HC Notes Talent 100

HSC Chemistry Production of Materials III © TALENT 100

TALENT 100: HSC SUCCESS. SIMPLIFIED. www.talent-100.com.au

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REVIEW OF LAST WEEK

ARGUMENTS FOR USING CELLULOSE FOR THE MANUFACTURE OF

PETROCHEMICALS

• Each glucose unit within cellulose has a basic 5 carbon atoms joined together which can be used

as the ……………………………………………………………………………………………………………

• Cellulose is a ………………………….. resource

• Cellulose can be broken down, that is it is…………………………. It can be used to make …………………….

ARGUMENTS AGAINST USING CELLULOSE FOR THE MANUFACTURE OF

PETROCHEMICALS

• Though cellulose is a cheap starting material, the conversion of cellulose to glucose using

enzymatic or acid digestion is very ……………… and ………………………………..

• The process also requires an energy input to drive machinery and heat reaction mixtures. This

typically comes from ……………………………………………………………………………….

• Producing petrochemicals from cellulose to fulfil commercial demand would require vast areas

of land to ……………………………………………………………………………….

• Cellulose has enormous potential as a raw material which can be fulfilled only with further

………………………….

PHB – A BIOPOLYMER

• The organism that produces PHB is ……………………………… .

• The desirable qualities of PHB include that it is a ………………………………………………………………………….

………………………………………………………………………………………………………… .

• Potentially PHB can be used in …………………………, ………………………………., …………………………,

………………………… and …………………………………………

• The use of PHB is limited because the process is far too ………………………….



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REACTIONS INVOLVING ETHANOL

• Ethanol can be dehydrated to form ethene. The equation for this reaction is:

�� + ��

• Ethene, on the other hand, can be hydrated to produce ethanol and is represented by the

following equation:

�� + ��

USES OF ETHANOL

• Ethanol is as an important solvent due to its ability to ………………………………………………………………….

……………………………………………………………. ..

• The hydroxy group is ………………… due to the high ……………………………………………………………………...

• The alkyl group is …………………

• Uses of ethanol include: …………………………………………………………………………………………… .

FERMENTATION

• Fermentation is the process in which ……………… is broken down to ………………… and …………………

………………… by the action of ……………………………………………………….

• The equation for fermentation is:

�!�"��!�#$�%&'() &*+,-&( �.,-'(&�� 2��0�� #$� + 2��

• The optimum temperature for fermentation is …………………

• An anaerobic environment is maintained, that means there is no …………………



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ADVANTAGES OF ETHANOL AS AN ALTERNATIVE FUEL

• It is a ………………… resource

• It produces less toxic …………………………………… during combustion

• Ethanol can be incorporated into …………………………………… such as E10

DISADVANTAGES OF ETHANOL AS AN ALTERNATIVE FUEL

• Large areas of land need to be devoted to the ……………………………………

• At concentrations of above 10%, engines ………………………………………………………

MOLAR HEAT OF COMBUSTION

• The molar heat of combustion of a substance is the heat liberated …………………………………………..

…………………………………………………………………………………………… at standard atmospheric pressure

with the final products being …………………………………………………………

• The molar heat of combustion is calculated by the following equation:

molar heat of combustion =



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This lesson begins with an analysis of the fermentation practical. Following that, we will explore the

conversion of chemical energy to electrical energy by learning about displacement reactions,

galvanic cells and calculations involving them.

PRACTICAL

FERMENTATION

Aim: To plan an investigation which monitors the changes in mass occurring during the fermentation

of glucose.

Risk Assessment: There was little to no risk involved in this experiment. The only risk that was

involved in the experiment was the possibility of breakage of glassware, and hence keep glassware

away from the edge of the bench.

Equipment: 250mL vacuum flask with rubber stopper and sidearm, 500mL beaker, 50mL measuring

cylinder, large test tube, 20mL glucose solution (10% w/v), 1g Sodium dihydrogen phosphate (yeast

nutrient), limewater, thermometer, electronic balance, equipment for control experiment.

