Chemistry 3.3 (AS 90696) - Describe oxidationreduction processes
Knowledge of appearance and state of the following
reactants and their products is expected.
Oxidants will be limited to: O2, Cl2, Fe3+, dilute acid(with
metals) H2O2, I2, MnO4, Cu2+, Cr2O72 /H+, OCl-,
concentrated HNO3, IO3, MnO2
Reductants will be limited to: Zn, Mg, Fe, Cu, C, CO, H2,
Fe2+, Br, I, H2S, SO2, SO32, S2O32, H2O2, H2C2O4
NOTE:
Calculations involving simple stoichiometry related to
oxidation-reduction reactions may be included.
Assigning Oxidation Numbers
Find the oxidation number for N in NH4NO3
In an ionic compound such as NH4NO3 the
oxidation numbers of N are determined by first
separating into ions NH4+ and NO3- and then
finding the oxidation numbers.
The values obtained are :
–3 in NH4+
and +5 in NO3
Common Reductants
Hydrogen gas, H2 - This colourless gas can be
used as a reductant, commonly combining with
oxygen to form water, H2O.
Oxalic acid, H2C2O4 – is a weak acid that is
oxidised to carbon dioxide CO2. Its anion is the
oxalate ion C2O42, found in salts such as sodium
oxalate.
ON C = +4
Workout the oxidation number for carbon in CO2
and C2O42- (oxalate ion)
ON C = +3
Redox Reactions

Complete each of the experiments in the
redox reactions experiment in your redox
booklet

Write your answers in each of the spaces
provided
Exercise - Balance each of the following equations.
By using half equations
1.
MnO4 + SO2
MnO4
x2
Mn2+ + SO42
Mn2+
reduction
  ++8H
+ H
+ 5e2+
2+ ++ 4H
2MnO
MnO
16
+
+
10

e
2
Mn
Mn
8H
44
2O
2O
SO2
x5



SO42
oxidation
2 2 ++ 4H
+ ++ +2e-10eSO
5SO
10H
5SO
20H
2 2 + + 2H
2O2O   SO
4 4
2 MnO4 + 5SO2 + 2H2O

2 Mn2+
5SO42  +4H+
20mls of 0.02molL-1 KIO3 was used in the following reaction;
known
IO3- + 6H+ + 5I- = 3I2 + 3H2O
The liberated iodine was titrated with unknown sodium thiosulfate.
(x3 to get moles of I2 the same in each)
6
3
2S2O3- + I2 =
unknown
6
3
2I- + S4O62-
The titre values of thiosulfate were
8.6, 9.2, 8.5, 8.5
Calculate the concentration of the unknown thiosulfate
You will be required to do a MnO4-/Fe2+ titration
Common Reductants
Metals, especially those high on the activity series,
are oxidised to metal ions. Common examples are
Zn (forms Zn2+), Mg (forms Mg2+) and Fe (forms
Fe2+)
Zn(s) 
Zn2+(aq) + 2e
Both Zn2+ and Mg2+ are colourless while Fe2+ is
pale green in solution.
Common Reductants
Iodide ions (I ) and bromide ions (Br) are
both colourless and are readily oxidised to the
halogens I2 (brown or black) and Br2
(orange) respectively
Thiosulfate ions, S2O32 - The thiosulfate ion is
oxidised to S4O62 (tetrathionate ion).
I2
+2
+ S2O32
+2.5

S4O62
+
2I
Assign oxidation numbers to S in both compounds
Common Reductants
Iron(II) ion, Fe2+, can be easily oxidised to
Fe3+. The colour change observed is from
pale green Fe2+ to orange Fe3+. This
occurs when a precipitate of dark green
Fe(OH)2 is exposed to air.
Sulfur dioxide (SO2) and sulfite ion (SO32) are
both oxidised to the sulfate ion (SO42).
All of these species are colourless.
Common Reductants
Carbon ( C) and carbon monoxide (CO) Carbon is commonly used as a reductant in the
form of graphite or coal (both black solids).
Fe2O3
+
3CO

