DP Chemistry
What is oxidation/reduction?
Oxidation/Reduction reactions, collectively known as redox reactions,
involve the transfer of electrons and always occur together
Oxidation is the loss of electrons
Reduction is the gain of electrons
An example:
CuO(aq) + Mg(s)  Cu + MgO(aq)
An easy way to
remember this is
the mnemonic:
‘’OIL RIG’’
Oxidation Is Loss
Reduction Is Gain
So if we look at the transfer of ein the ‘net ionic equation’
Cu has gained e-
As an ionic equation:
Cu2+ + O2- + Mg  Cu + Mg2+ + O2-
Cu2+ + Mg  Cu + Mg2+
Mg has lost e-
Agents of Redox
A substance that is oxidised
loses electrons to another
substance, so we call this
substance a reducing agent
A substance that is reduced,
gains electrons from another
substance, so we call this
substance a oxidising agent
Exercise
Considering the reaction between zinc and hydrochloric acid, write out:
1. A balanced equation
2. An ionic equation
3. A net ionic equation (excludes the spectator ions)
4. What are the oxidising and reducing agents
Exercise - Answers
Considering the reaction between zinc and hydrochloric acid, write out:
1. A balanced equation Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
2. An ionic equation Zn(s) + 2H+(aq) + 2Cl-(aq)  Zn2+(aq) + 2Cl-(aq) + H2(g)
3. A net ionic equation (excludes the spectator ions)
Zn(s) + 2H+(aq)  Zn2+(aq) + H2(g)
4. Two half reactions to identify the oxidation and reduction
Zn(s)  Zn2+(aq) + 2e- (loss of e-) - oxidation
2H+(aq) + 2e-  H2(g) (gain of e-) - reduction
5. What are the oxidising and reducing agents
r.a. – Zinc
o.a. - Hydrogen
How do we know it’s redox?
In every redox reaction there is an exchange of electrons, but since
these transfers are not shown in overall reaction equations, we need a
way to ‘see’ who is losing and who is gaining electrons.
Oxidation States (aka Numbers)
A set of rules applied to elements,
ions and compounds will allow us to
determine if we have a redox
reaction or not.
Oxidation – increase in number
Reduction – decrease in number
How do we know it’s redox?
Oxidation Number Rules
1.
2.
3.
4.
5.
6.
7.
The oxidation state of a free element (i.e. not part of a compound) is zero
(e.g. Zn(s), O2(g))
The oxidation state of an element in an ionic compound is equal the
electrical charge on its ion if monatomic. (e.g. Na+ = +1)
Oxidation states of elements in covalent compounds are calculated as if
they were ionic. The most electronegative atom (closest to F in the
periodic table) is assumed to gain electrons. (e.g. NH3; N = -3, H = +1)
The oxidation state of oxygen in a compound is normally -2, except for
peroxides, when it is -1. (e.g. H2O2; H = 1, O = -1)
The oxidation state of hydrogen in a compound is normally +1, except for
metal hydrides, when it is -1. (e.g. NaH; Na = 1, H = -1)
The sum of the oxidation states of all the elements in a neutral compound
= zero.
The sum of the oxidation numbers of all atoms in a polyatomic ion =
charge on the polyatomic ion.
Assigning Oxidation Numbers
When assigning oxidation numbers to the elements in a substance, take a
systematic approach. Ask yourself the following questions:
1. Is the substance elemental?
2. Is the substance ionic?
3. If the substance is ionic, are there any monoatomic ions present?
4. Which elements have specific rules?
5. Which element(s) do(es) not have rules?
Use rule 6 or 7 from above to calculate these.
Assigning Oxidation Numbers
Example - Na2SO4
1. Is the substance elemental? No, 3 elements are present
2. Is the substance ionic? Yes
3. If the substance is ionic, are there any monoatomic ions present? Yes,
Na+, so OS of Na = +1
4. Which elements have specific rules? Oxygen has a rule (#4) OS of O = -2
5. Which element(s) do(es) not have rules? Sulfur does not have a rule
Use rule 6 or 7 from above to calculate these.