Method

1. Weigh out 1g yeast and add it to the vacuum flask.

2. Measure 20mL of glucose solution using a measuring cylinder and add to the flask.

3. Insert the rubber stopper and weight the apparatus and record exactly the initial mass.

4. Place the reaction apparatus in a 500mL beaker containing water held at 30-35°C (see diagram)

5. Over successive days, re-weigh both flasks and record any mass changes that have occurred.

6. Connect the rubber hose to a glass test tube containing limewater. Observe and record any

changes in appearance.

7. Perform the same steps with the control experiment without the addition of the yeast.



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Rubber stopper

500 mL beaker

Water bath

(30- 35°C)

Yeast + Glucose

solution

250 mL vacuum flask

with side arm

Observations and Results

• The initial yeast solution was milky yellow in colour. It was opaque and its smell resembled

chicken broth.

• After 1 day, a creamy layer was on top of the mixture. An odour resembling that of wine was

noted.

• After 2 days, the wine odour was much stronger. When the flask was swirled, the creamy

substance on top of the mixture dissolved. Sediment was observed at the bottom of the flask.

• The limewater went cloudy.

Time Mass (g)

10:30 Thursday 26/10 238.410

10:30 Friday 27/10 235.217

1:30 Monday 30/10 234.217



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Calculations

Mass of CO�lost = 238.410 – 234.217

= 4.193g

Therefore,

NO��PQR =

4.193

44.01

= 0.095TP�

UP�VQ PW XRℎ#NP� = UP�VQ PW ��PQR

= 0.095

U#QQ&)Z'*[\ = 0.095 x 46.07 = 4.39g

Discussion

There were several assumptions made in this experiment. The first was that all of the carbon dioxide

gas escaped and all the mass that was lost was due to the escape of carbon dioxide (we verify that

the identity of the gas is CO2 by passing it through limewater and noting the subsequent cloudiness).

Therefore, it is assumed no oxygen or water vapour passed into or out of the flask. Another

assumption made was that all the glucose fermented and there was none left.

However, there were errors in this experiment. If the amount of carbon dioxide lost and the amount

of ethanol produced is added, this amount is more than the original mass of the glucose showing a

violation of the Law of Conservation of Mass. Sources of experimental error include:

• Evaporation of water which we might assume was due to carbon dioxide loss.

• Lack of sensitivity in the electronic scales (only 1 decimal place)

• As �� leaves the vessel, air enters and occupies mass.

• Using a small mass of glucose means the mass of CO2 produced is extremely small. This could be

improved by using more glucose.

A control is used to show that any mass loss in the catalysed reaction is due to the production of CO2

and not the evaporation of the reaction mixture.

Conclusion: We were successfully able to plan an investigation which monitors the changes in mass

occurring during the fermentation of glucose.



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Question 1 (7 marks)

If Alistair prepares the solution we prepared above in the experiment and measures it he finds it

weighs 356.32g. Two days later he finds that it weighs 351.17g. Assuming all the glucose has

fermented and the mass loss occurs only because of carbon dioxide escape. Find:

a) The mass of carbon dioxide lost 1

b) The moles of carbon dioxide lost 1

c) The volume of carbon dioxide lost 1

d) The mass of ethanol produced 2

e) The original mass of glucose 2

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METAL DISPLACEMENT REACTIONS

This occurs because an electron transfer occurs between the ions. One metal loses electrons while

another gains the electrons this metal has just lost. For example consider if we put zinc in a solution

of copper ions:

The zinc loses electrons:

_N�(� → _N�'a��b + 2Vc

Meanwhile the copper ion gains two elections:

�d�'a��b + 2Vc → �d�(�

Therefore, the overall reaction becomes:

�d�'a��b + _N�(� → _N�'a�

�b + �d�(�

The above example clearly demonstrates metal displacement reactions as electron transfer

reactions. The 2 electrons that the zinc loses are transferred to the copper ion which is converted

to a neutral atom. The copper is said to have been DISPLACED

A metal displacement reaction occurs when a more reactive metal converts the ………………… of a

less reactive metal metal into a …………………………….. through the ………………………………………… .