2Fe
+
3CO2
CO (made from combustion of coal) is used
as the reductant at the Glenbrook Steel Mill
where iron(III) oxide, Fe2O3 (from iron sand) is
reduced to Fe and CO is oxidised to CO2.
Common Reductants – remember these
donate electrons and are oxidised
Hydrogen sulfide, H2S – This gas can be used as
a reductant as it can be oxidised to a sulfur
species in a higher oxidation state eg SO2.
Hydrogen peroxide, H2O2 – When acting as a
reductant it is oxidised to O2. This may be seen
as bubbles of gas. Note that H2O2 can also act
as an oxidant
Common Oxidants remember these accept
electrons and are reduced
Oxygen gas, O2 is involved in all burning reactions
producing the oxide ion, O2. Both species are
colourless so no colour change is observed.
Hydrogen ions, H+ present in dilute acids is
reduced to hydrogen gas, H2. Metals above
hydrogen in the activity series will react with H+
in dilute acid or water.
Common Oxidants
Halogens - Chlorine, Cl2 (a yellow-green
gas), bromine, Br2 (an orange liquid), and
iodine, I2 (a shiny black solid) are all reduced
to their respective colourless halide ions, Cl,
Br, I.
Because of its oxidising properties Cl2
(added as a white solid of Ca(OCl)2 rather
than Cl2 gas) to sterilise swimming pools
Common Oxidants
Permanganate ion, MnO4- and manganese dioxide
MnO2 The purple ion MnO4- and the brown solid MnO2 are
both reduced to colourless Mn2+ ion if the reaction is
carried out in acidic solution :
MnO4
+
8H+ + 5e
MnO2 + 4H+
+ 2e


Mn 2+
+ 4H2O
Mn 2+ + 2H2O
Common Oxidants
Dichromate ion, Cr2O72 - This orange ion is
reduced to green Cr3+ the reaction requires an acid
catalyst
Cr2O7 2
+ 14H+
+ 6e 

2Cr 3+ + 7H2O
Hydrogen peroxide, H2O2 - When colourless H2O2 acts
as an oxidant it is reduced to water.
H2O2
+
2H+
+ 2e 

2H2O
Common Oxidants
Metal ions (Fe3+ and Cu2+) - A metal ion can be
displaced from a solution of its salt by a metal above
it in the activity series.
The orange Fe3+ ion usually undergoes reduction to
the pale green Fe2+ ion.
Common Oxidants
Concentrated nitric acid, HNO3 - When colourless
conc nitric acid is used as an oxidant. When reacted
with copper metal, the observed product is brown
nitrogen dioxide gas, NO2.
Cu + 2HNO3 + 2H+  2NO2 + Cu2+ + 2H2O
Iodate ion, IO3 - This ion is colourless and can be
reduced to form the halogen, I2 (a black solid or when
dissolved in water a brown aqueous solution).
2IO3 + 12H+
+ 10e 
I2
+
6H2O
Common Oxidants
Hypochlorite ion, OCl - A colourless ion that
can be reduced to form chlorine, Cl2, a yellowgreen gas.
2OCl + 4H+
+
2e

