9.1.1 Define oxidation and reduction in terms of electron loss and gain.
9.1.2 Deduce the oxidation number of an element in a compound.
9.1.3 State the names of compounds using oxidation numbers.
9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers.
Oxidation and reduction is applied to both ionic and covalent interactions during which electrons are transferred or lost/gained control over.
OILRIG : O
I
L
R
I
G
 reduction = gain of 1 or more electron; an atom has gained the control of electron(s) from another atom;

in a covalent bond, it results in the atom having a higher electron density around it;

during ionic bond formation, the atom becomes a negative ion;
 oxidation = loss of 1 or more electrons; an atom has lost control over 1 or more of its electrons;

in a covalent bond, it results in a lower electron density around the atom;

in ionic bond formation, the atom becomes a positive ion.
Recognising redox reactions: oxidation = happens when oxidation number of element increases
; reduction = when oxidation number of element decreases;
The oxidation number or the oxidation state of an element in a compound indicates the number of electrons over which an atom/ion of an element has gained or lost control during a reaction. Oxidation numbers are used to keep track of how many electrons are lost or gained by each atom or ion. They are used to identify oxidation and reduction reactions.
Rules for working out oxidation number:
1. The oxidation number of an atom in its element form or uncombined form is always 0.
For example the atoms in Na, He, N
2
and S
8
have oxidation numbers of 0.
2. The oxidation number of a monatomic ion equals the charge of the ion.
For example, the oxidation number of Na + is +1; the oxidation number of S 2 is -2 and of Al 3+ is +3.
3. The usual oxidation number of hydrogen is +1 (when bonded with a non-metal) but it is -1 when hydrogen is bonded with a metal (metal hydrides) e.g. as in CaH
2
.
IB chemistry topic 9 7 hours 1 | P a g e

4. The oxidation number of oxygen in compounds is usually -2.
Exceptions include OF
2
, since F is more electronegative than O, and H
2
O
2
, due to the structure of the peroxide particle which is [O-O] 2.
5. The oxidation number of a group 1 element in a compound is +1.
6. The oxidation number of a group 2 element in a compound is +2.
7. The oxidation number of a group 7 element in a compound is -1, except when that element is combined with one having a higher electronegativity.
For example in ICl, the oxidation number of I is +1, the oxidation number of Cl is -1 in HCl, but the oxidation number of Cl is +1 in HOCl.
8. The sum of the oxidation numbers in a neutral compound is 0.
Examples:
NaCl (oxidation number of Na) + (oxidation number of Cl) = 0
(+1) + (-1) = 0
Na
2
O 2(Ox. No. of Na) + (Ox. No. of O) = 0
2(+ 1) + (-2) = 0
CuS (Ox. No. of Cu) + (Ox. No of S) = 0
(+2) + (-2) = 0
CaBr
2
(Ox. No. of Ca) + 2(Ox. No. of Br) = 0
(+2) + 2(-1) = 0
9. The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
For example, the sum of the oxidation numbers for SO
4
2 is -2 because S = +6 and O = -2
so +6 + 4(-2) = -2
10. When elements show more than one oxidation state the oxidation number is shown using a
Roman numeral when naming the compound.
Examples: Fe (II)Cl
2
= iron (II) chloride, Fe(III)Cl
3
iron (III) chloride; Cu
2
O = copper (I) oxide; CuO =
copper (II) oxide
Examples
Example What is the oxidation number of thallium in TiCl
3
?
Method Chlorine always has the oxidation number -1.
Therefore (Ox. No. of Ti) + 3( -1) = 0 and the oxidation number of thallium is +3.
Example What is the oxidation number of Cl in Cl
2
O
7
?
IB chemistry topic 9 7 hours 2 | P a g e