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Thus, displacement reactions involve the transfer of electrons

But does this occur when any metal is place in a solution of metal ions? No. Only certain metals can

displace a particular metal ion from a solution; this is related to the reactivity of the two metals.

If we recall the metal activity series from the preliminary topic:

A metal further to the left will displace the ions of a metal to the right of it from solution. For

example consider if we added magnesium metal to cobalt ion solution. Since magnesium is more

reactive than cobalt, it will displace the cobalt ion and convert it to cobalt metal. Meanwhile, the

magnesium metal will be converted to U��b ion.

However, if we added iron to zinc ions in solution, nothing would happen. This is because iron is less

reactive than zinc, and therefore unable to displace it from solution.

The reaction of metals with dilute acids are also displacement reactions. The metal displaces

hydrogen gas from hydrogen ions in solution e.g.

U��(� + 2��'a�b → U��'a�

�b + ��e�

Talent Tip: If you are unsure of the reactivity of metals, use the standard table of potentials. Metals

higher on the table are more reactive, while those lower are less reactive.

Only a more active metal will ……………………………………………………………………………………………



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Worked Example 1

What occurs when copper is placed in silver ion solution? Write appropriate equations in your

answer.

Solution

Since copper is more reactive than silver it will displace the silver ions from solution. The copper

loses two electrons:

�d�(� → �d�'a��b + 2Vc

These electrons are transferred to the silver ion which gains electrons to be converted to neutral

silver metal

2f��'a�b + 2Vc → 2f��(�

This gives an overall reaction of:

2f��'a�b + �d�(� → �d�'a�

�b + 2f��(�

The copper has displaced the silver ion from solution.

Worked Example 2

What occurs when lead is placed nickel ion solution? Write appropriate equations in your answer.

Solution

Lead is less reactive than nickel so it will not be able to displace the nickel ion from solution. Hence

no reaction occurs.



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Determine what occurs when each of the following metals are placed in the particular metal ion

solutions specified (use equations where appropriate):

a) Magnesium in zinc ion 2

b) Nickel in cobalt ion 2

c) Tin in magnesium ion 2

d) Lead in silver ion 2

……………………………………………………………………………………………………………………………………………………………

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OXIDATION AND REDUCTION

Oxidation and reduction are chemical reactions that change the oxidation number of a particular

atom.

Oxidation State/Number essentially refers to the charge on the ion. The Oxidation State of an

element present in its stable element state is zero- regardless of the formula for that molecule or

element. This means that U�, ��, g�, ��, h, iV and �� all have an oxidation state of zero.

For ions, or when the element is part of a molecule, the oxidation number is the charge. So iV�b has

an oxidation number of +3, and ��c has an oxidation number of -2. In ��, oxygen has an oxidation

number of -2, and carbon has one of +4.

In a molecule, the oxidation number is equal to the sum of the oxidation numbers of the individual

atoms that make up that molecule.

Worked Example 3

What is the oxidation number of chromium in ��j��k��c ?

Solution

We know that the oxidation of ��j��k��c is equal to -2. This means if we add up the oxidation

numbers of each individual atom in this molecule, we should get an answer of -2. We know that

when it is within a molecule , oxygen has an oxidation number of -2.

So let the oxidation number of chromium = N. Note there are two chromium atoms here.

2g + 7l�−2� = −2.

2g − 14 = −2

2g = 12, g = 6.

The oxidation number of chromium is +6.



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Question 3 (1 mark)

What is the oxidation number of manganese in �UN�o�c"?

……………………………………………………………………………………………………………………………………………………………

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……………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………

A change in oxidation state of a species corresponds to a loss or gain of electrons

When these two reactions occur at the same time we get an overall REDOX reaction. Metal

displacement reactions are an example of redox reactions. The displaced metal gains electrons so it

is reduced. Meanwhile the metal that has done the displacing has lost electrons and been oxidised.

If we refer back to the activity series, we can see that as we move to the right, it becomes more

difficult to oxidise these metals. This also means it becomes easier to reduce them.

