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HSC Chemistry
Production of Materials
Week 6
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HSC Chemistry
Production of Materials
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1
Week 6 – Theory


Perform a first-hand investigation to identify the conditions under which a galvanic cell is
produced
Perform a first-hand investigation and gather first-hand information to measure the
difference in potential of different combinations of metals in an electrolyte solution
Conditions needed for a galvanic cell to work

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Separation of anode / cathode: before when we dipped a piece of zinc in CuSO4(aq) solution,
we observed a displacement reaction occurring. However we did not separate the anode
and cathode, and as a result, no electricity was produced. By separating the electrodes,
current is forced to travel through an external circuit where it can be made to do work.
Metals of different activities must be chosen: Recall earlier discussion about activity of
metals being the driving force behind displacement reactions. A galvanic cell produces
electricity through using a displacement reaction. If metals of the same activity are chosen
(or very close activities), no displacement occurs (or occurs very weakly) and no current can
be generated.
The less active metal must be in solution: e.g. in the cell Zn | Zn2+ || Cu2+ | Cu (discussed in
previous section), copper starts as an ion in solution, and zinc starts as a metal, because
copper is less active, and is to be displaced by zinc. If this was not the case, like Zn | Zn2+ ||
K+ | Cu, no electricity will be generated as solid copper cannot displace zinc ions in solution.
Need electrolyte: without an electrolyte at the anode and cathode to carry the metal ions to
be oxidised and reduced, the anode and cathode would not work.
Electrolyte concentrations: the anode electrolyte concentration should be minimised and
the cathode electrolyte should be maximised. The greater the difference in these
concentrations, the closer the generated potential will be to theoretical values.
Need a salt bridge: A galvanic cell would only work under certain conditions. It can be
shown that if the salt bridge was removed, the galvanic cell produced no voltage. This is
because removing the salt bridge opens the circuit of the cell and hence no current can flow.
Without a salt bridge, any electron flow will create an electric field opposing further current
flow due to the inability of the half cells to maintain electrical neutrality.
Salt bridge needs to be soaked in a salt solution: if the salt bridge was dry or soaked with
mere water, it would not be able to allow ions to migrate.
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HSC Chemistry
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2
Constructing a cell at school
The second dot-point in this section requires students to be able to recall that they constructed a cell
at school. The diagram below shows one possible cell construction using school materials:
This cell is noted as Cu | Cu2+ || Ag+ | Ag
In order to measure the produced potential (voltage) we attach a voltmeter across the external load.
The external load can be anything (students usually use a light-bulb), as long as it can take the
produced voltages (can be up to 2.5v using school materials).
In all cases, the salt bridge is soaked in KNO3 solution.
To satisfy the second dot-point, students need to measure the voltages across several cells. Each
time, either the anode and anode electrolyte or the cathode and cathode electrolyte was changed to
measure the potential difference between a new set of electrodes.
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HSC Chemistry
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3
Measured potentials
Cell
Measured potential
Theoretical potential
Cu | Cu2+ || Ag+ | Ag
0.3v
0.46v
2+
2+
Zn | Zn || Cu | Cu
0.7v
1.1v
Mg | Mg2+ || Cu2+ | Cu
1.8v
2.71v
2+
2+
Mg | Mg || Zn | Zn
1.1v
1.61v
Pb | Pb2+ || Zn2+ | Zn
0.4v
0.63v
The measured potential will always be significantly lower than the theoretical potential because
within the cell itself (the electrolyte, salt bridge and electrodes), there is internal resistance. If you
recall from Preliminary physics, this resistance will cause a voltage drop within the cell, making the
voltage output to be less than the theoretical amount.
The potential difference across electrodes of a galvanic cell depended upon the difference in activity
between the anode and cathode. For example, a cell using Zn and Pb would not produce as high a
potential as Mg and F2. This is because the difference in activity between the former is not as great
as the latter pair.
Increasing the ratio of cathode to anode electrolyte concentrations increases the potential
difference produced. This is because the potential difference is proportional to the ratio of
concentrations of electrolytes. A low anode electrolyte concentration and a high cathode electrolyte
concentration tend to favour the forward reaction and as equilibrium is approached, the potential
difference decreases to zero.