Cl2 + 2H2O
What is the oxidation number of Cl in
OCl- ?
Ans +1
QUESTION THREE: ANALYSIS OF COPPER IN BRASS
In an analysis of the amount of copper in a brass screw, the following series of
reactions were carried out.
Step 1 The brass screw was placed in concentrated nitric acid and left until the
reaction was complete.
Step 2 Aqueous potassium iodide was added. Reaction occurred to give a
white precipitate in a yellow-brown solution.
Step 3 The mixture was filtered to remove the white precipitate. The remaining
yellow-brown solution was titrated with sodium thiosulfate (using starch as an
indicator). At the end point of the titration, the solution was colourless.
(a) Describe the observations that would be made as step 1 is carried out.
(b) Write balanced half-equations for the reaction of copper with concentrated
nitric acid.
(c)Account for the observations at steps 2 and 3 by identifying the reactions
occurring. Include balanced equations for each reaction.
Step 2:
Potassium Permanganate
– a Strong Oxidising Agent
MnO4- can be reduced to different
substances depending on the pH of
the solution
Potassium Permanganate –
Strong Oxidising Agent
Conditions
O.N.
Acid
Neutral
Alkaline
Colour
MnO4-
MnO4-
MnO4-
Purple
+6
MnO42-
Green
+5
MnO43-
Blue
+7
+4
MnO2
Brown
+3
+2
Mn2+
Pink
Task A : write an ion half equation for the
reduction of MnO4- to Mn 2+ in acid
conditions
Acid Conditions
MnO4- + 8H+ + 5 e-
Mn2+ + 4H2O
We have Balanced redox
reactions in acid but what about
alkaline or neutral solutions?
Balancing Redox Reactions in Alkaline
or Neutral Solutions
For alkaline or neutral solutions use the
same method as in acid conditions – just
add OH- ions to both sides to balance
the H+ ions.
Task B : On a scrap of paper write
the half ion equation for the
reduction of MnO4- to MnO2 in a
neutral solution.
Balancing redox reactions in Alkaline or Neutral conditions
MnO4 -
MnO2
Balance the atoms that aren’t O or H
MnO4 -
MnO2
Balance the O by adding H2O
MnO4 -
MnO2 + 2H2O
Balance the H by adding H+
MnO4 -
+
4H+
MnO2
+
2H2O
Cancel out the H+ by adding OH- to both sides
MnO4- + 4H+ + 4OHMnO4- +
4H2O
MnO2 + 2H2O +
MnO2
+ 2H2O
+ 4OH-
Cancel out waters, etc
MnO4 -
+
2H2O
MnO2
+
4OH-
Balance the charge by adding electrons to most +ve side
MnO4- + 2H2O
+
3e -
MnO2
+
4OH-
4OH-
Task C : Balance MnO4- to MnO42- in an
alkaline solution
MnO4-
+
e-
MnO42-
Task D : Write an equation to
describe the thermal decomposition
of KMnO4 into oxygen + a green
product + a brown product
Heat
2KMnO4
K2MnO4 + MnO2 +
O2
Manganese and the Beginning of
Life?
Ecosystems 2.6 km below sea level were found to
consist of various species including tube worms, giant
clams and crabs.
How can an ecosystem such as this
function in the dark without photosynthesis?
Well it turns out that these ecosystems are situated
by hydrothermal vents with temperatures of around
350oC these vents eject an acidic solution into the
surrounding area.
Outside the vent in the cooler sea water Fe2+
forms as a black smoke of iron sulfide (FeS) while
Mn2+ ions stayed in solution.
Photosynthesis can’t occur in the dark yet energy
must be derived from somewhere to support these
organisms
It is thought that the following reaction occurs to
create energy producing carbohydrates
2H2S + CO2
(CH2O) + H2O + 2S
This reaction is catalysed by sulfide oxidising
bacteria
It is likely that a key role in the development of a
water splitting enzyme for photosynthesis was played
by a manganese (Mn 2+) ion.
In the water around the
vents Mn 2+ ions could have
reacted with dissolved CO2
to form Mn(HCO3)2 which
could have supplied the
sulfide oxidising bacteria
with the carbon they need
for the synthesis of
carbohydrates.
The overall Chemical Change is
+1
-2
-1
+1
+2
Mn(HCO3)2 + 4HS
+2
+4
-2
+1
+1
-2
MnO + 2(CH2O) + H2O + 2S
Sulfur
oxidising
0
-2
bacteria
Task : assign oxidation numbers to each atom
in the above reaction
0
Electrochemical Cells
Electrochemistry is the chemistry of reactions that
involve the transfer of electrons.
In spontaneous reactions, electrons are released and
used in electrochemical cells.
In non spontaneous reactions, electrons have to be
supplied to produce chemicals that are wanted in
electrolytic cells or electrolysis
Electrochemical Cells
If zinc powder is put into copper nitrate solution, a redox reaction
will occur:
Zn(s)
+
Cu(NO3)2 (aq)
Zn(NO3)2 (aq)
+
Cu(s)
Energy is lost in the form of heat
This redox reaction can be written as 2 half equations:
Zn(s)
Cu 2+ (aq) + 2e -
Zn 2+ (aq)
+
2e-
Cu(s)
These two half equations can occur in separate beakers
provided there is a path for the electrons to travel (a wire)
and a path for the ions to travel (a salt bridge)
The Electrolytic Cell
How does an electrochemical
cell actually work?
Standard Electrode Potentials
The voltage produced by a given combination
of half cells depends on the willingness for one
cell to lose electrons (be oxidised) and the
other cell to gain electrons (be reduced)
When the 2 half cells are connected with a
voltmeter the electromotive force (emf or E) of
the cell can be measured.
We can therefore use E figures to compare
the strength of oxidising agents
How to write Electrochemical Cells
The Zn/Cu cell is represented as:
Zn(s)/Zn 2+(aq)//Cu 2+(aq)/Cu(s)
 The double line ( // ) represents the salt bridge separating the
cells
 The single line ( / ) represents a change of phase (here a
solid metal and metal ion solution)
 Write the LH electrode first
Write the LH cell as oxidation (ie write reduced form first, then
the oxidised form)
Write the RH side as reduction (ie write the oxidised form first
then the reduced form)
Write the RH electrode last
Writing the cell in this way shows the electron flow from L to R
Standard Electrode Potentials
The equilibrium reaction in both cells is affected
by:
1. The redox power of the reagents
2. The concentration of the solutions
If we want to compare the oxidising strength of
oxidising and reducing agents we must test them
under the same conditions of temperature,
concentration and pressure.
An inert electrode must be used in cells in which both
species in a redox couple are in aqueous solution
(MnO4- and Mn2+).
The inert electrodes are commonly either platinum,
Pt(s) or graphite, C(s) electrodes. Since the two
species in the redox couple are in solution, they are
separated by a comma rather than a vertical line.
Eg
Cu(s) | Cu2+(aq) || MnO4(aq) , Mn 2+ (aq) | Pt(s)
The two half cells above are:
Cu(s)|Cu 2+(aq) and MnO 4 (aq), Mn 2+ (aq)|Pt(s).
Exercise
Write cell diagrams for each of the following redox reactions
(assuming they are set up as electrochemical cells). Where it is
needed, you must include an inert electrode.
a) Mg(s)
Sn2+(aq)
+