We can also tell this from the table of standard potentials, which is on your infocard. The meaning of

these numbers will become clearer later and for now this is a good indication of how easy a certain

species is to reduce or oxidise.

Talent Tip: Oxidation and reduction must occur simultaneously. A good mnemonic to remember the

difference between oxidation and reduction is OIL RIG: Oxidation Is Loss (of electrons); Reduction Is

Gain (of electrons)

………………… is a gain of electrons leading to a decrease (or reduction) in oxidation number

………………… is the loss of electrons leading to an increase oxidation number.



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A galvanic cell is a device that uses a spontaneous chemical reaction to …………………………………….

It converts ………………… energy into ………………… energy.

GALVANIC CELLS

A galvanic cell utilises electron transfer reactions to generate electricity by physically separating the

oxidation reaction and a reduction reactions. A galvanic cell consists of two-half cells. In one half-cell

oxidation occurs while in the other reduction occurs. These are connected by a conducting pathway

(such as an external wire or circuit) so that when electrons are released upon oxidation, they flow

through the circuit to the reduction half-cell where they are consumed. This electron flow

constitutes electrical energy and can be used to do work.

HOW GALVANIC CELLS OPERATE

The principle of a galvanic cell is best demonstrated through the use of an example. Consider we

have a piece of zinc metal dipped into zinc sulfate solution. This zinc metal is connected to a piece of

copper metal in copper sulfate solution by a wire. At the zinc half cell oxidation will occur:

_N�(� → _N�'a��b + 2Vc

Meanwhile at the copper half cell reduction will occur:

�d�(� → �d�'a��b + 2Vc



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Talent Tip: This cell can be represented using the shorthand Zn�q�/ Zn�st��b // Cu�st�

�b /Cu�q� where

the electrodes are written on the ends and the corresponding solutions in the middle. The left side

corresponds to the anode half cell and the right side corresponds to the cathode.

The electrons are released by oxidation travel from the zinc electrode to the copper electrode in the

reduction half cell which produces a current and voltage through the external wire.

The electrodes in the above cell are the zinc and copper metals. There are two electrodes in each

galvanic cell; an anode and a cathode.

In the case of our cell, the anode is the zinc metal as this is where oxidation occurs.

It this case the cathode is the copper metal as this is where reduction occurs. Since electrons flow

from the site of oxidation to the site of reduction, it follows that:

This is because electrons are liberated at the anode and consumed at the cathode. Another

important aspect of the galvanic cell is the electrolyte.

The electrolytes from the above cell are the copper sulfate and zinc sulfate solutions. Without the

electrolytes, the reactions would not be able to occur and there would be no voltage or current

generated.

Talent Tip: The Anode is where oxidation occurs, and reduction occurs at the cathode good way to

remember the definition of anode and cathode is by the mnemonics AN OX and RED CAT.

An electrolyte is a substance which in solution or molten state conducts electricity

In a galvanic cell, electron flow ALWAYS occurs from the ………………… to the ………………….

Electrodes: The metallic conducting plates

Anode: The electrode at which oxidation occurs. It is marked as negative (-) in galvanic cells.

Cathode: The electrode at which reduction occurs. It is marked as positive (+) in galvanic cells.



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THE SALT BRIDGE

Since we are removing electrons from the anode, the anode compartment becomes more and more

positively charged. Similarly, the cathode compartment becomes more and more negatively charged

due to an excess of electrons and because positive ions are being consumed in the reaction. Thus if

electrons were the only moving charged particles, there would be an electrical imbalance and the

cell would not function. Therefore, galvanic cells require salt bridges.

The salt bridge used is usually a piece of filter paper or chromatography paper soaked in an

electrolyte such as ug�� which joins the two half cells. The salt bridge ensures electrical neutrality

between the two half cells, allowing the cell to continue functioning. It provides anions �g��c�

which flow to the anode to make up for the loss of negative charge; and cations �ub� which flow to

the cathode to make up for the excess negative charge. It is easy to remember that:

From the salt bridge:

Anions flow to the anode

Cations flow to the cathode

The salt bridge maintains electrical neutrality and completes the circuit by allowing the migration

of ions to occur between the two half cells.