Method The exceptions to the rule that the oxidation number of Cl equals -1 are compounds with O and
F. Oxygen is the reference point with the oxidation number -2.
Therefore 2 (Ox. No. of Cl) + 7(-2) = 0 and the oxidation number of chlorine is +7.
Example What is the oxidation number of Cr in Cr(CN)
6
3 ?
Method The cyanide ion, CN , has a charge of - 1.
Therefore (Ox. No. of Cr) + 6( -1) = - 3 and the oxidation number of chromium is + 3.
Use http://www.occc.edu/kmbailey/chem1115tutorials/oxidation_numbers.htm
In exercises 1 to 3 the main emphasis is on the following elements and their oxidation states: Fe (II) and
(III); Mn (II) and (VII), Cr (III) and (VI) and Cu (I) and (II), halogens and their ions and S in its oxyacids.
1. Give the oxidation number of each of the elements in the following species:
a) H
2
O b) K
2
O c) AlCl
3
d) OF
2
e) MnO
2
f) MnO
4
g) ClO
3
h) CuO i) Cr
2
O
7
2-
j) CrO
4
2 k) S
2
O
3
2 l) PbO
2
m) K
2
C
2
O
4
2. Work out the oxidation states of nitrogen in: NH
3
, N
2
H
4
, N
2
, N
2
O, NO, NO
2
, NO
3
.
3. Give the oxidation numbers of the first element in each of the following compounds.
Remember the oxidation numbers of the elements in a compound add up to zero. Take oxidation
numbers for hydrogen (+1), oxygen (-2), fluorine (-1) and chlorine (-1) as reference points.
CuO, Cu
2
O, H
2
S, SO
2
, SO
3
, PbO, SF
6
, SCl
2
, TiCl
4
, V
2
O.
4. Say whether the underlined atom is oxidized, reduced or remains unchanged?
(a) PbCl
2
+ Cl
2

PbCl
4
(b) NaOH + HCl

NaCl + H
2
O
(c) 2IO
3
+ 5HSO
3

I
2
+ 5SO
4
2 + 3H + + H
2
O
(d ) Cu + ½ O
2

CuO
Oxidation states in names: some elements can have different oxidation states in different compounds.
The oxidation state is often indicated using Roman numerals or by the name, eg:

FeO and Fe
2
O
3
are called iron(II) oxide and iron(III) oxide. (Roman numerals are used; there is no
space between the numeral and the element.)

CuCl
2
and Cu
2
O: copper(II) chloride and copper(I) oxide.

The following ions, ClO , ClO
2
, ClO
3
, ClO
4
are called chlorate(I), chlorate(III), chlorate(V) and chlorate(VII).

sulphur: sulphuric acid, H
2
SO
4
, (S has oxidation state +6) and sulphurous acid, H
2
SO
3
, (S is +4);
IB chemistry topic 9 7 hours 3 | P a g e

nitrogen: nitric acid, HNO
3
, (N has +5) and nitrous acid, HNO
2
, (N has +3);

9.2.1. Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction.
9.2.2 Deduce redox equations using half equations.
9.2.3 Define the terms oxidizing agent and reducing agent .
9.2.4 Identify the oxidizing and reducing agents in redox equations.
A redox react ion is a reaction during which an oxidation and reduction happen at the same time; as a result an electron transfer takes place between the different reacting species in the reaction.
We can now distinguish two different species:

oxidizing agent= specie which oxidizes another substance during a redox reaction; the oxidizing agent itself is reduced i.e. it accepts electrons from the other substance;

reducing agent = specie which reduces another substance in a redox reaction; itself is oxidised i.e. it gives electrons to the other species;
We use reduction and oxidation half equations to describe each type of reaction (oxidation or reduction) in a redox reaction.
Reduction and oxidation half equation = stoichiometric equation showing the oxidation or reduction of a specie; such an equation involves electron symbols.
Examples: oxidation half equation (e
Fe 2+ (aq)
on product side !!!)

Fe 3+ (aq) + e - reduction half equation (e on reactant side !!!)
Cu 2+ (aq) + 2e - 
Cu (s)
2Cl (l)

Cl
2
(g) + 2e Zn 2+ (aq) + 2e - 
Zn (s)
Such equations must balance in respect to the charges and the number of atoms or ions.
When presented with an equation you need to be able to:

identify if it is a redox equation by using oxidation numbers

if it is, identify which species are reduced and which one is oxidized

deduce the oxidation and reduction half equations

identify the reducing and oxidizing agent
Example
Fe
2
O
3
(s) + 2Al (s)

Al
2
O
3
(s) + 2Fe (s)
As iron’s oxidation number changes from +3 to 0; aluminium’s oxidation number changes from 0 to +3; therefore it has been reduced . it has been oxidized.
Reduction half equation: Fe 3+ + 3e