Mg2+(aq)
+
Sn(s)
Mg(s) / Mg 2+(aq) // Sn 2+(aq)/ Sn(s)
b) Cu+(aq)
+
Fe3+(aq)

Cu2+(aq)
+
Fe2+(aq)
C(s) / Cu+ (aq), Cu 2+ (aq) // Fe 3+ (aq), Fe2+ (aq) / C(s)
c) 2Ag(s)
+
Ni2+(aq)
 2Ag+(aq)
+
Ni(s)
Ag (s) / Ag + (aq) // Ni 2+(aq) / Ni (s)
Copy out the cell
give the oxidation and reduction ½ equations then
write the cell diagram in the correct format
½ Eqns
Co(s)  Co2+ (aq) + 2e
Fe3+(aq) + e  Fe2+ (aq)
Cell diagram Co(s) / Cu 2+ (aq) // Fe 3+ (aq) , Fe2+ (aq) / Pt(s)
Standard Electrode Potentials
Electrode potentials (Eo values) are measured relative to a
particular electrode.
In this way, a scale of
relative values can be
established. The
standard hydrogen
electrode (SHE) is
used as the standard
reference electrode,
and it has arbitrarily
been given a value of
0.00 V.
Write and learn this!
Under standard conditions (when the pressure of
hydrogen gas is 1 atm, and the concentration of acid is 1
mol L-1) the potential for the reduction reaction is
assigned a value of zero.
2H+(aq) + 2e
H2(g)
Eo = 0.00 V
*The superscript o denotes standard state
conditions.
Thus using the SHE, the standard reduction potentials for many
half reactions have been measured under standard conditions (at
25 oC). A table of standard reduction potentials is given on page
62 in lab book
Expt : Making Electrochemical Cells
Turn to page 51 in lab book
Read the instructions carefully
We need to make up the 4 cells
between the class
Each group will have to work out their
own concentration and make up each
solution
Turn to making electrochemical cells page 51 in lab
book
Strongest
oxidising
agent
strongest
reducing
agent
Most
easily
reduced
Most
easily
oxidised
Oxidants
F2(g)
Reductants
+ 2e
H2O2(aq) + 2H+(aq) + 2e
Redox couple
•Eo /V

2F(aq)
F2 / F
+2.87

2H2O (l)
H2O2 / H2O
+1.78
MnO4 /Mn2+
+1.51
Cl2 / Cl
+1.36
MnO4 (aq) + 8H+(aq) + 5e

