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20-Jul-12
Chemsheets A2 045
1
REVISION – OXIDATION STATES

When using oxidation states, we effectively imagine everything to be an ion – the oxidation state is the charge it would
have if it was an ion.

Assigning oxidation states:
1) On simple ions, the oxidation state is the charge on the ion
2+
e.g. Cu (Cu +2);
Cl (Cl –1);
Al2O3 (Al +3, O -2)
2) In elements, the oxidation state is always zero
e.g. Cl2 (Cl 0)
3) The total of all the oxidation states must always equal the overall charge on the species.
4) In molecules and more complex ions, the more electronegative element is assumed to be the negative ion
5) H is nearly always +1 and oxygen –2
e.g. CH4 (C -4, H +1);
CO2 (C +4, O -2);
H2O (H +1, O -2)
! TASK 1 – OXIDATION STATES
1)
2)
Calculate the oxidation state of the stated element in the following species:
a) Fe in Fe ………
FeCl3 ………
b) Cl in Cl2 ………
ClO3 ………
-
ClO ………
K2FeO4 ………
Cl2O7 ………
[Fe(H2O)6]
2+
………
Cl2O3 ………
Calculate the oxidation state of each element in the following:
SO2
S
S
S
O
H2O2
H
O
NaH
Na
H
C
O
2-
SO3
S
O
CO3
H2S
H
S
Cr2O3
Cr
O
NH3
N
H
CrO3
Cr
O
NO2
N
O
KMnO4
K
Mn
O
N
K2MnO4
K
Mn
O
N
Cu2O
Cu
O
CuO
Cu
O
NaCuCl2
Na
Cu
N2
NO3
Cl
-
-
SO4
3)
-
FeCl2 ………
O
Cl
2-
S
O
Cl
Give the oxidation state of the transition metal in each of the following species.
a) [Co(H2O)6]
b) [CrCl6]
2+
3-
c) [Co(NH3)6]Cl2
Co ………
e) [FeO4]
2-
Cr ………
f) [Mn(CN)6]
Co ………
g) [Ni(CO)4]
d) [Co(NH3)5Cl]Cl2 Co ………
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4-
h) [Ni(EDTA)]
2-
Fe ………
i)
K2[CoCl4]
Co ………
Mn ………
j)
K3[AuF6]
Au ………
Ni ………
k) (NH4)2[IrCl6] Ir ………
Ni ………
l)
20-Jul-12
Na[Mn(CO)5] Mn ………
Chemsheets A2 045
2
REVISION – REDOX EQUATIONS
Writing half equations
1)
Work out oxidation states
2)
Balance the element which changes oxidation state
3)
Balance the electrons
4)
Balance O’s with H2O
5)
Balance H’s with H
6)
Check half equation by checking total charge on both sides of the equation is the same
+
e.g. Write a half equation for the conversion of Fe
2+
 Fe
3+
e.g. Write a half equation for the conversion of MnO4  Mn
-
2+
! TASK 2 – HALF EQUATIONS
1)
Write half equations for each of the following conversions.
a) I  I2
-
b) Fe
..........................................................................................................................................................
 Fe
2+
c) Fe  Fe
3+
..........................................................................................................................................................
2+
..........................................................................................................................................................
d) Cl  Cl2
-
..........................................................................................................................................................
 SO2
..........................................................................................................................................................
f) MnO4  Mn
..........................................................................................................................................................
e) SO4
2-
-
g) SO4
2-
2+
 H2S
..........................................................................................................................................................
h) H2O2  O2
3+
i)
Cr
j)
C2O4
k) Hg2
l)
 Cr2O7
2-
2+
2-
 CO2
 Hg
lO3  l2
-
..........................................................................................................................................................
..........................................................................................................................................................
..........................................................................................................................................................
..........................................................................................................................................................
..........................................................................................................................................................
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Chemsheets A2 045
3
Combining half equations
Add the half equations together by eliminating the electrons.
e.g. Write an overall equation for the oxidation of Fe
2+
to Fe
3+
-
+
by MnO4 /H
! TASK 3 – REDOX EQUATIONS
1)
Write redox equations for each of the following reactions, using your equations from TASK 2.
+
a) H /MnO4 + Cl
-
..........................................................................................................................................................
..........................................................................................................................................................
..........................................................................................................................................................
+
2-
b) H /MnO4 + C2O4 ..........................................................................................................................................................
..........................................................................................................................................................
..........................................................................................................................................................
+
c) H /SO4
2-
-
+ I
..........................................................................................................................................................
..........................................................................................................................................................
..........................................................................................................................................................
+
d) H /MnO4 + H2O2 ..........................................................................................................................................................
..........................................................................................................................................................
..........................................................................................................................................................
-
+
-
e) I + H /IO3
..........................................................................................................................................................
..........................................................................................................................................................
..........................................................................................................................................................
f) Fe
2+
+
+ H /Cr2O7
2-
..........................................................................................................................................................
..........................................................................................................................................................
..........................................................................................................................................................
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Chemsheets A2 045
4
Redox reactions
O xidation
I s
L oss
of electrons
R eduction
I s
G ain