Fe 3+ Oxidation half equation: Al

Al 3+ + e -
IB chemistry topic 9 7 hours 4 | P a g e

Fe
2
O
3 is the oxidizing agent as it causes the oxidation of aluminium. Aluminium is the best reducing agent as it reduces Fe 3+ to Fe.
1. For each of the redox equations below:

deduce the oxidation and reduction half equations

deduce the oxidizing and reducing agent
(a) Mg (s) + Cl
2
(g)

MgCl
2
(s)
(b) 2NaCl (aq) + F
2
(aq)

2NaF (aq) + CI
2
(aq)
(c) CuSO
4
(aq) + Zn (s)

ZnSO
4
(aq) + Cu (s)
(d) Cu
(s)
+ 2AgNO
3 (aq)
→ Cu(NO
3
)
2 (aq)
+ 2Ag
(s)
(e) CuO (s) + H
2
(g)

Cu (s) + H
2
O (l)
(f) 2Fe 2+ (aq) + I
2
(aq)

2Fe 3+ (aq) + 2I (aq)
(g) Sn 2+ (aq) + 2Fe 3+ (aq)  Sn 4+ (aq) + 2Fe 2+ (aq)
(h) 3I
2
(aq) + 3OH (aq)  IO
3
(aq) + 5I (aq) + 3H + (aq)
2. In the following reactions, identify which species have been odixized and which have been reduced.
Also
identify the reducing agent and the oxidizing agent. You could even write half-equations for each
reaction!!!
(a) Fe + S

FeS
(b) MnO
2
+ 4HCl

MnCl
2
+ Cl
2
+ 2H
2
O - a little tricky
(c) 2FeCl
2
+ Cl
2

2FeCl
3
(d) CuO + H
2
SO
4

CuSO
4
+ H
2
O
(e) PbO
2
+ 4HCl

PbCl
2
+ Cl
2
+ 2H
2
O
(f) 4Fe(OH)
2
+ O
2

2 H
2
O + 4Fe(OH)
3
(g) Zn + 2V 3+ 
Zn 2+ + 2V 2+
(h) 2Ca + O
2

2CaO
(i) CuO + H
2

Cu + H
2
O
(j) 2 AgNO
3
+ Cu

2 Ag + Cu(NO
3
)
2
IB chemistry topic 9 7 hours 5 | P a g e

Example:
Step 1: Write the unbalanced equation for the reaction in ionic form.
Fe 2+ + Cr
2
O
7
2- 
Fe 3+ + Cr 3+
Step 2: Separate the equation into two half-equations:
oxidation (loss of electrons) : Fe 2+ 
Fe 3+
reduction (gain of electrons): Cr
2
O
7
2- 
Cr 3+
Step 3: Balance the atoms other than O and H in each half-equation separately.
oxidation (loss of electrons) : Fe 2+ 
Fe 3+ (was already balanced)
reduction (gain of electrons): Cr
2
O
7
2-  2
Cr 3+
Step 4: For reactions in an acid medium, add H
2
O to balance the O atoms and H
+
to balance the H atoms that have been added after the addition of water.
Cr
2
O
7
2- 
Cr 3+ +
7H
2
O
(as there are 7 O atoms on the left side)
14H +
+ Cr
2
O
7
2- 
2Cr 3+ + 7H
2
O (to balance the 14 H atoms on the right side)
Step 5: Add electrons to one side of each half-reaction to balance the charges. If necessary, equalize the number of electrons in the two half-equations by multiplying one or both half-reactions by appropriate coefficients.
Fe 2+

Fe 3+ + e -
14H +
+ Cr
2
O
7
2- + 6e - 
2Cr 3+ + 7H
2
O
(the number of electrons makes sense
as Cr oxidation number changes from +6 to +3
per atom- there is 2)
To equalize the number of electrons in each half-equations, the oxidation equation needs to be
multiplied by 6.
6Fe 2+ 
6Fe 3+ + 6e -
Step 6: Add the two half-equations together and balance the final equation by inspection. The electrons on both sides must balance.
The two half-equations are added to give
IB chemistry topic 9 7 hours 6 | P a g e