Cl2(g) + 2e

Cr2O72(aq) + 14H+ + 6e
 2Cr3+(aq) + 7H2O
Cr2O72 /Cr3+
+1.33
MnO2(s) + 4H+(aq) + 2e

MnO2 /Mn2+
+1.23
O2(g) + 4H+(aq) + 2e

O2 / H2O
+1.23
2IO3 (aq) + 12H+ + 10e

IO3
+1.20
Br2(l) + 2e

NO3(aq) + 2H+(aq) + e

Mn
2+
(aq) + 4H 2O
2Cl-(aq)
Mn 2+ (aq) + 2H2O
2H2O(l)
I 2(s)
+
6H2O
/I
2
Br2 / Br
+1.07
NO2(g) + H2O(l)
NO3- / NO2
+0.94
2Br (aq)
Ag+(aq) + e

Ag(s)
Ag+ / Ag
+0.80
Fe3+(aq) + e

Fe2+(aq)
Fe3+ / Fe2+
+0.77
I2(s) + 2e

2I(aq)
I2 / I
+0.54
O2(g) + 2H2O(l) + 4e

4OH(aq)
O2 / OH
+0.40
Cu2+(aq) + 2e

Cu(s)
Cu2+ / Cu
0.34
SO42(aq) + 4H+(aq) + 2e
 SO2(g) + 2H2O(l)
SO42 / SO2
+0.20
Cu2+(aq) + e

Cu2+ / Cu+
+0.15
2H+(aq) + 2e

H2(g)
H+ / H2
0.00
Pb2+(aq) + 2e

Pb(s)
Pb2+ / Pb
-0.13
Fe2+(aq) + 2e

Fe(s)
Fe2+ / Fe
-0.44
Zn2+(aq) + 2e

Zn(s)
Zn2+ / Zn
-0.76
Al3+(aq) + 3e

Al(s)
Al3+ / Al
-1.66
Mg2+(aq) + 2e

Mg(s)
Mg2+ / Mg
-2.37
Ca(s)
Ca2+ / Ca
-2.87
K(s)
K+ / K
-2.93
Ca2+(aq) + 2e
K+(aq) + e


