Chemistry 3.3 (AS 90696) - Describe oxidation-reduction processes

1

• Chemistry 3.3 (AS 90696) - Describe
  oxidation-reduction 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.

2
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?
3
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
4
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

5
Exercise - Balance each of the following
equations. By using half equations 1. MnO4?
SO2 ? Mn2 SO42?
MnO4? ? Mn2 reduction
MnO4? 8H 5e- ? Mn2 4H2O
x 2
2 MnO4? 16 H 10 e- ? 2 Mn2
8 H2O
SO2 ? SO42? oxidation
SO2 2H2O ? SO42?
4H 2e-
x 5
5SO2 10H2O ? 5SO42?
20H 10e-
2 MnO4? 5SO2 2H2O ? 2 Mn2
5SO42 ? 4H
6
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 6 3
2S2O3- I2 2I- S4O62-
unknown
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
7
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.
8
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 S2O32? ? S4O62?
2I? Assign oxidation numbers to S in both
compounds
2.5
2
9
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.
10
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.
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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
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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.
13
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
14
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?
? Mn 2 4H2O MnO2 4H
2e? ? Mn 2 2H2O
15
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
16
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.
17
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
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Common Oxidants
Hypochlorite ion, OCl? - A colourless ion that
can be reduced to form chlorine, Cl2, a
yellow-green gas. 2OCl? 4H 2e?
? Cl2 2H2O
What is the oxidation number of Cl in OCl- ? Ans
1
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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 Step 3


20
Potassium Permanganate a Strong Oxidising
Agent
21
MnO4- can be reduced to different substances
depending on the pH of the solution
22
Potassium Permanganate Strong Oxidising
Agent
Conditions
O.N.
Acid
Neutral
Alkaline
Colour
Purple
MnO4-
MnO4-

MnO4-
7
Green
6
MnO42-
MnO43-
Blue
5
MnO2
Brown
4
3
Mn2
Pink
2
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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
24
We have Balanced redox reactions in acid but what
about alkaline or neutral solutions?
25
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.
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Task B On a scrap of paper write the half ion
equation for the reduction of MnO4- to MnO2 in a
neutral solution.
27
Balancing redox reactions in Alkaline or Neutral
conditions MnO4 -
MnO2 Balance the atoms that arent 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
4OH- MnO2 2H2O
4OH- MnO4- 4H2O
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-
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Task C Balance MnO4- to MnO42- in an
alkaline solution
MnO4- e- MnO42-
29
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
30
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?
31
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.
32
Photosynthesis cant 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
33
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.
34
The overall Chemical Change is
1
-1
2
-2
-2
-2
1
1
1
Mn(HCO3)2 4HS MnO 2(CH2O)
H2O 2S
Sulfur oxidising bacteria
4
0
-2
0
2
Task assign oxidation numbers to each atom
in the above reaction
35
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
36
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) Zn 2 (aq)
2e- Cu 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)
37
The Electrolytic Cell
38
How does an electrochemical cell actually work?
39
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
40
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

41
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.
42
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).
43

• 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.
• Mg(s) Sn2(aq) ? Mg2(aq)
  Sn(s)
•  
•  
• b) Cu(aq) Fe3(aq) ? Cu2(aq)
  Fe2(aq)
•  
•  
• c) 2Ag(s) Ni2(aq) ? 2Ag(aq)
  Ni(s)
•  