14H +
+ Cr
2
O
7
2- + 6Fe 2+ + 6e -

2Cr 3+ + 7H
2
O + 6Fe 3+ + 6e -
The electrons on both sides cancel, and we are left with a balanced net ionic equation:
Step 7: Check that the equation has the same types and numbers of atoms and the same charges on both sides of the equation.
A final check shows that the resulting equation is ‘atomically’ and ‘electrically’ balanced.
Exercise
Complete and balance the following equations (all in acidic solutions): In each redox reaction, identify the reducing agent and the oxidising agent and state how the oxidation numbers change for each of these reagents. a. Cu (s) + NO
3
(aq)

Cu 2+ (aq) + NO
2
(g) b. Cr
2
O
7
2 (aq) + CH
3
OH (aq)

HCOOH (aq) + Cr 3+ (aq) c. MnO
4
(aq) + Cl (aq)

Mn 2+ (aq) + Cl
2
(g) d. Cr
2
O
7
2 (aq) + NO
2
(aq)

NO
3
(aq) + Cr 3+ (aq) e. I - + NO
3

IO
3
+ NO
2 f. H
2
O
2
(aq) + Fe 2+ (aq)

H
2
O (l) + Fe 3+ (aq) g. Cr
2
O
7
2 (aq) + C
2
O
4
2 (aq)

CO
2
(g) + Cr 3+ (aq) h. MnO
4
(aq) + Fe 2+ (aq)

Mn 2+ (aq) + Fe 3+ (aq) i. MnO
4
(aq) + SO
3
2 (aq)

Mn 2+ (aq) + SO
4
2 (aq) j. H
2
O
2
+ I + H + (aq)  H
2
O + I
2
9.3.1 Deduce a reactivity series based on the chemical behaviour of a group of oxidizing and reducing agents.
9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series.
When metals react they lose electrons or are oxidized, when non-metals react they gain electrons or are reduced. Therefore the reactivity of a metal or non-metal is about how easily it is oxidized or reduced or how strong a reducing or oxidizing agent it is.
The strength of an oxidising or reducing agent can be found by using it in displacement reactions with other oxidising or reducing agents e.g. like what you have done with the halogens or metals in group 1 in topic 3 on periodicity.
Displacement reactions involving elements, metals or non-metals, are called single replacement reactions and these are all redox reactions as the reactant element is either oxidised (if it is a metal) or reduced (if it is a halogen).
Metals are reducing agents: displacing other metal ions
IB chemistry topic 9 7 hours 7 | P a g e

Metal displacement reactions involve placing one metal in a salt solution (e.g. a sulphate or nitrate) of another metal.
The metal which displaces the metal in the compound is the best reducing agent as it reduces the metal ion in the compound: it gives its electrons to the metal ion in the compound forcing it to come out of the solution as an atom. The reducing agent itself becomes an ion.
Example: CuCl
2
(aq) + Mg (s)

MgCl
2
(aq) + Cu (s)
This reaction is feasible as magnesium is the strongest reducing agent and reduces/displaces copper ions,
Cu 2+ , according to the following half-equations; the magnesium itself becomes a positive ion:
Half equation:

Cu 2+ (aq) + 2e 
Cu (s) = reduction

Mg (s)

Mg 2+ + 2e (aq) = oxidation
We state that magnesium is more reactive than copper; it loses its electrons more readily, it oxidizes more readily.
In this example, the magnesium is the reducing agent and copper (II) chloride is the oxidizing agent.
However, magnesium is a weaker reducing agent than sodium and can therefore never cause the reduction of sodium ions so the following reaction is impossible:
2NaCl (aq) + Mg (s)

MgCl
2
(aq) + 2Na (s)
The above reaction is not feasible because magnesium is not a strong enough reducing agent to cause the reduction of sodium ions.
However, if sodium atoms are placed in contact with magnesium ions than the following reaction will occur as sodium is the strongest reducing agent – this reaction now is feasible :
MgCl
2
(aq) + 2Na (s)

2NaCl (aq) + Mg (s)
By carrying out a series of displacement reactions, which follow the pattern shown below, between metal atoms and metal ions, a series of reactivity can be deduced with the strongest reducing agent at the top as the most reactive metal.
XCl (aq) + Y (s)