2(Na) + 4(O) + (S) = 0 
(+2) + (-8) + S = 0 
S = +6
Assigning Oxidation Numbers
Example - K2C2O4
1. Is the substance elemental? No, 3 elements are present
2. Is the substance ionic? Yes , metal + non-metal
3. If the substance is ionic, are there any monoatomic ions present? Yes, K+,
so OS of K = +1
4. Which elements have specific rules? Oxygen has a rule (#4) OS of O = -2
5. Which element(s) do(es) not have rules? Carbon does not have a rule
Use rule 6 or 7 from above to calculate these.
2(K) + 4(O) + (C) = 0 
(+2) + (-8) + 2(C) = 0 
C = +3
Exercises
Determine the oxidation number of each element in the following
compounds:
1. Ba(NO3)2
2. NF3
3. (NH4)2SO4
Exercises Answers
1. Ba(NO3)2
Is the substance elemental?
No, three elements are present.
Is the substance ionic?
Yes, metal + non-metal.
Are there any monoatomic ions?
Yes, barium ion is monoatomic.
Barium ion = Ba2+
Oxidation # for Ba = +2
Which elements have specific rules?
Oxygen has a rule....-2 in most compounds
Oxidation # for O = -2
Which element does not have a specific rule?
N does not have a specific rule.
Use rule 6 to find the oxidation # of N
Let N = Oxidation # for nitrogen
(# Ba) (Oxid. # of Ba) + (# N) (Oxid. # N) + (# of O) (Oxid. # of O)
=0
1(+2) + 2(N) + 6(-2) = 0
N = +5
Exercises Answers
2. NF3
Is the substance elemental?
No, two elements are present.
Is the substance ionic?
No, two non-metals.
Are there any monoatomic ions?
Since it is molecular, there are no ions present.
Which elements have specific rules?
F = -1
Which element does not have a specific rule?
N does not have a specific rule.
Use rule 6 to find the oxidation # of N
Let N = oxidation # of N
(# N) (Oxid. # N) + (# F) (Oxid. # F) = 0
1(N) + 3(-1) = 0
N = +3
Exercises Answers
3. (NH4)2SO4
Is the substance elemental?
No, four elements are present.
Is the substance ionic?
Yes, even though there are no metals present, the ammonium ion is a common
polyatomic cation.
Are there any monoatomic ions?
No, the cation and anion are both polyatomic.
Which elements have specific rules?
H = +1 because it is attached to a non-metal (N)
O = -2
Which elements do not have a specific rule?
Neither N nor S has a specific rule.
You must break the compound into the individual ions that are present and
then use rule 7 to find the oxidation numbers of N and S. Notice that if you try
to use rule 6, you end up with one equation with two unknowns: 2N + 8(+1) +
1S + 4(-2) = 0
The two ions present are NH4+ and SO42-.
N + 4(+1) = +1 so N = -3
S + 4(-2) = -2 so S = +6
Naming compounds
Recall, that many elements have multiple valencies, so they are
able to form a variety of compounds.
We name these compounds using their oxidation numbers.
Fe2+
Iron has two common oxidation states: +2 and +3
This means that iron can form two different compounds
with oxygen which has an oxidation state = -2
Fe3+
FeO – we call this iron(II) oxide (say, “iron 2 oxide”)
Fe2O3 – we call this iron(III) oxide (say, “iron 3 oxide”)
Using oxidation numbers to
identify redox in a reaction
Not all reactions are redox reactions
To determine if a redox reaction has occurred, assign oxidation
numbers to each element on both sides of the reaction and if:
• Oxidation # increases – that element has been oxidised
• Oxidation # decreases – that element has been reduced
• Oxidation # does not change – no oxidation has occured
Example:
CuO + H2  Cu + H20
Cu: +2  0 (reduced)
O: -2  -2 (no change)
H: 0  +1 (oxidised)
Redox has occurred
You try:
2CrO42- + 2H+  Cr2O72- + H20
Cr: +6  +6
O: -2  -2
H: +1  +1
No changes means no redox
Redox half-reactions
An ionic equation can be written as two half equations that show each process.