Mg(s) / Mg 2(aq) // Sn 2(aq)/ Sn(s)
C(s) / Cu (aq), Cu 2 (aq) // Fe 3 (aq), Fe2
(aq) / C(s)
Ag (s) / Ag (aq) // Ni 2(aq) / Ni (s)
44
Copy out the cellgive the oxidation and
reduction ½ equations then write the cell diagram
in the correct format
Co(s) ? Co2 (aq) 2e Fe3(aq) e ?
Fe2 (aq)
½ Eqns
Cell diagram
Co(s) / Cu 2 (aq) // Fe 3 (aq) , Fe2 (aq) /
Pt(s)
45
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.
46
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.
47
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  
48
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
49
Turn to making electrochemical cells page 51 in
lab book
50
Oxidants Reductants Redox couple Eo /V
F2(g) 2e? ? 2F?(aq) F2 / F? 2.87
H2O2(aq) 2H(aq) 2e? ? 2H2O (l) H2O2 / H2O 1.78
MnO4? (aq) 8H(aq) 5e? ? Mn 2 (aq) 4H 2O MnO4? /Mn2 1.51
Cl2(g) 2e? ? 2Cl-(aq) Cl2 / Cl? 1.36
Cr2O72?(aq) 14H 6e? ? 2Cr3(aq) 7H2O Cr2O72? /Cr3 1.33
MnO2(s) 4H(aq) 2e? ? Mn 2 (aq) 2H2O MnO2 /Mn2 1.23
O2(g) 4H(aq) 2e? ? 2H2O(l) O2 / H2O 1.23
2IO3? (aq) 12H 10e? ? I 2(s) 6H2O IO3? / I2 1.20
Br2(l) 2e? ? 2Br? (aq) Br2 / Br? 1.07
NO3?(aq) 2H(aq) e? ? NO2(g) H2O(l) NO3- / NO2 0.94
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? ? Cu (aq) 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
Ca2(aq) 2e? ? Ca(s) Ca2 / Ca -2.87
K(aq) e? ? K(s) K / K -2.93
Li(aq) e? ? Li(s) Li / Li -3.04
Most easily reduced
Strongest oxidising agent
Most easily oxidised
strongest reducing agent
51
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.  
52

• Exercise
• Identify the species, from the table of electrode
  potentials on page 62 in lab book, which fits
  each statement.
• The molecule which is the strongest oxidant.
  ____________
• The metal ion which is the weakest reductant.
  ___________
• The gas which is the strongest reductant.
  _______________
• The ion which is the strongest oxidant.
  ________________
• e) The gas which is the weakest oxidant.
  __________________

F2
Mn2
H2
MnO4 -
O2
53
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)
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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)
Eo -2.36 V
Write as a oxidation ½ Equation
Most negative therefore oxidation cell
Mg (s) Mg 2 (aq) 2e-

Pb 2 (aq) 2e- Pb (s)
Eo - 0.13 V
Write as a reduction ½ Equation
More positive (or less negative) therefore
reduction cell
Pb 2 (aq) 2e- Pb (s)
56
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
57
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)
 
58
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

59
E0 calculations oppositecalculate 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

60
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
61
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?
62
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
63
Expt Making Electrochemical Cells calculate
the Eo of each of the cells we made using our
method
64
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? Fe2 ?
Mn2 Fe3 Reduction reaction is 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
65
Calculating Ecell from a cell diagram.
For cells written using the IUPAC cell
convention, Ecell ERHE ELHE
RHE right hand electrode, LHE left hand
electrode
What is the Ecell for the cell below? C(s) /
C2O42(aq)/CO2(g) // Cr2O72(aq) , Cr3(aq)
/C(s) E(CO2/C2O42) 0.20 V
E(Cr2O72,Cr3) 1.33 V
Ecell ERHE ELHE 1.33 V (-0.20
V) 1.53 V
Do not reverse the polarity of the left hand
(oxidation) electrode thats done when we
subtract it.
66
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
67
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.
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69
Whats 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 cant
be recharged
70
The most common dry cell is the Leclanche cell.
This is commonly used in torches, radios etc. It
is a primary (non-rechargeable ) 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?
71
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)42.
72

• Questions
• 1. In the lead acid battery
• Which species is oxidised?
• Which species is reduced?
• 2. Why would the density of acid solution
  decrease as the battery discharges (look at the
  cell diagram carefully)?
• 3. In the common dry cell (Leclanche cell)
• What species is oxidised?
• What species is reduced?
• What is the function of the carbon?