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Galvanic cells have a number of components

a) Define an electrode 1

……………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………

b) Define an electrolyte 1

……………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………

c) Identify the purpose of the salt bridge? 1

……………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………

……………………………………………………………………………………………………………………………………………………………

All these features of galvanic cells are evident on the diagram on page 14.

Summary: How a Galvanic Cell operates



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The following combinations of half-cells are set up:

1. Copper in copper nitrate solution and silver in silver nitrate solution.

2. Zinc in zinc sulphate solution and cobalt in cobalt nitrate solution

3. Magnesium in magnesium nitrate solution and silver in silver nitrate solution

4. Iron (II) in Iron (II) nitrate and aluminium in aluminium nitrate solution

Draw a sketch of each of these combinations and answer the following questions:

a) Which is the anode? 1

b) Which is the cathode? 1

c) Write reaction half equations for each half cell and an overall reaction equation 3

d) Draw a labelled diagram of the galvanic cell showing all important features including the 4

direction of electron and ion flow

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B A

C V

PRACTICAL

INVESTIGATING THE CONDITIONS NEEDED FOR A GALVANIC CELL

Aim: To identify the conditions under which a galvanic cell is produced

Equipment list

• 2 pairs of copper electrodes.

• 2 pairs of zinc electrodes.

• A pair of graphite electrodes

• 1 mol / L of sodium chloride solution

• 1 mol/ L of sulfuric acid solution

• 150 ml Beakers

• Electrical leads

• Distilled water

• Safety glasses

• Voltmeter

• Alligator clips

Risk Assessment: Sulfuric acid is highly corrosive so it is advisable to wear safety goggle to avoid any

acid splash in the eyes. The voltages created are too low to cause a risk so this experiment would

therefore be a low risk experiment.

Method

1. Set up the apparatus as shown.



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2. Construct the following 6 galvanic cells and vary the electrodes A, B and the electrolytes A, B and

C in the following combinations.

3.

Experiment No Electrode A Electrode B Electrolyte C

1 Zn C Distilled Water

2 Zn Cu IM g#v�

3 Zn Cu 1M �2h�4

4 Zn Cu 1M �2h�4 , Touch the bottoms of A and

B together

5 Zn Zn 1M �2h�4

6 Zn C 1M �2h�4

4. Observe and record any signs of chemical reaction

5. Measure the voltage reading

Results

Experiment No Electrode A Electrode B Electrolyte C Results

1 Zn C Distilled Water Nothing occurs

2 Zn Cu 1M g#�� 0.75V reading on voltmeter

3 Zn Cu 1M �2h�4 1V on voltmeter which falls to

0.95 over time. Zinc bubbles

violently

4 Zn Cu 1M �2h�4 , Touch

the bottoms of A

and B together

Voltage drops to 0 as there is a

short circuit

5 Zn Zn 1M �2h�4 Nothing occurs

6 Zn C 1M �2h�4 Reading on Voltmeter



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Discussion

Cell 1 did not function like Cells 2 and 3 because the electrolyte in Cell 1 was weak. This meant that

there were no free ions so there was no reduction occurring at the copper electrode. There also was

no oxidation of zinc at the zinc electrode and no voltage could be generated. It can be concluded

that a strong electrolyte is required between the electrodes for an effective cell.

At Cell 4, the electrodes are touching each other creating a short circuit. This means that the current

does not go through the external circuit. It can be concluded that in a galvanic cell the electrodes

cannot be touching and they must be in contact with an electrolyte between them

Cell 5 had copper electrodes in both compartments. This meant that the electrodes were at the

same potential and there was no potential difference between them so there was no current or

voltage. It can be concluded that a potential difference between the electrodes is required for an

effective galvanic cell.

It can be concluded from these experiments that a galvanic cell forms when two electrodes of

different potential are placed in a strong electrolyte. The electrolyte causes reduction at the cathode

and oxidation at the anode and thus a net flow of charge between the electrodes.