A redox reaction is one in which one or more elements change oxidation state.
CuSO4 + Zn → Cu + ZnSO4
e.g.
Cu
S
O
Zn

+2
0
+6
-2
+6
-2
+2
0
redox reaction
Cu reduced from +2 to 0; Zn oxidised from 0 to +2
A disproportionation reaction is one in which an element is both oxidised and reduced.
Cl2 + H2O → HCl + HOCl
e.g.
Cl
H
O
0
-1
+1
+1
-2
+1
Cl reduced from 0 to -1 and oxidised from 0 to +1
+1
-
redox reaction
(disproportionation)

An oxidising agent (oxidant) is the substance that does the oxidising (i.e. takes away the electrons from the substance
that is oxidised, and so the oxidising agent is itself reduced).

An reducing agent (reductant) is the substance that does the reducing (i.e. donates the electrons to the substance that
is reduced, and so the reducing agent is itself oxidised)
! TASK 4 – REDOX REACTIONS?
Redox
reaction?
Equation
Disproportionation
reaction?
Species
oxidised
Species
reduced
Oxidising
agent
Reducing
agent
Fe2O3 + 3 CO  2 Fe + 3 CO2
Mg + 2 HCl  MgCl2 + H2
MgO + 2 HCl  MgCl2 + H2O
Al2O3 + 2 Fe  2 Al + Fe2O3
[Co(H2O)6]
2+
+ 4Cl  [CoCl4] + 6H2O
-
2-
Na2O + H2O  2 NaOH
2 H2O2  2 H2O + O2
2 NaBr + H2SO4  Na2SO4 + 2 HBr
(NH4)2Cr2O7  Cr2O3 + N2 + 4 H2O
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Chemsheets A2 045
5
! TASK 5 – REDOX BASICS
1)
a)
Write balanced half-equations for each of the following oxidation or reduction processes.
i)
-
Br to Br2
iv)
H2SO4 to S
-
v)
NO3 to NH4
2+
vi)
BrO3 to Br2
ii) N2 to NO3
3+
iii) V
to VO
-
+
-
(6)
-
b) Write a redox equation for the reaction of H2SO4 with Br ions, producing S and Br2, using the half-equations
from (a). Also state which species is oxidised, which is reduced, and the oxidising and reducing agents.
c)
(3)
When concentrated nitric acid is added to copper metal, the copper is oxidised to oxidation state +2 and the
nitric acid is reduced to nitrogen (IV) oxide. Derive half-equations and then write an equation for the reaction.
(3)
(Total 12)
2)
State whether the following three reactions are redox reactions or not. For those that are redox reactions,
clearly indicate any changes in oxidation state.
a) Zn + 2 HCl  ZnCl2 + H2
b) CuO + 2 HCl  CuCl2 + H2O
c) MnO2 + 4 HCl  MnCl2 + Cl2 + 2 H2O
d) Cl2 + H2O  HCl + HOCl
(7)
(Total 7)
3)
a) Explain, in terms of electrons, what happens to oxidising and reducing agents in reactions.
(1)
b) Using these definitions, explain which species is oxidised and which is reduced, and the oxidising and
reducing agent in reaction below.
(2)
Mg + H2SO4  MgSO4 + H2
c) What is the oxidation state of each element in the following species:
Na2SO4
Na2S2O3
Na2SO3
S8
KClO3
KClO
KH
Na2O2
Na2O
(9)
(Total 12)
4)
a) For each of the following reduction or oxidation processes write a half equation.
i)
+
conversion of H to H2
2-
ii) conversion of SO4 to SO3
2-
iii) conversion of H2O2 to O2
-
iv) conversion of IO3 to I2
-
v) conversion of I to I2
b)
(5)
-
-
Use your half-equations to write a redox equation for the reaction of IO3 with I to form I2.
(1)
(Total 6)
5)
Write redox equations for the
following reactions using the halfequations provided.
-
-
-
3+
MnO4 + 8 H + 5 e  Mn + 4 H2O
2 Cl  Cl2 + 2 e
3+
2+
2 Cr + 7 H2O  Cr2O7 + 14 H + 6 e
-
+
-
2+
a) Reaction of acidified MnO4 and Cl .
(1)
b) Reaction of acidified MnO4 and Cr .
(1)
(Total 2)
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Chemsheets A2 045
6
ELECTRODE POTENTIALS
1) The potential of an electrode