NaY (aq) + X
If the reaction above is feasible than Y is more reactive than X as it is a better reducing agent. It loses its electrons more readily.
You are already familiar with some parts of this series which is called the reactivity series:
K Na Li Ca Mg Al Zn Fe H Cu Ag Au strongest reducing agent; weakest reducing agent its atoms will displace any ions its ions will be displaced by any atom from elements more to the right from elements more to the left
K(s)

K + (aq) + e Au + (aq) + e 
Au (s)
IB chemistry topic 9 7 hours 8 | P a g e

As the potassium atom, K, is a strong reducing agent, its ion, K + , is therefore a very weak oxidising agent. Potassium atoms will displace/reduce ions of less reactive metals.
Also, in the series above Au + (gold ion) is the strongest oxidising agent as Au (gold atom) is the weakest reducing agent.
Non-metals
As non-metals usually gain electrons during reactions they are usually oxidising agents. Just like with metals, single displacement reactions can also be used to place non-metals into a reactivity series – those non-metals with the greatest tendency to accept electrons are the most reactive so the most reactive will be the best oxidising agent as shown by the equation below:
When fluorine is added to a solution of potassium chloride, chloride ions are displaced by fluorine atoms.
2KCl (aq) + F
2
(g)

2KF (aq) + Cl
2
(g)
reducing agent oxidising agent
In the above example, the following half equations can be deduced:
oxidation: 2Cl (aq)

Cl
2
(g) + 2e -
reduction: F
2
(g) + 2e -

2F (aq)
redox: 2Cl (aq) + F
2
(g)

Cl
2
(g) + 2F (aq)
From experimental evidence from displacement reactions involving halogens the following series, which you again are familiar with, has been deduced:
F
2
Cl
2
Br
2
I
2
At
2
Fluorine atoms strongest oxidizing agent
Fluoride ions, F ,
Astatine atoms weakest oxidizing agent
Astatide ions, At , weakest reducing agents strongest reducing agent
The oxidizing agents to the left in the series gain electrons from ions from the halogens more to the right.
Describe the experiments you would do to confirm experimentally which halogen is the strongest oxidising agent. Describe observations for each tests. Write equations for those combinations in which reactions take place.
IB chemistry topic 9 7 hours 9 | P a g e

All common reducing and oxidizing reagents are in the table below which shows part of the electrochemical series.
Lithium
Rubidium
Potassium
Sodium
Strongest reducing agent, greatest tendency to lose electrons and form positive ions in solutions.
Calcium
Magnesium
Aluminium
Zinc
Iron
Lead
HYDROGEN
Copper
Silver
Iodine
Bromine
Chlorine
Elements above hydrogen should displace the hydrogen ion from acids. Elements near the top can also displace hydrogen from water.
Fluorine Strongest oxidizing agent. Greatest tendency to gain electrons and to form negative ions in solution.
Prediction of feasibility of redox reactions
These reactivity series can also be used to predict the feasibility of redox reactions:

Atoms of weaker reducing agents can never displace ions from stronger reducing agents.

Atoms of weaker oxidising agents can never displace ions from stronger oxidizing agents.
Exercises on predicting the feasibility of the following reactions.
1. Decide which ones of the following reactions are feasible; if the reaction is feasible, write the half-
equations and complete the equation.
1. Cu (s) + HCl (aq)