Cu+ (aq)
This table on page 62 can also be used to decide the relative
strength of species as oxidants or reductants.
The species on the left in the couple with the most positive
reduction potential, will be the strongest oxidising agent or
oxidant. In our table it is F2(g) (NOT F2 / F).
This means F2 has the greatest tendency to gain electrons.
As the electrode potential decreases, the strength as an
oxidant decreases.
Conversely the strongest reducing agent or reductant
would be Li(s). This means Li has the greatest
tendency to lose electrons.
Exercise
Identify the species, from the table of electrode potentials on page
62 in lab book, which fits each statement.
F2
a) The molecule which is the strongest oxidant. ____________
2+
Mn
b) The metal ion which is the weakest reductant. ___________
H2
c) The gas which is the strongest reductant. _______________
MnO4 d) The ion which is the strongest oxidant. ________________
O2
e) The gas which is the weakest oxidant. __________________
Predicting Reactions
Redox potentials can be used to predict
whether or not a redox reaction will occur.
The half cell equations are arranged in
order of half reactions of increasing Eo
values.
(ie the most negative Eo half cell is the
oxidation reaction and is written first)
Example : will Mg metal react with lead (II) nitrate solution?
Eo (Mg 2+ /Mg) = -2.36 V and Eo (Pb 2+/Pb) = -0.13 V
The first step is to arrange the ½ equations with the more
negative (or less positive value) on top.
Mg 2+ (aq) + 2e-
Mg (s)
Write as a oxidation ½ Equation
Mg (s)
Pb 2+ (aq) + 2e-
Mg 2+ (aq) + 2ePb (s)
Write as a reduction ½ Equation
Pb 2+ (aq) + 2e-
Pb (s)
Eo = -2.36 V
Most negative therefore
oxidation cell
Eo = - 0.13 V
More positive (or less negative)
therefore reduction cell
Writing the cell equation
Mg (s)/Mg 2+ (aq) // Pb 2+(aq)/Pb(s)
Writing the full redox equation
Pb 2+ (aq) + Mg(s)
Pb (s) + Mg 2+
Using reduction potentials to determine Eocell
In any electrochemical cell, the standard cell potential (voltage),
E0cell, is the difference between the reduction potentials of the two
redox couples involved.
The couple with the most positive reduction potential will be the
reduction half-cell (ie the oxidant) (cathode). This means that
the Eocell for any combination of electrodes can be predicted using
the relationship
Eocell = Eo(reduction half-cell) - Eo(oxidation half-cell)
OR
Eocell = Eo(cathode) - Eo(anode)
OR
Eocell = Eo(RHE) - Eo(LHE)
E0 calculations
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
E0(cell)
= E0(RHE)-E0(LHE)
= +0.34 – (-0.76)
= 1.10 V
A positive E0 value indicates a spontaneous
reaction and 1.10 V is generated under
standard conditions
E0 calculations – opposite
calculate the E0 value for this cell
-state whether it will occur
Cu(s) | Cu2+(aq) || Zn2+(aq) | Zn(s)
E0(cell)
= E0(RHE)-E0(LHE)
= -0.76 – (+0.34)
= -1.10 V
A negative E0 value indicates a
non - spontaneous reaction and the reaction
would require 1.10 V to occur under standard
conditions
Calculate the cell EMF or Eo
Mg (s)/Mg 2+ (aq) // Pb 2+(aq)/Pb(s)
E0(cell)
= E0(RHE)-E0(LHE)
= -0.13 – (-2.38)
= 2.25 V
A positive E0 value indicates a spontaneous
reaction and 2.25 V is generated under
standard conditions
Exercise
Calculate the standard cell potential for each of the
following electrochemical cells. Write each of the full
redox equations. Use potentials on pg 62.
a)
Zn(s) | Zn2+(aq) || Cr2O72-(aq) , Cr3+(aq) | C(s)
b)
Cu(s) | Cu2+(aq) || Fe3+(aq), Fe2+(aq) | Pt(s)
c)
A cell made up using the couples Pb2+/Pb and I2/I
Task : complete calculate your
potential on page 63 lab book - read
the instructions carefully
In the exam be prepared for questions
to be asked in slightly different ways
Expt : Making Electrochemical Cells
calculate the Eo of each of the cells we
made using our method
Can a solution of acidified potassium permanganate oxidise the
Fe2+ present in a solution of iron (II) nitrate?
(Note in questions such as this you will have to recognise that
ions such as sodium and potassium are spectator ions.) The
unbalanced equation for the reaction would be:
MnO4
+
Reduction reaction is
Fe2+

Mn2+
MnO4  Mn2+
Eo (MnO4 /Mn2+) = +1.51 V
Oxidation reaction is
Fe2+
 Fe3+
Eo (Fe3+/Fe2+) = +0.77 V
Eo cell
= Eo (MnO4 /Mn2+) - Eo (Fe3+/Fe2+)
= +1.51 - (+0.77) = 0.74 V
+
Fe3+
Calculating Ecell from a cell diagram.
For cells written using the IUPAC cell convention,
E°cell = E°RHE – E°LHE
What is the Ecell for the cell below?
RHE = right hand
electrode,
LHE = left hand
electrode
C(s) / C2O42–(aq)/CO2(g) // Cr2O72–(aq) , Cr3+(aq) /C(s)
E°(CO2/C2O42–) = –0.20 V
E°(Cr2O72–,Cr3+) = 1.33 V
E°cell = E°RHE – E°LHE
= 1.33 V – (-0.20 V)
= 1.53 V
Do not reverse the polarity of the left hand (oxidation)
electrode – that’s done when we subtract it.
Spontaneous redox reactions make it impossible for
some compounds to exist. For example given the
following Eo values determine whether or not Fel3 will
form?
Eo(Fe3+ /Fe2+ ) = +0.77 V and Eo (I2 / I) = +0.54 V
which means Eocell = 0.77 – 0.54 = 0.23 V and since it is
positive a reaction occurs spontaneously because Fe3+ will
be reduced to Fe2+ oxidising I to I2 not allowing Fel3 to
form
Some common cells
Note that you are not expected to remember the details of any
specific cell. Rather resource information for any cell will be
included in the exam paper. The following gives you some typical
information with regard to a variety of common cells.
What’s Unique about the lead acid cell
The feature that distinguishes the lead acid battery
from most other cells is that the above reactions can
be reversed. By passing an electric current back
through the cell the Pb 2+ can be converted to Pb
and PbO 2
After many charge/discharge cycles some of the
PbSO4 falls to the bottom and the H2SO4
concentration becomes low – and can’t be recharged
The most common “dry cell” is the Leclanche cell. This is
commonly used in torches, radios etc. It is a primary (nonrechargeable ) cell in which the major reactants are the zinc case
(the anode), which is separated from the MnO2 paste by a porous
partition. The cathode is a carbon (graphite) rod. The MnO2 is
mixed into an electrolytic paste with ammonium chloride and
water to form a conducting medium. The oxidation reaction is
Zn(s)  Zn2+(aq) + 2e
The Leclanche cell continued.
The reduction reaction is complicated but can be
represented as
MnO2(s) + H2O + e