Pb (s)
PbO2 (s)
Because water is produced in the reaction and
this dilutes the acid
Zn (s)
MnO2 (s)
It acts as a positive electrode (cathode)
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74
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
75
Potassium permanganate solution is added to
acidified hydrogen peroxide solution. Observation
s 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 H2O2 ? O2 2H 2e Overall 2MnO4
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

76
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.
Carbon or platinum (A)
(b) Explain why this substance can be used as an
electrode.
unreactive and conducts electricity (M) for both
77
Using standard notation, write the cell diagram
for the cell above.
PtFe3, Fe2Co3, Co2Pt (M) C could
replace Pt
78
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)
79
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.
80
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)
81
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)
82
Discovering Electrochemical Cells
PGCC CHM 102 Sinex
83
A Picture of a Car Battery
84
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

85
The Alkaline Battery
86
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-

87
Fuel Cells
88
Part II Galvanic Cells
Batteries and corrosion
89
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
90
What about half-cell reactions?
What about the sign of the electrodes?
-

Why?
cathode half-cell Cu2 2e- ? Cu
anode half-cell Zn ? Zn2 2e-
Cu plates out or deposits on electrode
Zn electrode erodes or dissolves
What happened at each electrode?
Cu
Zn
1.0 M CuSO4
1.0 M ZnSO4
91
Galvanic cell

• cathode half-cell ()
• REDUCTION Cu2 2e- ? Cu
• anode half-cell (-)
• OXIDATION Zn ? Zn2 2e-
• overall cell reaction
• Zn Cu2 ? Zn2 Cu

Spontaneous reaction that produces electrical
current!
92
Now for a standard cell composed of Cu/Cu2 and
Zn/Zn2, what is the voltage produced by the
reaction at 25oC?
Standard Conditions Temperature - 25oC All
solutions 1.00 M All gases 1.00 atm
93
Now replace the light bulb with a volt meter.
-

1.1 volts
cathode half-cell Cu2 2e- ? Cu
anode half-cell Zn ? Zn2 2e-
Cu
Zn
1.0 M CuSO4
1.0 M ZnSO4
94
We need a standard electrode to make measurements
against!
The Standard Hydrogen Electrode (SHE)
H2 input 1.00 atm
25oC 1.00 M H 1.00 atm H2
Pt
Half-cell 2H 2e- ? H2
inert metal
EoSHE 0.0 volts
1.00 M H
95
Now lets combine the copper half-cell with the
SHE
Eo 0.34 v

0.34 v
cathode half-cell Cu2 2e- ? Cu
anode half-cell H2 ? 2H 2e-
H2 1.00 atm
KCl in agar
Pt
Cu
1.0 M CuSO4
1.0 M H
96
Now lets combine the zinc half-cell with the SHE
Eo - 0.76 v
-
0.76 v
cathode half-cell 2H 2e- ? H2
anode half-cell Zn ? Zn2 2e-
H2 1.00 atm
KCl in agar
Pt
Zn
1.0 M ZnSO4
1.0 M H
97
Assigning the Eo
Write a reduction half-cell, assign the voltage
measured, and the sign of the electrode to the
voltage.

• Al3 3e- ? Al Eo - 1.66 v
• Zn2 2e- ? Zn Eo - 0.76 v
• 2H 2e- ? H2 Eo 0.00 v
• Cu2 2e- ? Cu Eo 0.34
• Ag e- ? Ag Eo 0.80 v

98
The Non-active Metals
Metal H ? no reaction since Eocell lt 0
99
Comparison of Electrochemical Cells
galvanic
electrolytic
need power source
produces electrical current
two electrodes
anode (-) cathode ()
anode () cathode (-)
conductive medium
salt bridge
vessel
DG gt 0
DG lt 0