2. I
2
(s) + NaBr (aq)

2. Mg (s) + CuSO
4
(aq)

3. Cl
2
(g) + KBr (aq)

4. At
2
(s) + MgCl
2
(aq)

IB chemistry topic 9 7 hours 10 |
P a g e

6. Cu (s) + Fe 2+ (aq)

7. Cl
2
(aq) + 2I - (aq)

8. Mg (s) + Fe 2+ (aq)

9. Cu (s) + Ag + (aq)

10. F
2
(aq) + 2 Cl - (aq)

2. For each of the combination of species below:
(i) write half equations and ionic equations for all possible reactions
(ii) list the species which do not react and state why a b
K (s) , Mg (s), Al (s), Cu (s), Ag (s) K + (aq), Mg
F
2
(aq) , Cl
2
(aq), Br
2
(aq), I
2
(aq), At
2
(aq) F -
2+ (aq), Al 3+ (aq), Cu 2+ (aq), Ag + (aq),
(aq), Cl 2 (aq), Br -(aq), I (aq), At (aq),
9.4.1 Explain how a redox reaction is used to produce electricity in a voltaic cell.
9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode).
Spontaneous redox reactions can be used to produce an electric current by using difference in reactivity or oxidising ability between metals.
A voltaic or electrochemical cell is made by connecting two half-cells using an external circuit and a salt bridge as shown below. The external circuit allows electrons to be transferred from one half cell to the other.
A half cell is a piece of a metal immersed in an aqueous salt solution of that metal, usually a nitrate or a sulphate.
A salt bridge is a piece of filter paper soaked in a salt solution, usually potassium nitrate, and it allows the flow of ions between both half cells; this is necessary to ensure each half cell remains electrically neutral.
(
Image from http://www.saskschools.ca/curr_content/chem30_05/6_redox/redox2_2.htm
)
IB chemistry topic 9 7 hours 11 |
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In the half cell of the more reactive metal (better reducing agent), the metal atoms oxidise releasing electrons. This makes this half cell the negative electrode as a current always flows from the negative to positive electrode. Because oxidation occurs it is also called the anode.
In the electrochemical cell above, zinc is more reactive than copper. As a result oxidation takes place in the zinc half cell and it becomes the negative electrode/anode.
In the half cell of the less reactive metal, the metal ions gain the electrons lost by the oxidation in the other half cell and as a result these ions are reduced. This makes this half cell the positive electrode. Because reduction occurs it is also called the cathode.
In the electrochemical cell above, reduction occurs in the copper half cell and as a result it becomes the positive electrode/cathode.
Equations of reactions taking place in the zinc-copper voltaic cell:
At the cathode Zn (s)

Zn 2+ (aq) + 2e - oxidation
At the anode Cu 2+ (aq) + 2e - 
Cu (s) reduction
Overall Zn (s) + Cu 2+ (aq)

Zn 2+ (aq) + Cu (s) redox
See animation on http://www.saskschools.ca/curr_content/chem30_05/6_redox/redox2_2.htm
Reactivity difference of metals
The greater the difference in reactivity between the metals, the greater the voltage produced.
The redox reaction in an electrochemical cell is a spontaneous reaction.
Other examples of voltaic cells
For each of the voltaic cells, deduce the half equations at each electrode and identify the reducing and oxidising agent.
(a) iron and magnesium
(b) iron and copper
(c) Pb (s) + PbO
2
(s) + 2H
2
SO
4
(aq)

2PbSO
4
(aq) + 2H
2
O (l)
9.5.1 Describe, using a diagram, the essential components of an electrolytic cell.
9.5.2. State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode
(cathode).
9.5.3. Describe how current is conducted in an electrolytic cell.
9.5.4. Deduce the products of the electrolysis of a molten salt.
Electrolysis reaction is a non-spontaneous redox reaction (endothermic) which is made to happen using electrical energy from an outside source.
To be able to carry out an electrolysis reaction an electrolytic cell is needed. An electrolytic cell is a piece of apparatus which converts chemical energy into electrical energy (opposite to an electrochemical reaction). It needs an external source of electrical energy, two electrodes and an electrolyte.
IB chemistry topic 9 7 hours 12 |
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An electrolyte is a liquid containing ions which allows a current to be conducted through it.
During an electrolysis reaction…..
At the cathode/negative electrode
Reduction of positive metal ions to atoms using the electrons supplied by the external source
At the anode/positive electrode
Oxidation of negative ions releases electrons which return to the external source
(from http://www.ausetute.com.au/elecysis.html
)
The diagram below shows an electrolytic cell used to electrolyse molten sodium chloride.
2Cl - (l)

Cl
2
(g)+ 2e -
2Na + (l) + 2e -

2Na (s)
During electrolysis, the current in the electrolytic cell is carried by:

the movement of ions in the electrolyte

by electrons in the wires and electrodes
The flow of electrons is achieved in the following way:
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Electrons from the negative terminal of the power pack accumulate in the negative electrode
(=cathode). The negatively charged field attracts the positive salt ions (cations) which accept the free electrons available at the cathode. This removal keeps the flow of electrons going in that part of the circuit. At the cathode a reduction of positive metal ions occurs.
At the positive electrode (=anode) there is a shortage of electrons creating a positively charged field which attracts the anions in the solution. These negative ions move towards the anode and become oxidised; they release their electrons to the anode. The released electrons are attracted to the source/power pack, maintaining the flow of electrons in that part of the circuit
External source of electric current positive terminal; source of electric current negative terminal; source of electric current electrons anode – oxidation of negative ions electrons are added to circuit electrons cathode – reduction of positive ions; electrons are removed from circuit negative ions at anode, negative ions are attracted and lose their electrons – the negative ions are oxidised positive ions at the cathode the positive ions are attracted and gain electrons from the cathode – they become reduced
Products of electrolysis of molten salts
General half-equations at each electrode:
 reduction of metal ion at cathode: M + + e -