Increasing the surface area of the electrodes to the electrolytes did not affect potential difference,
but increased current flow. This is because with more contact, more space is available for the
reaction to occur, thus increasing current flow.
Risk management
Some solutions are mildly toxic, such as CuSO4 or Pb(NO3)2. Students should wear gloves when
handling these solutions. Hands should be fully washed after handling each solution, and great care
should be taken when pouring and disposing of these solutions.
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HSC Chemistry
Production of Materials
4
Gather and present information on the structure and chemistry of a dry cell or lead-acid
cell and evaluate it in comparison to one of the following:
o button cell
o fuel cell
o vanadium redox cell
o lithium cell
o liquid junction photovoltaic device (e.g. the Gratzel cell)
in terms of:
o chemistry
o cost and practicality
o impact on society
o environmental impact
This dot-point requires students to compare one of either a lead-acid cell or dry cell, with one of the
more modern cells such as the vanadium redox cell. Exam questions dealing with this dot-point will
ask along the lines if “In your study of HSC Chemistry, you have come across various modern cells,
compare / assess / analyse / describe it in relation to a dry cell / lead acid cell”, so you should study
one cell of each category in good detail.
We will be discussing the lead-acid cell and comparing it to the vanadium redox cell.
The Lead Acid battery
The lead acid battery typically consists of six lead-acid cells connected in series. Each cell contains a
lead anode and a lead (IV) oxide cathode. Each electrode is immersed in a highly acidic sulfuric acid
solution (pH as low as 0).
The electrode reactions are:
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Anode: Pb(s) + H2SO4(aq)  PbSO4(aq) + 2H+(aq) + 2eCathode: PbO2(s) + H2SO4(aq) + 2H+(aq) + 2e-  PbSO4(aq) + 2H2O(l)
(Don’t worry if you find the chemistry of this battery difficult, you won’t get something as hard as
this as a calculation question in exams)
Each cell has a cell voltage of about 2v, and combining 6 in series gives the standard structure used
in cars, a typical 12 volt battery.
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HSC Chemistry
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5
This battery is able to be recharged, with electrons being supplied to the anode and taken from the
cathode. The reactions are reversed by applying an external potential difference which is greater
than the potential produced while the battery is being discharged. Because of this, the lead
accumulator is said to be a storage battery. The recharging equations are as follows:
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Anode: PbSO4(aq) + 2H+(aq) + 2e-  Pb(s) + H2SO4(aq)
Cathode: PbSO4(aq) + 2H2O(l)  PbO2(s) + H2SO4(aq) + 2H+(aq) + 2e-
As you can see, the more charged the battery is, the higher the concentration of sulfuric acid. As the
battery discharges, sulfuric acid concentration falls and lead sulfate concentration increases.
Each cell is constructed with the same principles as a normal galvanic cell. The anode and cathode
are separated by a microporous separator which acts as a salt bridge, allowing H+ ions to move from
the anode to the cathode.
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HSC Chemistry
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6
Vanadium Redox Cell
The vanadium redox cell in comparison works in a very different way. This cell was recently invented
and patented by UNSW in 1986. It is a ‘flow battery’, meaning it stores electrolytes in a separate
tank from the battery itself. Another example of a flow battery is the fuel cell.
The battery generates electricity by using different vanadium oxide solutions (with different
oxidation states) at the anode and cathode. Their reactions are:
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Anode: V2+  V3+ + eCathode: VO2+ + 2H+ + e-  VO2+ + H2O
Overall reaction: V2+ + VO2+ + 2H+  V3+ + VO2+ + H2O
To make it easier to understand, we can express this as word equations:
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Anode: vanadium (II)  vanadium (III) + eCathode: vanadium (V) + e-  vanadium (IV)
As you can see, within the anode electrolyte (the anolyte), we have vanadium (II) being oxidised to
vanadium (III) and emitting an electron. Within the cathode electrolyte (the catholyte), we have
vanadium (V) being reduced to vanadium (IV). Thinking about it this way, it is easier to understand
how this type of cell works.
When electricity is needed, the electrolytes are pumped into the cell stack where an ion selective
membrane allows hydronium ion to pass, but not electrolyte ions, allowing electrical neutrality to be
preserved. This membrane therefore functions like a salt bridge. The necessary transfer of electrons
to allow this is made to flow through an external circuit due to the ion selective membrane
disallowing direct transfer through the electrolytes themselves.