Cu2+ + Mg  Cu + Mg2+
Oxidation Half-reaction
Reduction Half-reaction
Mg  Mg2+ + 2e-
Cu2+ + 2e-  Cu
Note: the number of atoms and charge must
balance in half-reactions just as in full reactions
More complex half-reactions
With simple monatomic ions, half-reactions are simple as in the previous
example. However, with polyatomic ions, the process is a bit more complex.
Some require acidic conditions to proceed
Consider the reduction of dichromate (VI) ion to chromium (III)
Cr2O72-  Cr3+
Action
Equation
Balance the # of Cr atoms
Cr2O72-  2Cr3+
Add H2O molecules to balance
oxygen atoms
Cr2O72-  2Cr3+ +7H2O
Add H+ ions to balance hydrogen
atoms
Cr2O72- + 14H+  2Cr3+ +7H2O
Add e- so electrical charges
balance
Cr2O72- + 14H+ + 6e- 2Cr3+ +7H2O
More complex half-reactions
Now you try…
Consider the oxidation of sulfur dioxide to sulfate
SO2  SO42-
Action
Equation
More complex half-reactions
Answer
Consider the oxidation of sulfur dioxide to sulfate
SO2  SO42-
Action
Equation
Balance S atoms (done)
SO2  SO42-
Balance O atoms w/ water
SO2 + 2H2O  SO42-
Balance H atoms w/ H+ ions
SO2 + 2H2O  SO42- + 4H+
Add e- to balance charge
SO2 + 2H2O  SO42- + 4H+ + 2e-
Adding redox half-reactions
Consider the two half reactions from the previous example in acidic solution:
SO2(aq) + 2H2O(l)  SO42-(aq) + 4H+(aq) + 2e- (oxidation)
Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq) +7H2O(l) (reduction)
We can add the two half reactions to get the net ionic reaction, but notice that
adding these two half reactions results in an imbalance in the number of
electrons. In this case, we must multiply the first by 3 to get the electrons to
cancel. In addition some of the waters and H ions also cancel.
3SO2(aq) + Cr2O72-(aq) + 2H+(aq)  3SO42-(aq) + 2Cr3+(aq) +H2O(l)
What is the oxidising agent? Reducing agent?
Adding redox half-reactions
Consider the two half reactions from the previous example in acidic solution:
SO2(aq) + 2H2O(l)  SO42-(aq) + 4H+(aq) + 2e- (oxidation)
Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq) +7H2O(l) (reduction)
We can add the two half reactions to get the net ionic reaction, but notice that
adding these two half reactions results in an imbalance in the number of
electrons. In this case, we must multiply the first by 3 to get the electrons to
cancel. In addition some of the waters and H ions also cancel.
3SO2(aq) + Cr2O72-(aq) + 2H+(aq)  3SO42-(aq) + 2Cr3+(aq) +H2O(l)
What is the oxidising agent? Reducing agent? Cr2O72-, SO2
Reactions of metals
Reactions with oxygen (combustion)
All metals form oxides except Ag, Au and Pt
heat
Metal + oxygen  metal oxide
e.g. 2Mg + O2  2MgO
Tendency to form metal oxides:
• Li, Na, K, Ca, Ba (react at room temp)
• Mg, Al, Fe, Zn (react slowly at room temp, vigorously when heated)
• Sn, Pb, Cu (react slowly and only when heated)
Reactions of metals
Reactions with water
Reactive metals react with water or steam
Metal + water  metal hydroxide + hydrogen gas
e.g. Na + 2H2O  2NaOH + H2
Metal + steam  metal oxide + hydrogen gas
e.g. Zn + H2O  ZnO + H2
Relative reactivity:
• Li, Na, K, Ca, Ba (react with water at room temp)
• Mg, Al, Zn, Fe (react with steam at high temp)
• Sn, Pb, Cu, Ag, Au, Pt (do not react)
Reactions of metals
Reactions with dilute acid
More metals react with acid than water
Metal + acid  salt + hydrogen gas
Zn + 2HCl  ZnCl2 + H2
Relative reactivity:
• Li, Na, K, Ca, Mg, Al, Zn, Fe, Co, Ni (react readily)
• Sn, Pb (slow to react without heat)
• Ag, Hg, Pt, Au (do not react)
Reactivity Series
Based on the ease of reactions with oxygen, water
and acids, metals can be organised in order of
reactivity, known as an activity series.