n+
When a piece of metal (M) is dipped into a solution of its metal ions (M ), an equilibrium is set up. There is a tendency
for the metal to form positive ions and go into solution. However, there is also a tendency for the metal ions in solution
to gain electrons and form metal.
n+
M (aq) + n e
-
2+
e.g. Zn (aq) + 2 e
M(s)
-
Zn(s)

If this equilibrium lies to the left, then the metal acquires a negative charge due to a build up of electrons on the metal
(the electrode has a negative potential).

If the equilibrium lies to the right, then a positive charge builds up on the metal as electrons have been used up to form
metal from the metal ions (the electrode has a positive potential).

The position of the equilibrium (and so the charge on the metal) depends on the metal. For example, reactive metals
n+
tend to form M ions, so negative charge builds up on the metal. The more unreactive metals tend to have positive
charge on the metal.

A metal dipping into a solution of its ions is called a half-cell or an electrode.

There are other types of half-cell where there is no solid metal involved in the half-equation. For these half-cells, a metal
electrode is required and usually platinum is used as it is so unreactive (an inert electrode).

Here are three general types of electrode.
i)
Metal electrodes
These are the type met above, which consist of a metal surrounded by a solution of its ions,
2+
e.g. Zn(s) | Zn (aq).
ii) Gas electrodes
This is for a gas and a solution of its ions. Here an inert metal (usually platinum) is the actual
+
electrode, e.g. Pt(s) | H2(g) | H (aq)
iii) Redox electrodes
This is for two different ions of the same element (e.g. Fe and Fe ), where the two types of
ions are present in solution with an inert electrode (usually Pt).
2+
3+
e.g. Pt(s) | Fe (aq), Fe (aq)
2+
3+
2) Measuring the potential of an electrode using electrochemical cells

The actual potential (E) of a half-cell cannot be measured directly.

To measure it, it has to be connected to another half-cell of known potential, and the potential difference between the
two half-cells measured.

Combining two half-cells together produces an electrochemical cell.

Before the potential of any half-cells could be measured, a potential had to be assigned to one particular half-cell (then
the potential of all the other electrodes could be measured against it).
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Chemsheets A2 045
7

The electrode chosen was the standard hydrogen electrode (SHE) and this electrode is assigned the potential of 0
volts.

The SHE is known as the primary standard as it is the potential to which all others are compared.
Note 

When two half cells are joined together, a salt bridge is used to join them.

A salt bridge completes the circuit and allows ions to move from one half-cell to the other (but allows
the solutions to remain separate).

A salt bridge is often a tube containing a saturated solution of KCl or KNO3 in agar gel.
There are several factors which affect the potential of a half-cell, so they are measured under standard conditions.
-3
cell concentration
1.0 mol dm of the ions involved in the half-equation
cell temperature
298 K
cell pressure
100 kPa (only affects half-cells with gases)

The potential should also be measured under zero-current conditions [to measure the full potential difference (emf), no
current must be drawn from the cell - this is achieved by using a high resistance voltmeter.