MnO(OH)(s) + OH (aq)
Furthermore the formation of OH- ions results in the NH4+
ions reacting to form NH3. This is not released as gaseous
ammonia, but forms a complex ion with the Zn2+ ions,
[Zn(NH3)4]2+.
Questions
1. In the lead acid battery
a. Which species is oxidised?
Pb (s)
b. Which species is reduced?
PbO2 (s)
2. Why would the density of acid solution decrease as the
battery discharges (look at the cell diagram carefully)?
Because water is produced in the reaction and this dilutes the acid
3. In the common dry cell (Leclanche cell)
a. What species is oxidised?
Zn (s)
b. What species is reduced?
MnO2 (s)
c. What is the function of the carbon?
It acts as a positive electrode (cathode)
QUESTION ONE: Describe what would be observed as each of
the following reactions proceeds. Explain these observations in
terms of the chemical reactions occurring. For each reaction write
balanced ion-electron half-equations and a balanced overall net
ionic equation.
1. Concentrated nitric acid is added to copper metal.
Observations and explanation
Copper dissolves. Solution turns green/blue as the
copper is oxidised. Brown gas given off (NO2) as HNO3 is
reduced.
Equation
Cu  Cu2+ + 2e
NO3– + 2H+ + e  NO2 + H2O
OR
Overall: 2NO3– + 4H+ +Cu  Cu2+ + 2NO2 + 2H2O
Potassium permanganate solution is added to acidified hydrogen
peroxide solution.
Observations and explanation
Colour of permanganate changes from pink to colourless (very pale
pink) as it is reduced to Mn2+. Bubbles of colourless gas given off
(O2) as H2O2 is oxidised.
Equations
MnO4– + 8H+ + 5e  Mn2+ + 4H2O
2MnO4–
H2O2  O2 + 2H+ + 2e
Overall:
+ 6H+ + 5H2O2  2Mn2+ +8H2O + 5O2
3.Describe how the reaction in Q2 above would change if the
solution is not acidified.
A brown/black precipitate (of MnO2) will be formed
Electrochemical cells
The diagram shows the experimental set up for a cell made
up of the Fe3+,Fe2+ and Co3+,Co2+ half cells.
(a) Name a substance that could be used for the electrodes.
(b)
Carbon or platinum (A)
Explain why this substance can be used as an electrode.
unreactive and conducts electricity (M) for both
Using standard notation, write the cell diagram for
the cell above.
Pt|Fe3+, Fe2+||Co3+, Co2+|Pt (M)
C could replace Pt
Write the equation for the cell reaction.
Fe2+ + Co3+  Fe3+ + Co2+ (A)
How will the colour of the left hand half cell change
as the cell is operated for some time?
Becomes more yellow in colour (A)
Which half cell would have the negative electrode?
Fe3+/Fe2+ is negative (A)
Give a reason for your answer
Oxidation occurs in the left-hand half-cell. Oxidation is loss of
electrons, so the Fe3+/Fe2+ has a surplus of electrons making it
negative.
Explain the function of the salt bridge in the cell.
The salt bridge completes the circuit:
allowing ions to flow from one half-cell to the other
without introducing further potentials
To maintain a balance of ions/charge. (M)
Calculate the potential of the cell using the E° values;