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Recharging this cell reverses the cathode and anode reactions by forcing electricity to flow the
opposite way via an external power source. Doing this will oxidise vanadium (IV) into vanadium (V)
at the cathode, and reduce vanadium (III) into vanadium (II) at the anode.
Comparion of cells
The vanadium redox cell holds many advantages over the lead accumulator in terms of chemistry,
cost, practicality and impacts on society and the environment.
Advantages of the vanadium redox cell
 Unlimited capacity, since electrolytes
are stored in external storage tanks. If
more capacity is needed, just use bigger
storage tanks
 Can be recharged instantly simply by
topping up electrolyte (similar to how a
car tops up on petrol). Lead acid cells
don’t have this luxury of instant
recharging
 No memory effect from long term use,
unlike lead acid batteries
 No adverse effects from being
completely discharged, unlike lead acid
batteries
 Simpler cell design means longer life.
Lead acid batteries’ electrode plates
eventually warp and touch each other,
shorting out the anode and cathode.
 Does not use dangerous and
environmentally harmful chemicals like
lead and sulfuric acid, unlike lead-acid
batteries
 Electrolytes can be reused indefinitely,
meaning no waste is produced
 Recharging produces little hydrogen
unlike lead-acid cells which can become
filled with explosive hydrogen gas if
overcharged
Advantages of the lead acid cell
 Higher peak current delivery, makes this
battery suitable for high current draw
applications, such as starting a car
engine
 Highest power to mass ratio (due to
high peak current) out of any battery
available today
 Higher energy to mass ratio (40Wh/kg)
than vanadium redox (only 25Wh/kg)
 Cheap to produce, since materials
needed are very common and cheap
 Smaller, lighter since there’s no need for
a separate reaction cell and storage
tanks
 Familiar technology, easily repaired and
used since lead-acid batteries have been
around for a long time
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HSC Chemistry
Production of Materials
8
Summary of evaluation
Chemistry
Vanadium cells are superior, as their energy conversion is more efficient, reliable, do not
involve harmful substances like lead and sulfuric acid, and can be reused indefinitely.
Cost and practicality
Lead acid cells are superior, as they are an old and familiar technology, meaning they are
easily repaired and replaced. They are mass manufactured for a very low cost compared to
vanadium cells.
Impact on society
Lead acid cells have had a longer, more profoundly positive impact on society. Vanadium
cells, being a new technology, is still prohibitively complex, expensive and unfamiliar to be
widely applied, bringing any significant good to society. Meanwhile, lead acid batteries have
been starting our cars, powering our UPS (uninterruptible power systems) for hospitals,
datacentres etc, and involved in countless other applications for decades already.
However, in the foreseeable future, we may see increased usage of vanadium cells in more
general applications such as electric or hybrid cars, power storage and UPS applications,
which will have a positive impact on society.
Impact on the environment
Vanadium cells definitely have a less negative impact on the environment than lead acid
cells because it does not use any harmful chemicals (like lead and sulfuric acid) in its
chemistry, and its electrolytes can be recycled indefinitely, producing little waste.
However as discussed above, more widespread adoption and application of vanadium cells
would need to happen before there is any noticeably positive impact on the environment.
Conclusion
Vanadium cells are still very new technology and are not widely implemented enough to
have a significant positive impact on society and the environment. However, its superior
chemistry, structure and design give it much potential to contribute positively to society and
the environment in the future.

Solve problems and analyse information to calculate the potential E° requirement of
named electrochemical processes using tables of standard potentials and half-equations
The E° symbol denotes the standard electrode potential of a certain reaction.
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HSC Chemistry
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9
By definition, this is the potential of that particular electrode reaction when put against a hydrogen
electrode under standard conditions (gas pressure at 1 atm with concentrations of electrolytes at
1M)
The hydrogen electrode refers to the reduction of hydrogen ions: 2H+(aq) + 2e-  H2(g)
By having all electrode potentials relative to this standard electrode, cell potentials of any
combination of electrodes can be calculated. (This is like defining a parametric variable in Maths
Extension 1)
To calculate any galvanic cell’s potential (voltage), we use the table of standard electrode potentials.