Standard conditions are required because the position of the redox equilibrium will change with conditions.
example, in the equilibrium:
n+
M (aq) + n e
M(s)
an increase in the concentration of M

n+
For
would move the equilibrium to the right, so making the potential more positive.
A standard potential is written as E .
Standard hydrogen electrode (SHE) - the primary standard


The potential of all electrodes are measured by comparing their potential to
that of the standard hydrogen electrode.
+
(aq)
+ 2e
-
H2(g)
E
o
o
o
= +0.00 V
o
o
o
o
o
o
o
Pt(s) | H2(g) | H (aq)
o
Pt
1.0 M H+(aq)
versus the SHE
For any cell, its emf (E
cell)
is the potential of the right hand electrode minus the potential of the left hand electrode:
emf = E

o
temperature
= 298 K
+
The cell notation is:
Measuring E

salt bridge
This is therefore called the primary standard as it is the standard to which all
other potentials are compared.
2H

H2 at 100 kPa
cell=
E
right
- E
left
When finding the potential of a half-cell under test, the standard electrode is always the left hand electrode, which in the
case of the SHE gives:
E
cell
= E
test
- E
 E
SHE
cell
= E
V
2+
e.g. measuring E for the Cu /Cu half-cell
against the SHE.
E
cell
test
high resistance
voltmeter
2+
= E (Cu /Cu)
H2 at 100 kPa
salt bridge
= + 0.34 V
o
o
o
o
temperature
= 298 K
o
o
o
o
o
o
o
Pt
o
1.0 M H+(aq)
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20-Jul-12
Cu
1.0 M Cu2+(aq)
Chemsheets A2 045
8
E
cell
2+
= E (Cu /Cu) = + 0.340 V
+
E
2+
cell
2+
= E (Zn /Zn) = – 0.763 V
+
Pt(s) | H2(g) | H (aq) || Cu (aq) | Cu(s)
2+
Pt(s) | H2(g) | H (aq) || Zn (aq) | Zn(s)
Secondary standards

The SHE is difficult to use, so often a different standard is used which is easier to use.

These other standards are themselves calibrated against the SHE.

This is known as using a secondary standard - i.e. a standard electrode that has been calibrated against the primary
standard.

The common ones are:
silver / silver chloride
E
= +0.22 V
calomel electrode
E
= +0.27 V
3) Cell notation (cell diagrams)

Rather than drawing complicated diagrams of electrochemical cells, a shorthand form is written.
e.g.
in the cell on the page before last with Cu
two half-cells connected by a salt bridge):
2+
/ Cu as the right hand half-cell and SHE as the left hand half-cell (the
+
2+
Pt(s) | H2(g) | H (aq) || Cu (aq) | Cu(s)
where
|
||
represents a phase boundary (i.e. between species in different states)
represents a salt bridge

Note that the actual solid electrodes are written at the two ends.

Note that the most oxidised species are near the middle (ROOR – reduced, oxidised, oxidised, reduced).