M
 oxidation of non-metal ion at anode: X - 
X + e -
Examples of half equations at electrodes in different electrolytic cells
molten KCl reaction at cathode
2K + + 2e -

2K reaction at anode 2Cl -

Cl
2
+ 2e -
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molten MgI
2 molten Al
2
O
3 overall equation - redox reaction at cathode reaction at anode overall equation – redox reaction at cathode reaction at anode overall equation - redox
2K + + 2Cl - 
Cl
2
+ 2K
Mg 2+ + 2e - 
Mg
2I - 
I
2
+ 2e -
Mg 2+ + 2I - 
I
2
+ Mg
2Al 3+ + 6e - 
2Al
3O 2-  1½ O
2
+ 6e -
2Al 3+ + 3O 2- 
O
2
+ 2Al
Overall pattern: metal at cathode and non-metal at anode!!!!!
Write both half-equations and overall equation for the electrolysis of the following salts: a) molten sodium fluoride b) molten calcium oxide c) molten copper chloride
Electrochemical reactions V electrolysis
Electrochemical reactions are spontaneous redox reactions carried out in voltaic cells. They take place whenever a more reactive metal is placed in contact by an electrical wire with a less reactive metal dipped in a solution of one of its salts.
Electrolysis is the decomposition of compounds, which can ionise or dissociate using electricity, as electrolysis reactions are non-spontaneous.
energy conversions spontaneity apparatus reaction at negative electrode reaction at positive electrode
1. (M07)
converts chemical energy into electrical energy spontaneous redox reaction voltaic cells: two separate solutions and a salt bridge oxidation reduction
converts electrical energy into chemical energy non-spontaneous redox reaction electrolytic cell + power supply; one solution and no salt bridge reduction oxidation
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2. (N07)
3. (N07)
4. (N07)
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5. (N06) Which are examples of reduction?
I. Fe 3+ becomes Fe 2+
II. Cl becomes Cl
2
III. CrO
3
becomes Cr 3+
A. I and II only B. I and III only C. II and III only D. I, II and III
6. (N06) Which statement is correct for the electrolysis of a molten salt?
A. Positive ions move toward the positive electrode.
B. A gas is produced at the negative electrode.
C. Only electrons move in the electrolyte.
D. Both positive and negative ions move toward electrodes.
7. (N06) Which statement about the following reaction is correct?
2 Br (aq) + Cl
2
(aq)

Br
2
(aq) + 2 Cl - (aq)
A. Br (aq) is reduced and gains electrons. B. Cl
2
(aq) is reduced and loses electrons.
C. B r (aq) is oxidized and loses electrons. D. Cl
2
(aq) is oxidized and gains electrons.
8. (M05) What happens when molten sodium chloride is electrolysed in an electrolytic cell?
A. Chlorine is produced at the positive electrode.
B. Sodium ions lose electrons at the negative electrode.
C. Electrons flow through the liquid from the negative electrode to the positive electrode.
D. Oxidation occurs at the negative electrode and reduction at the positive electrode.
9. (N05) Which equations represent reactions that occur at room temperature?
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A. I and II only B. I and III only C. II and III only D. I, II and III
10. (N05) Which equation represents a redox reaction?
11. (N05) The following information is given about reactions involving the metals X, Y and Z and solutions
of their sulfates.
When the metals are listed in decreasing order of reactivity (most reactive first), what is the correct
order?
A. Z > Y > X B. X > Y > Z C. Y > X > Z D. Y > Z > X
12. (N02) In the reaction
A. Br
2
is only oxidised.
B. Br
2
is only reduced.
C. Br
2
is neither oxidised nor reduced.
D. Br
2
is both oxidised and reduced.
13. (N02) Consider the following statements regarding electrolysis of molten lead(II) bromide.
I. Oxidation takes place at the anode where lead ions gain electrons.
II Reduction takes place at the cathode where lead ions gain electrons.
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III Oxidation takes place at the anode where bromide ions lose electrons.
IV. Reduction takes place at the cathode where bromide ions lose electrons.
Which of the above statements are correct?
A. I and II only B. I and IV only C. II and III only D. II and IV only
14. (M02) Which of the following changes represents a reduction reaction?
15. (M02) During the electrolysis of a molten salt, the cation moves toward the ..
I
.. and undergoes .
II
..
I II
A. negative electrode reduction
B. negative electrode oxidation
C. positive electrode oxidation
D. positive electrode reduction
16. (N01) What is the oxidation number of phosphorus in NaH
2
PO
4
?
A. +3 B. -3 C. + 5 D. –5
17. (N01) Which product is formed at the cathode (negative electrode) when molten MgCl
2
is electrolysed?
A. Mg 2+ B. Cl C. Mg D. Cl
2
18. (N00)
2AgNO
3
(aq) + Zn (s)