This table will be given to you in exams, so don’t worry about memorising the numbers.
Note that these electrode potentials are measured values and can differ from one source to another.
Therefore in the exam they will give you the electrode potential the examiner wants you to use. A
table of the more common electrodes is below.
Reduction reaction
F2 + 2e-  2FAu+ + e-  Au
Cl2 + 2e-  2ClO2 + 4H+ + 4e-  2H2O
Br2 + 2e-  2Br2Hg2+ + 2e-  Hg22+
NO3- + 2H+ + e-  NO2 + H2O
Ag+ + e-  Ag
Fe3+ + e-  Fe2+
MnO4- + 2H2O + 3e-  MnO2 + 4OHO2 + 2H2O + 4e-  4OHCu2+ + 2e-  Cu
Sn4+ + 2e-  Sn2+
2H+ + 2e-  H2
Pb2+ + 2e-  Pb
Sn2+ + 2e-  Sn
Ni2+ + 2e-  Ni
Co2+ + 2e-  Co
PbSO4 + 2e-  Pb + SO42Cd2+ + 2e-  Cd
Fe2+ + 2e-  Fe
Zn2+ + 2e-  Zn
Al3+ + 3e-  Al
Mg2+ + 2e-  Mg
Na+ + e-  Na
Ca2+ + 2e-  Ca
Ba2+ + 2e-  Ba
K+ + e -  K
E° (v)
+2.87
+1.69
+1.36
+1.23
+1.09
+0.92
+0.80
+0.80
+0.77
+0.60
+0.40
+0.34
+0.15
0.00 (by definition)
-0.13
-0.14
-0.26
-0.28
-0.36
-0.40
-0.45
-0.76
-1.66
-2.37
-2.71
-2.87
-2.91
-2.93
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Notice that this table lists the reduction standard potentials. The voltage of the oxidation reaction
(reverse process) is simply the negative of the reduction potential. For example, the potential of K 
K+ + e- is +2.93v. Similarly, the potential of Fe2+  Fe3+ + e- is -0.77v. You have to be careful what the
potentials are listed for. If they are listed for the reduction potentials, take your reduction potential
as positive and add the negative of your oxidation reaction.
Worked examples
Example 1: before we setup the cell:
Zn | Zn2+ || Cu2+ | Cu

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Anode: Zn  Zn2+ + 2e- = +0.76 (take the negative of the table value, because this is
oxidation, and table lists reduction potentials)
Cathode: Cu2+ + 2e-  Cu = +0.34v (straight off the table)
Add the two to get the cell potential = 1.1v
Example 2: in the school experiment, we tested Mg | Mg2+ || Cu2+ | Cu
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Anode: Mg  Mg2+ + 2e- = 2.37 (negative value taken, since oxidation reaction)
Cathode: Cu2+ + 2e-  Cu = 0.34 (positive value taken, since reduction reaction)
Cell voltage is therefore 2.37 + 0.34 = 2.71𝑣
The easy way
Sometimes you don’t know which is the anode and cathode, but you just know the reactants. In this
case, simply take the higher potential, and subtract the other from it. (This can be interpreted as
taking the absolute value of the difference of their electrode potentials) The electrode potentials are
for the reduction reaction (cathode) and so the electrode with the higher of the two will be the
cathode. This allows us to deduce direction of current, as well as cell voltage.
Standard electrode potentials also allow us to calculate the theoretical minimum input potentials
required for recharging (if possible) that particular cell. However in reality, due to activation energies
resistance and other complications, a much higher potential is needed.
Example 3: we have the following cell: Pb | Pb2+ || Fe2+ | Fe and we have no clue which is
the anode and which is the cathode.