Note the convention that the half-cell with the more positive potential is set up as the right hand electrode, and so cell
diagrams are drawn that way (except when measuring potentials against primary or secondary standards when the
standard electrode is always the left hand electrode).
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Chemsheets A2 045
9
! TASK 6 – CELL NOTATION
Write the cell notation for each of the following cells.
2+
Zn (aq) + 2 e2+
Cu (aq) + 2 e-
Zn(s)
3+
Al (aq) + 3 e-
Cu(s)
2+
-
Pb (aq) + 2 e
Al(s)
Pb(s)
Cr2O72Cr3+
hydrogen ion
H+
+
2 H (aq) + 2 e
2Cr2O7 (aq)
+
+ 14 H + 6 e
-
-
H2(g)
3+
2 Cr (aq) + 7 H2O
2+
Fe (aq) + 2 eMnO4 (aq)
2+
Ni (aq) + 2 e+
2 H (aq) + 2 e-
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Mn2+
+
+8H +5e
-
2+
Mn (aq) + 4 H2O
+
Ag (aq) + e-
Ni(s)
2+
Zn (aq) + 2 e
H2(g)
20-Jul-12
Fe(s)
Ag(s)
-
Zn(s)
Chemsheets A2 045
10
! TASK 7 – ELECTRODE POTENTIALS
+
1)
Calculate the standard electrode potential of the Na /Na electrode given that when it was joined to the standard
hydrogen electrode, the cell emf was -2.71 volts.
2)
Calculate the emf of a cell with the standard AgCl/Ag electrode (E
2+
Fe /Fe (E = -0.44 V) electrode as the right hand one.
3)
Calculate the emf of a standard cell:
given that:
Calculate E
values.
-
Zn(s)
E
= -0.76 V
2+
-
Pb(s)
E
= -0.13 V
2+
of the following cells using the E
cell
4+
-
2+
2+
-
Ag (aq) + e
Br2(l) + 2 e
c) Pt(s)  Cl (aq)  Cl2(g)  Br2(l), Br (aq)  Pt(s)
-
2 I (aq)
+
+
-
Sn (aq)
-
I2(s) + 2 e
b) Pt(s)  I (aq)  I2(s)  Ag (aq)  Ag(s)
-
Ag(s)
-
-
2 Br (aq)
-
-
Cl2(g) + 2 e
2 Cl (aq)
E
= -0.25 V
E
= +0.15 V
E
= +0.54 V
E
= +0.80 V
E
= +1.07 V
E
= +1.36 V
2+
Calculate the E of the Cu /Cu couple given:
2+
-
Cu(s) | Cu (aq) || Cl2(g) | Cl (aq) | Pt(s)
-
-
Cl2(g) + 2 e
2 Cl (aq)
E
E
cell
= +1.02 V
= +1.36 V
Calculate the standard electrode potentials of the half-cells for which the potential is not given. Write your answer by
writing the half equation with its potential.
a) Mg(s)  Mg (aq)  Ti (aq), Ti (aq)  Pt(s)
2+
3+
2+
Mg (aq) + 2 e
-
2+
Mg(s)
2+
Mg (aq) + 2 e
-
2+
E
Mg(s)
E
c) Pt(s)  Hg(l)  Hg2Cl2(aq), KCl(aq)  Rb (aq)  Rb(s)
+
-
-
Hg2Cl2(aq) + 2 e
E
E
b) K(s)  K (aq)  Mg (aq)  Mg(s)
+
7)
Ni(s)
-
Sn (aq) + 2 e
a) Ni(s)  Ni (aq)  Sn (aq), Sn (aq)  Pt(s)
6)
-
Ni (aq) + 2 e
4+
2+
5)
2+
Zn(s) | Zn (aq) || Pb (aq) | Pb(s)
2+
Zn (aq) + 2 e
Pb (aq) + 2 e
4)
2+
= +0.22 V) as the left hand electrode and the
2 Hg(l) + 2 Cl (aq)
E
E
cell
= +2.00 V
= -2.38 V
cell
= +0.54 V
= -2.38 V
cell
= -3.19 V
=+0.27 V
For each of the following questions,
i)
Draw the cell notation for the cell produced when the two half cells are joined via a salt bridge.
ii) Calculate the cell emf.
Remember the cell emf should be positive, although it may not be if the SHE is involved.
2+
-
Cr(s)
E
= -0.91 V
2+
-
Zn(s)
E
= -0.76 V
2+
-
Cu(s)
E
= +0.34 V
E
= +0.77 V
E
= +1.51 V
E
= +1.36 V
a) Cr (aq) + 2 e
Zn (aq) + 2 e
b) Cu (aq) + 2 e
3+
-
Fe (aq) + e
-
2+
Fe (aq)
+
c) MnO4 (aq) + 8 H (aq) + 5 e
-
Cl2(g) + 2 e
-
2+
Mn (aq) + 4 H2O(l)
-
2 Cl (aq)
4) The redox process in the electrochemical cell
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Chemsheets A2 045
11

Metal atoms lose electrons at the one electrode (oxidation), making it -ve, which travel through the wire to the other
electrode, adding to ions to produce metal atoms (reduction).

Remember: oxidation / anode / negative.

Remember also that usually the positive electrode is the right hand electrode.
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Chemsheets A2 045
12
THE ELECTROCHEMICAL SERIES

The electrochemical series is a list of electrode potentials in order of decreasing (or increasing) potential.

