ where oxidation happens
Contains the more reactive metal
The negative electrode: produces electrons
REMEMBER
 AnOx
Cathode-Reduction
Drawing a cell

Draw and fully label a zinc/iron cell. Include:




Labels for cathode and anode
Labels for positive and negative
Each half-equation
Arrow showing direction of electron flow
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Constructing Voltaic Cells

You will need to build and measure the potential of
voltaic cells comprising various combinations of the
following:






Cu/Cu2+
Fe/Fe2+
Mg/Mg2+
Sn/Sn2+
Zn/Zn2+
Follow the instructions here
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Key Points

Voltaic cells extract electrical energy from redox
reactions by separating each half

At the anode, the more reactive of two metals is oxidised

At the cathode, the less reactive of two metals is reduced
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Lesson 5-6
Electrolytic Cells
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Refresh
A particular voltaic cell is made from magnesium and iron halfcells. The overall equation for the reaction occurring in the cell is
Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)
Which statement is correct when the cell produces electricity?
A.
B.
C.
D.

Magnesium atoms lose electrons.
The mass of the iron electrode decreases.
Electrons flow from the iron half-cell to the magnesium half-cell.
Negative ions flow through the salt bridge from the magnesium
half-cell to the iron half-cell.
For each incorrect statement, explain why it is wrong.
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Lesson 5-6: Electrolysis

Objectives:

Describe electrolytic cells

Identify at which electrode oxidation and reduction takes place

Understand how current is conducted in electrolytic cells

Deduce the products of electrolysis of a molten salt
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Electrolytic Cells

Electrolytic cells use electricity to provide the energy for an endothermic redox reaction

Electrolysis of lithium chloride:
Li+(l) + Cl-(l)  Li(s) + ½Cl2(g)

Current is carried by moving ions



As ions need to be able to move, the ionic compound must be either:



Cations (+) move to the cathode (negative electrode)
Anions (-) move to the anode (positive electrode)
Molten
Dissolved in solution
The opposite of a voltaic cell:


Voltaic cells turn stored chemical energy into electrical
Electrolytic cells turn electrical energy into stored chemical energy
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An Electrolytic Cell:
e-
eBubbles of gas
formed
ANODE
(+)
CATHODE
(-)
X X
X X
MOLTEN SALT
or
SALT SOLUTION
M+
M+
X-
M
Layer of metal
formed
M
M+
M
Anions move to
anode
X-
X-
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Cations move to
cathode
Products of Electrolysis

Assuming a molten simple metal salt involving only monatomic ions

Cathode – metal


Anode – non-metal


Typically as bubbles of gas
For example, electrolysis of molten magnesium bromide:



Deposited on the surface of the electrode
Cathode: a layer of magnesium metal (Mg2+ + 2e-  Mg)
Anode: bubbles of bromine gas (2Br-  Br2 + 2e-)
More complicated systems


Aqueous solutions – a range of possibilities depending on the stability of the ions
relative to water
Polyatomic ions – a mixture of various products
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By the end of the next lesson you should:

Complete the electrolysis of molten zinc chloride in the fume hood.
Instructions here:

http://www.nuffieldfoundation.org/practical-chemistry/electrolysis-zinc-chloride

Produce an animation showing how electrolysis happens. Could be
PowerPoint, flicker book, smart phone animation app, stop-motion
(see: http://www.wikihow.com/Create-a-Stop-Motion-Animation )...be
creative.

Draw a Venn diagram comparing and contrasting electrolytic and
voltaic cells

Write three potential internal-assessment research questions and
on electrolytic or voltaic cells, select one and start producing a plan
for it.
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Key Points

Electrolysis uses electricity to drive endothermic redox
reactions

Metal salts must be molten or dissolved so ions can move
and carry charge

The negative cathode reduces positive metal ions

The positive anode oxidises negative non-metal ions
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Lesson 7
HL Only
Standard Electrode Potentials
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Refresh
Which processes occur during the electrolysis of molten sodium
chloride?
I.
II.
III.
A.
B.
C.
D.