The reduction potential for Pb is -0.13v and for iron(II) is -0.45v. Since iron(II) is the higher
value, it will be the cathode. Therefore the cell will work like this:
Anode: Pb2+ + 2e-  Pb = 0.45v (negative, since oxidation reaction)
Cathode: Fe  Fe2+ + 2e- = -0.13v (right off the table, since cathode)
Therefore the cell potential is the sum of anode + cathode = 0.45 − 0.13 = 0.32𝑣
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HSC Chemistry
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11
Week 6 – Homework

Perform a first-hand investigation to identify the conditions under which a galvanic cell is
produced
 Perform a first-hand investigation and gather first-hand information to measure the
difference in potential of different combinations of metals in an electrolyte solution
1. Outline the conditions required to produce a galvanic cell. [4 marks]
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2. Predict what would happen if we removed the salt bridge. [2 marks]
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3. Explain why galvanic cells run flat after letting them run for enough time. [2 marks]
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
HSC Chemistry
Production of Materials
12
Gather and present information on the structure and chemistry of a dry cell or lead-acid
cell and evaluate it in comparison to one of the following:
o button cell
o fuel cell
o vanadium redox cell
o lithium cell
o liquid junction photovoltaic device (e.g. the Gratzel cell)
in terms of:
o chemistry
o cost and practicality
o impact on society
o environmental impact
1. Explain how a lead-acid cell works, and write a balanced equation for the anode and cathode
reactions. Include a diagram in your answer. [6 marks]
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HSC Chemistry
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13
2. Explain how a vanadium-redox cell works, and write redox half-equations for the anode and
cathode. Include a diagram in your answer. [4 marks]
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
HSC Chemistry
Production of Materials
14
Solve problems and analyse information to calculate the potential E° requirement of
named electrochemical processes using tables of standard potentials and half-equations
Table of standard electrode potentials
(all solutes at 1.00 mol/L, all gas pressures at 1 atm)
Reduction reaction
E° (v)
F2 + 2e-  2F+2.87
+
Au + e  Au
+1.69
Cl2 + 2e-  2Cl+1.36
+
O2 + 4H + 4e  2H2O
+1.23
Br2 + 2e-  2Br+1.09
2+
2+
2Hg + 2e  Hg2
+0.92
NO3- + 2H+ + e-  NO2 + H2O
+0.80
+
Ag + e  Ag
+0.80
Fe3+ + e-  Fe2+
+0.77
MnO4 + 2H2O + 3e  MnO2 + 4OH
+0.60
O2 + 2H2O + 4e-  4OH+0.40
2+
Cu + 2e  Cu
+0.34
Sn4+ + 2e-  Sn2+
+0.15
+
2H + 2e  H2
0.00 (by definition)
Pb2+ + 2e-  Pb
-0.13
2+
Sn + 2e  Sn
-0.14
Ni2+ + 2e-  Ni
-0.26
2+
Co + 2e  Co
-0.28
PbSO4 + 2e-  Pb + SO42-0.36
Cd2+ + 2e-  Cd
-0.40
2+
Fe + 2e  Fe
-0.45
Zn2+ + 2e-  Zn
-0.76
3+
Al + 3e  Al
-1.66
Mg2+ + 2e-  Mg
-2.37
+
Na + e  Na
-2.71
Ca2+ + 2e-  Ca
-2.87
2+
Ba + 2e  Ba
-2.91
K+ + e -  K
-2.93
1. Using the table of standard electrode potentials above, calculate the cell voltages for the
following cells: [2 marks each]
a. Fe | Fe2+ || Cu2+ | Cu
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b. Ag | Ag+ || Br-, Br2 | Pt
…………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………….
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HSC Chemistry
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15
c. Ni | Ni2+ || Cl- | Cl2, Pt
…………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………….
d. Ag | Ag+ || Cl- | Cl2, Pt
…………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………….
e. Pt, H2 | H+ || Cl- | Cl2, Pt
…………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………….
2. Draw a fully labelled diagram for each of these cells. [2 marks each]
a. Fe | Fe2+ || Cu2+ | Cu
b. Pt, Cl2 | Cl- || Zn2+ | Zn
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HSC Chemistry
Production of Materials
DUX
16
3. Explain why there is a need to specify that the cell solutes are at 1.00 mol/L for the table of
standard potentials to be meaningful. [1 mark]
…………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………….
4. Explain the need to define a particular cell voltage (2H+ + 2e-  H2) to equal exactly zero. [1
mark]
…………………………………………………………………………………………………………………………………………………………….
…………………………………………………………………………………………………………………………………………………………….
End of homework
© Dux College 2015 | Do not distribute
Need help?
Visit the student forums and ask our tutors.
www.duxcollege.com.au/forums
T: (02) 8007 6824
E: [email protected]
W: www.duxcollege.com.au