Conclusion: We were successfully able to identify the conditions under which a galvanic cell is

produced



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CALCULATING THEORETICAL POTENTIALS PRODUCED BY

GALVANIC CELLS

Recall the table of standard electrode potentials we first encountered on page 7. Here it is again:

1

1 Table of Potentials from Board of Studies NSW © 2007

Ea

se o

f re

du

ctio

n i

ncr

ea

ses

Ea

se o

f ox

ida

tion

incre

ase

s



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This table shows the standard electrode potential for each reduction reaction. It essentially shows

the voltage that would be generated if we placed that electrode in a cell in which the standard

hydrogen electrode is the other electrode. More about the standard hydrogen electrode follows on

the next page.

A positive reduction potential indicates that that process occurs spontaneously and thus liberates

energy. Conversely, a negative standard reduction potential indicates that that process requires an

input of energy to occur (i.e. it does not occur spontaneously).

A STANDARD HYDROGEN ELECTRODE CONSISTS OF:

• A piece of Platinum metal immersed in a 1.000mol/L solution of HCl (note the significant

figures).

• Hydrogen gas being bubbled through at pressure of 100.0kPa

• Temperature of 25°C

It is used as a reference electrode to measure the potential of other electrodes in their standard

state. This is because cell emf depends upon variables such as concentration of species present,

temperature and pressure.

Also, note that the term “standard electrode potential” may only be applied to reduction half

equations and never oxidation half reactions.

The table of standard potentials is very useful in determine the theoretical voltages/emf generated

by galvanic cells. In order to do this we must know the two half equations. The following is the

procedure for determining the voltage- or cell emf (if we do not know which species undergoes

oxidation and which undergoes reduction):

The ……………………………………………………… of an electrode refers to the potential of that electrode

in its standard state relative to the standard hydrogen electrode



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1. Identify the half-cell equations and their standard reduction potentials:

f��'a�b + Vc → f��(�, Xw = +0.80 x

_N�'a�b� + 2Vc → _N�(�, Xw = −0.76 x

2. Select the half-cell reaction with the most positiveXw, and write the half-equation and Xw value.

This half-reaction is the reduction reaction (this occurs because as we go down the left hand side

of the table, the ease of reduction for each ion increases. The ion that is easier to reduce will,

logically, be reduced):

f��'a�b + Vc → f��(�, Xw = +0.80 x

3. Reverse the direction of the other half-cell reaction and change the sign of Xw. This half-reaction

is the oxidation reaction (this occurs because the reduction of the other ion is more favourable

so this reaction will become an oxidation):

_N�(� → _N�'a�b� + 2Vc, Xw = 0.76 x

The balanced equation is:

2f��'a�b + _N�(� → 2f��(� + _N�'a�

b�

Talent Tip: To get the oxidation equation, we reverse the reduction half equation. In doing so, we

must reverse the sign of the Xw value

4. Calculate the cell voltage by adding the half-cell potentials.

Xw = 0.80 + 0.76

= 1.56 x

Talent Tip: From the balanced equation we saw that two times the silver ions are required to react

with zinc. When calculating the cell voltage, the ratio does not affect the calculated EMF and thus we

do not need to multiply 0.80 V by 2 for silver. The cell voltage of a galvanic cell must always be

positive. If we already know which is the reduction half equation and oxidation half equation we can

skip to step 3 and merely reverse the standard emf for the oxidation equation.

EMF�{|{s}� = EMF�~��{�|� �t�s{�|�� + EMF�|��s{�|� �t�s{�|��



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We can use our knowledge of the standard table to explain the observations we made on page 7 and

8 about of oxidation and reduction:

1. As we move down the table, the reduction potentials become more positive. Therefore, the

reduction reactions become more spontaneous and the ease of reduction for the species on the

left hand side increases.

2. As we move up the table, the oxidation potentials become more positive. Therefore, the

oxidation reactions become more spontaneous and the ease of oxidation for the species on the

right hand side increases.