Sodium and chloride ions move through the electrolyte.
Electrons move through the external circuit.
Oxidation takes place at the anode.
I and II only
I and III only
II and III only
I, II and III
Justify your answer.
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Lesson 7: Standard Electrode Potentials

Objectives:

Describe the standard hydrogen electrode

Define the term standard electrode potential

Use standard electrode potentials to calculate the potential of
a cell

Use standard electrode potentials to determine the feasibility
of a reaction
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Half-cell potential

When you set up a half-cell, it has a certain ‘POTENTIAL’.

When two half-cells are joined, electrons flow away from the
half-cell with more negative potential towards the half-cell
with more positive potential

You can think of this a little like enthalpy



There is no absolute measure of potential, only relative
You can only measure potential differences (voltage) between two
half-cells
The potential of a half-cell is always measured relative to the
standard hydrogen electrode.
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The Standard Hydrogen Electrode

Standard electrode potential defined as 0.00 V
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Standard Electrode

Temperature = 298K

Pressure = 1 atm
Connecting wire
Metal Electrode
Aqueous solution of
metal ions:
[M+] = 1.0 mol dm-3
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Standard Electrode Potential, Eo


This is the potential of a
standard electrode relative
to the standard hydrogen
electrode.
Always measure the
potential of the reduction

Measured in Volts,V

Full table in the data
booklet
Half Cell
Standard Electrode
Potential, Eo / V
H+(aq) + e- ⇌ ½ H2(g)
0.00
Li+(aq) + e- ⇌ Li(s)
-3.04
Mn2+(aq) + 2e- ⇌ Mn(s)
-1.19
Cu2+(aq) + 2e- ⇌ Cu(s)
+0.34
½ Br2(l) + e- ⇌ Br-(aq)
+1.07

Look at the table in the data booklet:
 What trends do you notice?
 How do the values relate to your ideas
of reactivity?
 How do the values compare to the
reactivity series you constructed
earlier?
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The Potential of a Cell, Eocell


To make a cell, two half-cells are connected
by a salt-bridge and a volt-meter
The potential of a cell is easy to calculate:




0.00
Li+(aq) + e- ⇌ Li(s)
-3.04
Mn2+(aq) + 2e- ⇌ Mn(s)
-1.19
Cu2+(aq) + 2e- ⇌ Cu(s)
+0.34
½ Br2(l) + e- ⇌ Br-(aq)
+1.07
Lithium: -3.04 V, Manganese: -1.19V
Eocell = -1.19 – -3.04 = 1.85 V
What is the potential of a Manganese-Bromine cell?


Manganese: -1.19 V, Bromine: 1.07 V
Eocell = 1.07 – -1.19 = 2.26 V
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Standard Electrode
Potential, Eo / V
H+(aq) + e- ⇌ ½ H2(g)
What is the potential of a Lithium-Manganese cell?


Subtract the potential of the more negative ½cell from the potential of the more positive
This gives the potential difference:
Half Cell
Determining the Reaction in a Cell:

A little more tricky, but still OK

The more negative half-cell moves to the left (gets
oxidised)

The more positive half-cell moves to the right
(gets reduced)

Determine the reaction for each half cell and then
combine them (making sure they balance)
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Determining the Reaction in a Cell: Example 1
Half Cell

The Lithium-Manganese cell?

H+(aq) + e- ⇌ ½ H2(g)
0.00
Li+(aq) + e- ⇌ Li(s)
-3.04
Mn2+(aq) + 2e- ⇌ Mn(s)
-1.19
Cu2+(aq) + 2e- ⇌ Cu(s)
+0.34
½ Br2(l) + e- ⇌ Br-(aq)
+1.07
Lithium: -3.04 V, Manganese: -1.19V



Standard Electrode
Potential, Eo / V
Lithium: Li(s)  Li+(aq) + e- ….more negative so goes left
Mn2+(aq) + 2e-  Mn(s) …more positive so goes right
2 Li(s) + Mn2+(aq)  2 Li+(aq) + Mn(s)
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Determining the Reaction in a Cell: Example 2
Half Cell

The Manganese-Copper cell?