Activity series for metals:
K>Na>Ba>Ca>Mg>Al>Zn>Fe>Sn>Pb>Cu>Ag>Hg>Pt>Au
most reactive
least reactive
Grp 1>Grp 2> Grp 3>some TM (Zn, Fe)>Grp 4>more TM
N.B. TM = transition metals
Metal Displacement
The reactivity series of metals gives us a guide
as to how readily reactions will take place
among metals
For example, Mg is more
reactive than Cu, so Mg will
displace Cu from solution. Cu
will precipitate and Mg will
dissolve into solution
www.bbc.co.uk
Voltaic Cell – metal equilibrium
Recall the half-equations that we used
to show oxidation and reduction
reactions.
Consider a solid metal in a solution of
it’s component ions (see diagram).
This shows there is an equilibrium
between the metal and its ion.
If we connect these two metals together, something interesting happens….
Voltaic cell- Danielle Cell
Because there is a difference in reactivity between these two metals, ewill flow from the more reactive (anode) where oxidation occurs, to the
less reactive (cathode) where reduction occurs through an external wire.
Voltmeter
A salt bridge filled
with an electrolyte
such as KCl allows
the ions in solution
to flow, completing
the circuit.
Recall that Zn was found to
be more reactive than Cu.
This means Zn is more
likely to be oxidised.
So if we connect these two
metals together, a
spontaneous reaction
occurs and we can get
electrons to flow in a closed
circuit called a voltaic cell
(aka Galvanic Cell)
+
-
The cathode is
the site of
reduction and is
positive and
attracts
electrons
The anode is
the site of
oxidation and is
negative due to
the production
of electrons
Oxidation: e- loss
Cu2+(aq)
Another mnemonic:
Overall reaction
+ Zn(s)  Cu(s) + Zn2+(aq)
An Ox
Red Cat
Reduction: e- gain
Quantifying the Reactivity Series (AHL)
The reactivity series of metals can be
estimated using the reactions previously
described.
However, a more precise method can
allow us to quantify the differences in
reactivity and can include more than just
metals.
If we connect a half-cell to the standard
hydrogen half-cell, there are 2 possibilities:
A hydrogen half-cell (left) is used
as a standard to which all other
substances are compared.
This cell consists of a platinum
electrode over which hydrogen gas
is bubbled under standard
conditions:
•
•
•
Standard H2 half-cell
1 mol dm-3 [H+]
298 K
1 atm (H2)
1.
2.
The substance is a stronger oxidising agent
than H+ (gets reduced) – see Cu above
The substance is a weaker oxidising agent
than H+ (gets oxidised) – see Zn below
Standard Reduction Potentials
Notice these reactions
are all written as
reductions , by
convention
With the standard hydrogen
half-cell as a reference, we
are able to measure standard
electrode potentials (E0)
measured in volts (V).
E0 values are either:
• Negative – if the
substance is a stronger
reducing agent or;
• Positive – if the substance
is a stronger oxidising
agent
Notice
hydrogen is
set to 0.00V.
All above are
+ and all
below are –
compared to
this
reference
cell
Cell Potential
Considering the example containing Cu and Zn half-cells, we can
measure the potential difference by placing a voltmeter across the
external wire connecting the two.
Voltmeter
(1.10V)
The voltmeter for this
pair of half-cells
“shows” that 1.10V is
produced.
Since e- flow from one
half-cell to the other, we
can measure the energy
difference between the
two.
Potential difference is
the measure of the
energy difference
between the two halfcells in volts (V) and is
also called the cell
potential or emf.
-
+
Cu2+(aq)
Overall reaction
+ Zn(s)  Cu(s) + Zn2+(aq)
(AHL) The potential difference can be
estimated using the Standard Reduction
Potentials Table. The substance higher on
the table (more negative E0) will be oxidised
(Zn), while the lower (Cu) is reduced
Predicting redox reactions
Oxidising agents get
stronger as we move
down the left side of the
Standard Reduction
Potentials Table
This means a substance
such as Cl2 will oxidise
any of the reductants
above it and itself be
reduced
Example:
Cl2 will oxidise both Pb(s)
or Fe(s) which are higher
in the table
This means for example:
Pb(s)  Pb2+ + 2eNotice the reaction has
been reversed for
oxidation
Reducing agents get
stronger as we move up
the right side of the
table
This means a
substance such as Zn
will reduce any oxidants
below it and itself be
oxidised. This means
the reaction will be
reversed (oxidation)
Example:
Zn(s) will reduce Cu2+
which is lower in the
table.