The electrochemical series for some common potentials is shown.
Standard electrode potentials
Increasing
oxidising
power
F2(g) + 2 e-

2 F-(aq)
+ 2.87
+
-

MnO2(s) + 2 H2O(l)
+ 1.55
+
-

-

-
+ 14 H (aq) + 6 e

Br2(g) + 2 e-

MnO42-(aq)
+ 4 H (aq) + 2 e
MnO4-(aq)
+ 8 H (aq) + 5 e
Cl2(g) + 2 e
Cr2O72-(aq)
+
2+
+ 1.51
-
+ 1.36
3+
2 Cr (aq) + 7 H2O(l)
+ 1.33
2 Br-(aq)
+ 1.09
Ag(s)
+ 0.80
Mn (aq) + 4 H2O(l)
2 Cl (aq)
+
-

3+
-

Fe (aq)
+ 0.77
-

MnO42-(aq)
+ 0.56
Ag (aq) + e
Fe (aq) + e
MnO4-(aq)
+ e
2+
-

-
-
2 I (aq)
+ 0.54
Cu (aq) + 2 e

Cu(s)
+ 0.34
Hg2Cl2(aq) + 2 e-

2 Hg(l) + 2 CI-(aq)
-

Ag(s) + Cl (aq)
+
-

H2(g)
0.00
2+
-

Pb(s)
- 0.13
2+
-
I2(g) + 2 e
2+
AgCl(s) + e
2 H (aq) + 2 e
Pb (aq) + 2 e
-
+ 0.27
+ 0.22

Sn(s)
- 0.14
-
V (aq) + e

2+
V (aq)
- 0.26
Ni2+(aq) + 2 e-

Ni(s)
- 0.25
Sn (aq) + 2 e
3+
2+
-

Fe(s)
- 0.44
2+
-

Zn(s)
- 0.76
3+
-

Al(s)
- 1.66
2+
-

Mg(s)
- 2.36
-
Na (aq) + e

Na(s)
- 2.71
Ca2+(aq) + 2 e-

Ca(s)
- 2.87

K(s)
- 2.93
Fe (aq) + 2 e
Zn (aq) + 2 e
Al (aq) + 3 e
Mg (aq) + 2 e
+
+
-
K (aq) + e
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E/V
20-Jul-12
Chemsheets A2 045
Increasing
reducing
power
13
Using the electrochemical series
The more positive half
equation gains the electrons
Example question 1
-
-
What reaction will take place when the Cl2/Cl half cell is joined to the Br2/Br half cell?
Cl2 + 2 e
-
2 Cl
-
2 Br
Br2 + 2 e
-
E
= +1.36 V
this is more positive so gains electrons (and so goes in the forwards direction)
-
E
= +1.09 V
this therefore goes backwards
 Cl2 + 2 e  2 Cl
-
-
2 Br  Br2 + 2 e
-
and
-
so overall:
Cl2 + 2 Br  2 Cl + Br2
-
-
Example question 2
2+
2+
The Cu /Cu and Zn /Zn half cells were joined. Calculate the emf, write an equation for the chemical reaction that takes
place, state which is the positive electrode and where oxidation takes place.
–
0
+
Cell emf = +1.10 V
– electrode
+ electrode
e
2+
Positive electrode = Cu /Cu
–
Oxidation at negative electrode
+1.10 V
+0.34 V
Cu
2+
–
+ 2 e  Cu
2+
–
Reactions: Zn → Zn + 2 e
2+
–
Cu + 2 e → Cu
Equation:
2+
Zn + Cu
→ Zn
2+
+ Cu
–0.76 V
Zn
2+
–
+ 2 e  Zn
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20-Jul-12
Chemsheets A2 045
14
! TASK 8 – REDOX EQUILIBRIA PROBLEMS
2+
2+
1)
Write an equation for the reaction that would take place and calculate E
half-cells were connected together.
2)
What reaction would take place if a piece of silver and copper was placed in a solution containing a mixture of silver
nitrate and copper sulphate. Justify this using E data.
3)
Which of the species MnO4 (aq), Cr2O7 (aq) and Fe (aq) are able to liberate Cl2 from an acidic solution of sodium
chloride? Explain your reasoning.
4)
Calculate standard electrode potentials for the electrodes shown. Also, write an equation to show the reaction that
takes place in the cell.
-
Be(s) Be (aq) Cu (aq)Cu(s)
E
cell
= +2.19 V
4+
Pt(s)  H2(g)  H (aq)  Th (aq)Th(s) E
cell
= -1.90 V
2+
2+
+
b) Th (aq)/Th(s)
if the Zn (aq)/Zn(s) and Ni (aq)/Ni(s)
3+
2+
a) Be (aq)/Be(s)
5)
2-
cell
4+
For each of the following cells with the redox equilibria for each electrode shown:
i)
Give the cell diagram.
ii) Calculate the cell emf.
iii) Write a balanced equation for the reaction that will take place in the cell.
a) Br2(l) + 2 e
-
-
2 Br (aq)
2+
-
b) Cu (aq) + 2 e
6)
Cu(s)
+
-
2 H (aq) + 2 e
3+
-
Fe (aq) + e
H2(g)
2+
Fe (aq)
For each of the following questions, predict whether the reaction will take place or not in aqueous solution. Give
clear reasons for your prediction. If the reaction does occur, write the representation of a standard cell in which
the reaction would occur and determine the emf.
a) Will Mg reduce V
-
3+
2+
to V ?
3+
b) Will Cl reduce Fe
2+
to Fe ?
-
c) Will Cl2 oxidise Br to Br2?
d) Will Sn reduce Fe
2-
3+
2+
to Fe ?
+
-
e) Will Cr2O7 /H oxidise Cl to Cl2?
-
+
-
f) Will MnO4 /H oxidise Cl to Cl2?
+
2+
g) Will H oxidise Fe to Fe ?
+
2+
h) Will H oxidise Cu to Cu ?
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20-Jul-12
Chemsheets A2 045
15
EXAMPLES OF ELECTROCHEMICAL CELLS
1) Primary Cells