H+(aq) + e- ⇌ ½ H2(g)
0.00
Li+(aq) + e- ⇌ Li(s)
-3.04
Mn2+(aq) + 2e- ⇌ Mn(s)
-1.19
Cu2+(aq) + 2e- ⇌ Cu(s)
+0.34
½ Br2(l) + e- ⇌ Br-(aq)
+1.07
Copper: +0.34 V, Manganese: -1.19 V



Standard Electrode
Potential, Eo / V
Manganese: Mn(s)  Mn2+(aq) + e- ….more negative so goes left
Copper: Cu2+(aq) + 2e-  Cu(s) …more positive so goes right
Mn(s) + Cu2+(aq)  Mn2+(aq) + Cu(s)
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Drawing and Labelling a Cell

Example: The manganese-copper cell
V
electron
flow
electron
flow
Cu2+(aq)+ Mn(s)  Cu(s) + Mn2+(aq)
Manganese
Salt
Bridge
Copper
1.00 mol dm-3
Mn2+(aq)
1.00 mol dm-3
Cu2+(aq)
ANODE, (-)
oxidation occurs:
CATHODE, (+)
reduction occurs
Mn(s)  Mn2+(aq) + e-
Cu2+(aq) + 2e-  Cu(s)
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Write the equation for, and calculate the potential of
the following cells. Draw and fully label two of them*.
1.
Copper and Zinc
2.
Silver and Lead
3.
Hydrogen and Nickel
4.
Iodine and Iron
5.
Iron and Iron (II)
*Note: If you don’t have a solid metal, use platinum for your electrode,
and make sure it is in contact with both phases of the ½-cell.
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Predicting The Feasibility of a Reaction

Will iron (II) react with copper?

The only possible reaction between Cu and Fe2+ is:
Cu(s) + Fe2+(aq)  Cu2+(aq) + Fe(s)

Select relevant ½-cells:



Cu2+(aq) + 2e- ⇌ Cu(s)
Fe2+(aq) + 2e- ⇌ Fe(s)
+0.34 V
-0.45 V
Compare the ½-cell potentials:



The more negative ½-cell, (Fe2+(aq) + 2e- ⇌ Fe(s)), always gets oxidised
(moves to the left)
This reaction shows it being reduced (moving to the right)
Therefore this reaction can’t happen
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Predicting The Feasibility of a Reaction

Will zinc ions react with magnesium?

The only possible reaction between Zn2+ and Mg is:
Mg(s) + Zn2+(aq)  Mg2+(aq) + Zn(s)

Select relevant ½-cells:



Zn2+(aq) + 2e- ⇌ Zn(s)
Mg2+(aq) + 2e- ⇌ Mg(s)
-0.76 V
-2.37 V
Compare the ½-cell potentials:



The more negative ½-cell, (Mg2+(aq) + 2e- ⇌ Mg(s)), always gets oxidised
(moves to the left)
This reaction shows it being oxidised (moving to the left)
Therefore this reaction will happen
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Determine whether each of the following reactions is
feasible, stating why/why not in each case.

Pb2+ ions reacting with calcium

Zn2+ ions reacting with iron

Cl- ions reacting with fluorine

Silver reacting with chlorine

Silver reacting with iodine
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Key Points:

Eocell is the difference between the two Eo values

The more negative half-cell gets oxidised

The more positive half-cell gets reduced

A reaction will only be feasible if it results in the more
negative species being oxidised and the more positive
being reduced
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Lesson 8
HL Only
Eocell and non-standard conditions
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Refresh
Consider the following standard electrode potentials.
Zn2+(aq) + 2e– Zn(s)
Cl2(g) + 2e– 2Cl–(aq)
Mg2+(aq) + 2e– Mg(s)

What will happen when zinc powder is added to an aqueous solution of
magnesium chloride?
A.
B.
C.
D.

Eo = –0.76 V
Eo = +1.36 V
Eo = –2.37 V
No reaction will take place.
Chlorine gas will be produced.
Magnesium metal will form.
Zinc chloride will form.
Justify your answer.
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Lesson 8: Deviations from Standard Conditions*

Objectives:

Understand the effects of non-standard conditions on the potential
of a cell

Use the Nernst equation to calculate the effect of non-standard
conditions on cell potential
*Note: this lesson does not lie within the syllabus but is useful
to deepen your understanding and is a rich source of potential
IA questions!
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The Nernst Equation

You have learnt to calculate the potential of a cell in standard conditions

Cells are rarely under standard conditions

We can still calculate the cell potential using the Nernst Equation
E  Eo 

RT
ln Q
nF
Where:







E is the non-standard cell potential
E is the standard cell potential
R is the gas constant, 8.31
T is the temperature in Kelvins
Walther Nernst
N is the number of electrons transferred in the reaction
F is the Faraday constant, 96,500
Q is the reaction quotient: concentration of products divided by concentration of
reactants
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What does the Nernst Equation mean?
RT
EE 
ln Q
nF
o


From the maths of the Nernst equation we can draw three important
conclusions
1.
Increasing the temperature will decrease the cell potential
2.
Increasing the concentration of the anode (the oxidised species, our ‘product’)
will reduce cell potential
3.
Increasing the concentration of the cathode (the reduced species, our
reactant) will increase cell potential
Note: effects two and three will be relatively small as they are
logarithmic
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Testing the Nernst Equation

In this experiment you will design and plan experiments
to confirm the findings of the Nernst equation.

You will be using a magnesium-copper cell as this gives
about the largest potential we can feasibly investigate with
our facilities

Follow the instructions here
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Key Points

The Nernst Equation describes how Ecell changes under
non-standard conditions

Temperature reduces Ecell

Increasing the anode concentration reduces Ecell

Increasing the cathode concentration increases Ecell
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Lesson 9
HL Only
Advanced Electrolysis
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Refresh
Consider
the following cell:
Pb(s) + 2 Ag+(aq)  Pb2+(aq) + 2 Ag(s)
Describe
and explain what happens to cell potential upon:
Increasing
Reducing
the temperature
the concentration of Pb2+ ions
Increasing
the concentration of Ag+ ions
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Lesson 9: Advanced Electrolysis

Objectives:

Describe the products of electrolysis in aqueous solution

Explain the uses of electrolysis for electroplating

Complete an experiment to investigate electroplating
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Electrolysis of Water

Water itself can be readily electrolysed


Anode (oxidation, moves to left):


½ O2(g) + 2 H+(aq) ⇌ H2O(l)
Eo = +1.23V
Cathode (reduction, moves to right):


Since Kw of water is very low, H+ is added to help carry current
2 H2O(l) + e- ⇌ H2(g) + 2 OH-(aq)
Eo = -0.83 V
Overall*:

2 H2O(l)  2 H2(g) + O2(g)
*Note: the reverse of this is the reaction that takes place in a
hydrogen fuel cell and has an Eocell of 2.06 V.
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Electrolysis of Aqueous Solutions

Electrolysis of molten salts is simple:


Cathode  Metal
Anode  Non Metal

Electrolysis of aqueous solutions is less straightforward due to the fact that the water can
compete with the salts to undergo electrolysis.

At the Anode:



If an anion has an Eo more positive than +1.23 V, water will be oxidised instead of the anion
Bubbles of O2 gas will be formed and H+ ions will enter solution
This happens for anions including (but not only): fluoride and sulphate


Note: Although chlorine has a potential of +1.36 V, it WILL be discharged in aqueous solution., contrary to
what you might expect. The reasons for this are complex and are to do with changes in Cl- concentrations
at the anode once current starts to flow.
At the Cathode:



If a cation has an Eo more negative than -0.83 V, water will be reduced instead of the cation
Bubbles of H2 gas will be formed and OH- ions will enter solution
This happens for cations including (but not only): lithium, potassium, sodium and magnesium
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Example 1: What are the products of the
electrolysis of aqueous copper sulphate?

Consider the standard electrode potentials:

At the cathode:

2 H2O(l) + e- ⇌ H2(g) + 2 OH-(aq)
Cu2+(aq) + 2e- ⇌ Cu(s)

Copper less negative than water so Cu2+ is reduced, cathode gains coating of metallic copper


At the anode:




-0.83 V
+0.34 V
½ O2(g) + 2 H+(aq) ⇌ H2O(l)
S2O82–(aq) + 2e– ⇌ 2 SO42–
+1.23 V
+2.01 V
Sulphate is more positive than water so water is oxidised, bubbles of O2 gas produced and solution
becomes acidic
Overall reaction:

Cu2+(aq) + SO42-(aq) + 2H2O(l)  Cu(s) + SO42-(aq) + 2H+(aq) + O2(g)
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Example 2: Electrolysis of Sodium Chloride

Consider the standard electrode potentials:

At the cathode:




-0.83 V
-2.71 V
Sodium more negative than water so water is reduced, cathode produces bubbles of H2 gas and solution
becomes more alkaline
At the anode:




2 H2O(l) + e- ⇌ H2(g) + 2 OH-(aq)
Na+(aq) + e- ⇌ Na(s)
½ O2(g) + 2 H+(aq) ⇌ H2O(l)
½ Cl2(g) + e–
⇌ Cl-
+1.23 V
+1.36 V
Chlorine is more positive than water, so you expect it water to be oxidised, but chloride is due to reasons
stated earlier.
Overall reaction:

2Na+(aq) + 2Cl-(aq) + 2H2O(l)  2Na+(aq) + 2OH-(aq) + H2(g) + Cl2(g)
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Predict the products at the anode and cathode for electrolysis of
the following aqueous solutions. For each one, write the overall
equation for the reaction.

KCl

NiSO4

PbI2

ZnCl2

LiOH
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Electroplating

Electroplating uses electrolysis to coat metal objects in a
fine layer of another metal, by setting it as the cathode.

Examples include:



Coating things in nice shiny chrome
Coating with hold to improve conductivity
Coating with sacrificial metal for protection
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Investigating Electroplating

In this short experiment, you will electroplate some small
metal objects with a variety of different metals

Follow the instructions here
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Key Points

In electrolysis of aqueous solutions, it is possible for water to
be oxidised and reduced.

At the cathode:

If a metal has a more negative Eo than water:



Water will be reduced
Hydrogen gas and OH- ions will be produced
At the anode:

If a species has a more positive Eo than water:


Water will be oxidised
O2 gas and H+ ions will be produced
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Lesson 10
HL Only
Quantitative Electrolysis
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Refresh

Sodium metal can be obtained by the electrolysis of
molten sodium chloride.
a.
Explain why it is very difficult to obtain sodium from
sodium chloride by any other method.
b.
Explain why an aqueous solution of sodium chloride
cannot be used to obtain sodium metal by electrolysis.
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Lesson 10: Advanced Electrolysis

Objectives:

Describe the effect of time on the amount of amount of
product formed during electrolysis

Describe the effect of ionic charge on the amount of product
formed during electrolysis

Complete an experiment to determine Avogadro’s constant by
electrolysis of copper sulphate solution.
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First some physics

The ‘amount’ of electricity (charge, Q) is measured in
Coloumbs, C

Current (I), the flow of electricity, is measured in Amperes, A


A current of 1.00 A means 1.00 C of charge flows every second
So, the total amount of charge flowing through a circuit is
given by:


Q = I.t
Charge = Current x time
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And a bit more physics

The amount of charge on a single electron, e, is:

e = 1.602x10-19

If you If you divide 1.00 C (charge) by the charge on a
single electron (e), you can determine the number of
electrons that make up a Coloumb:

Electrons per Coulomb = 1.00 / 1.602x10-19
= 6.24x1018
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The Effect of Time

The greater the time of electrolysis, the more products
formed:

This stands to reason but is clear if you look at the
equation for charge and currents:

Q=Ixt



The longer the time, the bigger Q
The bigger Q, the bigger the number of electrons that have flowed
The bigger the number of electrons, the more reduction/oxidation
can take place
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The effect of ionic charge

Of three solutions of equal concentration, electrolysed at equal current for
an equal time, which would form the greatest quantity (in mol) of metal at
the cathode?




Answer:


PbCl2
PbCl4
CrCl3
PbCl2
Reason:


The same current for the same time means the same number of electrons flows
PbCl contains Pb2+ ions which requires the fewest electrons (2) to reduce them,
this means more ions can be reduced for the same number of electrons:


PbCl4 contains Pb4+ ions – needs 4 electrons
CrCl3 contains Cr3+ ions – needs 3 electrons
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Calculating Avogadro’s Constant

Avogadro’s constant can be determined by electrolysis

Follow the instructions here

This goes beyond the requirements of the syllabus but is
interesting and will deepen your understanding
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Key Points

The longer the electrolysis, the more products will form

All else being equal, metal ions with a lower charge will
produce more metal during electrolysis than those with a
higher charge
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Lesson 11-12
Internal Assessment
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Internal Assessment

You should design and conduct and internal assessment
on an aspect of oxidation and reduction.

Electrolysis and voltaic cells both provide rich hunting
grounds.
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