This means:
Zn(s)  Zn2+ + 2e- (ox)
Cu2+ + 2e-  Cu(s)(red)
Predicting redox reactions
Predicting a redox reaction using the table:
Will Cl2 react with Zn(s)?
If so, what is the cell potential?
Cl2 will oxidise Zn(s) which is higher in the table, so yes,
a reaction will occur.
This means:
Zn(s)  Zn2+ + 2e- (ox)
1/2Cl2 + e-  Cl- (red)
E0 = +0.76V (rvrsd)
E0 = +1.36V
Added together:
Zn(s) + Cl2  Zn2+ + 2ClE0cell = E0 (red) - E0 (ox)
E0cell = 1.36 – (-0.76)
E0cell = 2.12V (this is the predicted cell potential)
Note: spontaneous reactions always have a positive E0
value.
Exercises
Using the table to the right, predict if
the following substances will result in a
reaction. If there is a reaction, write
out the half-equations and determine
the cell potential.
1.
2.
3.
Cu2+ , Ba(s)
F2, Pb(s)
Na+ , Fe(s)
Exercises - Answers
Using the table to the right, predict if
the following substances will result in a
reaction. If there is a reaction, write
out the half-equations and determine
the cell potential.
1.
2.
3.
Cu2+ , Ba(s)
F2, Pb(s)
Na+ , Fe(s)
1.
Cu2+ + 2e-  Cu(s)
0.34V
Ba(s)  Ba2+ + 2e2.90V
Ecell = 2.90 + 0.34 = 3.24V
1/2F2 + e-  F2.87V
2+
Pb(s)  Pb + 2e0.13V
Ecell = 2.87 + 0.13 = 3.00V
No rxn. (Na+ is a weaker oxidising
agent than water (E0 = -0.83V) and
has a less positive E0 value than the
reducing agent Fe(s)
2.
3.
Exercise
In your notebook, draw the
voltaic cell to the right and:
1. Label the anode and
cathode including signs
2. Label the salt bridge
3. Indicate the direction of eflow in the cell
4. Write out the oxidation
and reduction half
reactions
5. Predict the cell potential
using the Standard
Reduction Potentials Table
Exercise - Answers
In your notebook, draw the
voltaic cell to the right and:
1. Label the anode and
cathode including signs
2. Label the salt bridge
3. Indicate the direction of eflow in the cell
4. Write out the oxidation
and reduction half
reactions
5. Predict the cell potential
using the Standard
Reduction Potentials Table
-
+
Oxidation
Cu(s)  Cu2+ + 2e(E0 = -0.34V)
Reduction
Ag+ + e-  Ag(s)
(E0 = +0.80V)
Ecell = Eox + Ered = -0.34V + 0.80V = +0.46V
Electrolytic Cells
Previously, we have seen that voltaic
cells generate electric current in an
external circuit by spontaneous redox
reactions.
There is a second type of
electrochemical cell where we input
electrical energy into a system to
cause a non-spontaneous reaction to
occur.
These reactions are called electrolysis
reactions. Some examples:
• Electrolysis of water
• Electrolysis of molten NaCl
• Electroplating
Voltaic cell:
Chemical energy  Electrical energy
Electrolytic cell:
Electrical energy  Chemical energy
Electrolytic Cell Components
The electrolytic cell is different to the voltaic cell
in operation and structure. The electrolytic cell
contains:
•
•
•
•
•
•
Power source
One container (instead of two)
Electrolyte to allow conduction of current in
the solution
Reduction at the Cathode (-)
Oxidation at the Anode (+)
Inert electrodes (e.g. carbon)
Note the polarities of the electrodes have changed from the voltaic cell and are determined by which terminal of
the power supply they are connected to. Reduction still occurs at the cathode and oxidation at the anode.