A battery is more than one cell joined together.

In non-rechargeable (primary) cells, the chemicals are used up so the voltage drops
Non-rechargeable (primary) cells – Zinc-carbon
-0.80 V
+0.70 V
Zn(NH3)2
2+
+ 2e
-
Zn + 2 NH3
+
2 MnO2 + 2 H + 2 e
-
Mn2O3 +
H2O
emf =
overall reaction during discharge =
comments
Non-rechargeable (primary) cells – alkaline
2+
+ 2e
-
-0.76 V
Zn
Zn
+0.84 V
MnO2 + H2O + e
-
MnO(OH) +
-
OH
emf =
overall reaction during discharge =
comments
Non-rechargeable (primary) cells – lithium
comments
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20-Jul-12
Chemsheets A2 045
16
2) Secondary Cells

In rechargeable (secondary) cells the reactions are reversible – they are reversed by applying an external current.

It is important that the products from the forward reaction stick to the electrodes and are not dispersed into the
electrolyte.
Rechargeable (secondary) cells – Li ion
+0.60 V
+
+
-3.00 V
-
Li + CoO2 + e
Li + e
-
LiCoO2
Li
emf =
overall reaction during discharge =
overall reaction during re-charge =
comments
Rechargeable (secondary) cells – lead-acid
+
-
+1.68 V
PbO2 + 3 H + HSO4 + 2 e
-0.36 V
PbSO4 + H + 2 e
+
-
-
PbSO4
+ 2 H2 O
Pb + HSO4
-
emf =
overall reaction during discharge =
overall reaction during re-charge =
comments
Rechargeable (secondary) cells – nickel-cadmium
-
+0.52 V
NiO(OH) + 2 H2O + 2 e
-0.88 V
Cd(OH)2 + 2 e
-
Ni(OH)2 +
2 OH
Cd + 2 OH
-
emf =
overall reaction during discharge =
overall reaction during re-charge =
comments
3) Fuel Cells
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Chemsheets A2 045
17

In a fuel cell the reactants (hydrogen and oxygen) are supplied from an external source rather than having a finite
amount within the cell.
+0.40 V
O2 + 2 H2O + 4 e
-0.83 V
2 H2 O + 2 e
-
-
-
4 OH
H2 + 2 OH
-
emf =
overall reaction during discharge =
comments
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20-Jul-12
Chemsheets A2 045
18
4) Pros and cons of using cells
Pros
Cons
Using cells
Using non-rechargeable
(primary) cells
Using re-chargeable
(secondary) cells
Using fuel cells
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Chemsheets A2 045
19