E° (Co3+, Co2+) = 0.82 V; E°(Fe3+, Fe2+) = 0.77 V above.
E = 0.82 – 0.77 = 0.05 V (A)
Discovering
Electrochemical Cells
PGCC CHM 102 Sinex
A Picture of a Car Battery
An Alkaline Battery
• Anode: Zn cap:
Zn(s)  Zn2+(aq) + 2e• Cathode: MnO2, NH4Cl and carbon paste:
2 NH4+(aq) + 2 MnO2(s) + 2e-  Mn2O3(s) +
2NH3(aq) + 2H2O(l)
• Graphite rod in the center - inert cathode.
• Alkaline battery, NH4Cl is replaced with KOH.
• Anode: Zn powder mixed in a gel:
The Alkaline Battery
Fuel Cells
• Direct production of electricity from
fuels occurs in a fuel cell.
• H2-O2 fuel cell was the primary source
of electricity on Apollo moon flights.
• Cathode: reduction of oxygen:
2 H2O(l) + O2(g) + 4e-  4OH-(aq)
• Anode:
2H2(g) + 4OH-(aq)  4H2O(l) + 4e-
Fuel Cells
Part II – Galvanic Cells
Batteries and corrosion
Cell
Construction
Salt bridge –
KCl in agar
Provides conduction
between half-cells
Observe the
electrodes
to see what
is occurring.
Cu
Zn
1.0 M CuSO4
1.0 M ZnSO4
What about half-cell
reactions?
What about the sign
of the electrodes?
-
+
cathode half-cell
Cu+2 + 2e-  Cu
Cu
plates out
or
deposits
on
electrode
Cu
1.0 M CuSO4
Why?
anode half-cell
Zn  Zn+2 + 2e-
What
happened
at each
electrode?
Zn
1.0 M ZnSO4
Zn
electrode
erodes
or
dissolves
Galvanic cell
• cathode half-cell (+)
REDUCTION
Cu+2 + 2e-  Cu
• anode half-cell (-)
OXIDATION
Zn  Zn+2 + 2e-
• overall cell reaction
Zn + Cu+2  Zn+2 + Cu
Spontaneous reaction that produces electrical current!
Now for a standard cell composed of
Cu/Cu+2 and Zn/Zn+2, what is the voltage
produced by the reaction at 25oC?
Standard Conditions
Temperature - 25oC
All solutions – 1.00 M
All gases – 1.00 atm
Now replace the light bulb with a volt meter.
+
cathode half-cell
Cu+2 + 2e-  Cu
1.1 volts
anode half-cell
Zn  Zn+2 + 2e-
Cu
Zn
1.0 M CuSO4
1.0 M ZnSO4
We need a standard electrode
to make measurements against!
The Standard Hydrogen Electrode (SHE)
25oC
1.00 M H+
1.00 atm H2
H2 input
1.00 atm
Half-cell
2H+ + 2e-  H2
EoSHE = 0.0 volts
Pt
1.00 M H+
inert
metal
Now let’s combine the copper half-cell with the SHE
Eo = + 0.34 v
+
0.34 v
cathode half-cell
Cu+2 + 2e-  Cu
anode half-cell
H2  2H+ + 2eH2 1.00 atm
KCl in agar
Cu
Pt
1.0 M CuSO4
1.0 M H+
Now let’s combine the zinc half-cell with the SHE
Eo = - 0.76 v
-
0.76 v
anode half-cell
Zn  Zn+2 + 2e-
cathode half-cell
2H+ + 2e-  H2
H2 1.00 atm
KCl in agar
Zn
Pt
1.0 M ZnSO4
1.0 M H+
Assigning the Eo
Al+3 + 3e-  Al
Eo = - 1.66 v
Zn+2 + 2e-  Zn
Eo = - 0.76 v
2H+ + 2e-  H2
Eo = 0.00 v
Cu+2 + 2e-  Cu
Eo = + 0.34
Ag+ + e-  Ag
Eo = + 0.80 v
Increasing activity
Write a reduction half-cell, assign the voltage
measured, and the sign of the electrode to the
voltage.
The Non-active Metals
Metal + H+  no reaction
since Eocell < 0
105
Db
107
Bh
Comparison of Electrochemical Cells
galvanic
electrolytic
need
produces
power
electrical
two
current electrodes source
anode (-)
conductive anode (+)
cathode (+) medium
cathode (-)
salt bridge
DG < 0
vessel
DG > 0