At the cathode, e- are supplied from the power source, so a reduction of the strongest oxidising agent in solution
occurs. (positive ions or water)
At the anode, e- are removed, so oxidation of the strongest reducing agent occurs here. (negative ions or water)
Electrolysis of molten NaCl
Molten NaCl consists of Na+ and Cl- ions that
migrate towards the electrodes of opposite charge
in an electrolytic cell completing the circuit
The following equation represents the breaking apart of NaCl(l):
2NaCl(l) → 2Na(l) + Cl2 (g)
The half-reactions involved in this process are:
reduction
oxidation
2e- → Na(s)
(l) → Cl2 (g) + 2e-
2Na+
Cl-
(l) +
E°
-2.71 V
-1.36V
________________________________________
net voltage required
- 4.07V
The negative voltage (-4.07V) that results when we add up the half-reactions indicates
that the overall reaction will not be spontaneous. An EMF of more than 4.07 volts will be
required for this reaction to occur.
Electrolysis of Water (AHL)
This is one of the most common
demonstrations of a simple electrolysis.
It involves the oxidation and reduction of
water. Overall:
2 H2O(l)  2 H2(g) + O2(g)
The half reactions are:
2H2O + 2e-  H2 + 2OH-
Eored = -0.83 V
2H2O  O2 + 4H+ + 4e-
Eoox = -1.23 V
Note: in any aqueous
solution, these two reactions
are possible during
electrolysis
Electrolysis of Aqueous NaCl –
competing reactions (AHL)
At the anode the possible reactions are:
Cl-(l) → Cl2 (g) + 2 eE0oxid = -1.36V
2H2O(l) → O2 (g) +4H+(aq) + 4eE0oxid = -1.23V
At the cathode the possible reactions are:
2H2O(l) + 2e-→ +H2(g) + 2OH- (aq)
E0red = - 0.83V
Na+(l) + e- → Na(s)
E0red = - 2.71V
How to predict?
Oxidation: most - E0 on the reduction potential table (E0oxid most +)
Reduction: most + E0 on the reduction potential table (E0red most +)
So, which two reactions are predicted?
See further examples in Derry, “Chemistry Higher Level”, pp 257-262
Electrolysis of a concentrated salt solution (brine)
Source: www.answers.com/topic/electrolysis
These are the products unless the
solution is highly concentrated. In this
case, we will start to see the evolution
of chlorine gas as in the diagram.
Electroplating (AHL)
Electroplating is an application of
electrolysis where a thin layer of a metal
such as silver or nickel is plated over
another metal such as iron or copper to
provide a bright shiny finish to the
object. Some examples: food cans (Sn),
tools (Ni, Cr), galvanising (Zn).
Silver plating
Objects such as flatware are sometimes electroplated
using silver. The reactions are:
Ag(s)  Ag+(aq) + eAg+(aq) + e-  Ag(s)
Identify the anode and cathode in this example.
Factors affecting amount of
products formed (AHL)
1. Charge on the ion –
2. Current –
3. Duration of Electrolysis –
How do you think these factors will affect the products formed in an
electrolysis reaction?
Factors affecting amount of
products formed (AHL)
1. Charge on the ion – assuming the same amount of applied current
a M+ ion will produce 2x the amount of atoms as M2+ which requires
twice the number of e-.
2. Current – if the current in an electrolytic cell increases, then the
amount of electrons increases and so does the amount of product
3. Duration of Electrolysis – assuming there are enough reactants, the
longer current is applied, the more products will be produced
Note: Each e- carries a charge of 1.602 x 10-19 C (coulombs)
Inert vs Active electrodes (AHL)
Inert
Inert electrodes (C and Pt are the most common) do not take part in
the reactions at the surface of the electrode.
They simply provide a surface for the electron transfer reactions to
take place.
Active
Active electrodes are ones that may take part in a reaction at their
surface. Fe and Ag are more reactive substances and are more
likely to be oxidised than water. See values below.
Ag(s)  Ag2+(aq) + 2eFe(s)  Fe2+(aq) + 2e2H2O(l)  O2(g) + 4H+(aq) + 4e-
E0 = -0.80V
E0 = +0.44V
E0 = -1.23V
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