2Ag(s) + Zn(NO
3
)
2
(aq)
Zn(NO
3
)
2
(aq) + Co (s)

no reaction
2AgNO
3
(aq) + Co (s)

2Ag(s) + Co(NO
3
)
2
(aq)
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Using the above information, the order of increasing reactivity of the metals is
A. Ag

Zn

Co B. Co

Ag

Zn C. Co

Zn

Ag D. Ag

Co

Zn
19. (N00) The oxidation number of sulfur in HS
2
O
5
is
A. –1 B. +3 C. +4 D. +5
PAPER 2
1. (M08/TZ1)
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2. (N06)
(a) For each of the following reactions in aqueous solution, state one observation that would
be made, and deduce the equation.
The reaction between chlorine and sodium iodide. [2] (i)
(ii) The reaction between silver ions and chloride ions. [2]
(b) Deduce whether or not each of the reactions in (a) is a redox reaction, giving a reason in each case.
[4]
(c) (i) Draw a diagram of apparatus that could be used to electrolyse molten potassium bromide. Label
the diagram to show the polarity of each electrode and the product formed. [3]
(ii) Describe the two different ways in which electricity is conducted in the apparatus. [2]
(iii) Write an equation to show the formation of the product at each electrode. [2]
3. (M04) Consider the following redox equation.
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5Fe
2+
(aq) +MnO

4
(aq) +8H(aq) → 5Fe 3+
(aq) + Mn
2+
(aq) + 4H
2
O(l)
(a) (i) Determine the oxidation numbers for Fe and Mn in the reactants and in the products.
(2)
(ii) Based on your answer to (i), deduce which substance is oxidized.
(1)
(iii) The compounds CH
3
OH and CH
3
O contain carbon atoms with different oxidation numbers. Deduce the oxidation numbers and state the kind of chemical change needed to make CH
2
O from CH
3
OH.
(3)
(b) A part of the reactivity series of metals, in order of decreasing reactivity, is shown below. magnesium zinc iron lead copper silver
If a piece of copper metal were placed in separate solutions of silver nitrate and zinc nitrate
(i) determine which solution would undergo reaction.
(1)
(ii) identify the type of chemical change taking place in the copper and write the half-equation for this change.
(2)
(iii) state, giving a reason, what visible change would take place in the solutions.
(2)
(c) (i) Solid sodium chloride does not conduct electricity but molten sodium chloride does.
Explain this difference, and outline what happens in an electrolytic cell during the
electrolysis of molten sodium chloride using carbon electrodes.
(ii) State the products formed and give equations showing the reactions at each electrode.
(iii) State what practical use is made of this process.
(4)
(4)
(1)
4. For the following reaction 2Cu + 
Cu + Cu 2+
(a) state the oxidation number of each species. [1]
(b) write a balanced half-reaction for the oxidation process. [1]
(c) write a balanced half-reaction for the reduction process. [1]
Mark scheme
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PAPER 1
1
5
9
13
PAPER 2
1. (M08)
B
B
B
C
2
6
10
14
A
D
B
C
3
7
11
15
B
C
A
4
8
12
16
D
A
D
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2. (N06)
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3.
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4.
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