Worked Example 4

Find the voltage generated when a galvanic cell is made of magnesium in magnesium nitrate

solution and zinc in zinc nitrate solution.

Solution

The two equations are:

_N�'a�b� + 2Vc → _N�(�, Xw = −0.76 x

U��'a��b + 2Vc → U��(� Xw = −2.36 x

From the equations we can see that zinc ion is more easily reduced so therefore the magnesium is

oxidised rather than reduced and we reverse the equation and the voltage:

U��(� → U��'a��b + 2Vc Xw = 2.36 x

Therefore this gives a net equation of:

U��(� + _N�'a�b� → _N�(� + U��'a�

�b

Therefore, the cell voltage is:

Xw = 2.36 + −0.76

= 1.6x



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Calculate the voltage generated by a galvanic cell with the following combinations of half-cells.

Assume standard conditions:

a) magnesium and magnesium nitrate with copper (II) and copper (II) nitrate 1

b) aluminium and aluminium nitrate with and silver and silver nitrate 1

c) iron and iron (II) nitrate with silver and silver nitrate 1

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PRACTICAL

POTENTIALS OF DIFFERENT METAL COMBINATIONS IN ELECTROLYTE

Aim: To measure the difference in potential of different combinations of metals in electrolyte

solution.

Equipment list

• Pair of zinc electrodes

• Pair of copper electrodes

• Pair of magnesium electrodes

• Pair of lead electrodes

• 1M zinc sulfate

• 1M copper sulfate

• 1M magnesium sulfate

• 1M lead nitrate

• 150mL beakers

• Chromatography paper

• Potassium nitrate solution

• Voltmeter

• Electrical Leads

• Alligator clips

• Emery paper

Risk Assessment: Lead nitrate is toxic so we must ensure we wash our hands after this experiment

and clean up any spills of lead nitrate immediately. Do not dispose of lead solutions down the sink.

Overall, this is a low risk experiment.



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V

S

B A

A B

Method

1. Clean the metal electrodes with emery paper.

2. Set up the galvanic cell as follows:

3. Construct the following galvanic cells

Cell Beaker A Beaker B

1 Cu + 1M CuSO4 Pb + 1M Pb(NO3)2

2 Mg + 1M MgSO4 Pb + 1M Pb(NO3)2

3 Zn + 1M ZnSO4 Pb + 1M Pb(NO3)2

4 Cu + 1M CuSO4 Mg + 1M MgSO4

5 Zn + 1M ZnSO4 Cu + 1M CuSO4

6 Zn + 1M ZnSO4 Mg + 1M MgSO4

4. Record the measured net voltage and the polarity of the electrodes



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Results

Cell Beaker A Beaker B Result

1 Cu + 1M CuSO4 Pb + 1M Pb(NO3)2 0.48V

2 Mg + 1M MgSO4 Pb + 1M Pb(NO3)2 -1.4V

3 Zn + 1M ZnSO4 Pb + 1M Pb(NO3)2 -0.6V

4 Cu + 1M CuSO4 Mg + 1M MgSO4 1.95V

5 Zn + 1M ZnSO4 Cu + 1M CuSO4 -0.8V

6 Zn + 1M ZnSO4 Mg + 1M MgSO4 0.9V


For all the above cells:

a) Find the cathode and anode 1

b) Write the oxidation and reduction half equations 2

c) Write the net ionic equation 1

d) Calculate the theoretical cell voltage 1



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Discussion

The cell voltages we obtained were lower than the theoretical cell voltages we calculated above.

This occurred for a number of reasons:

• We used low quality voltmeters which created a load on the circuit causing the voltage to drop

significantly

• While the concentrations were kept at approximately to 1M (1 significant figure), we could not

ensure standard conditions were maintained i.e. solute concentration of 1.000mol/L (4

significant figures), temperature of 25°C, and pressure of 100.0kPa. This may have affected the

voltages slightly

• The voltmeter had markings only at every 0.1V increment so there may have been errors reading

the values of the analogue voltmeter. This could be improved by having a more sensitive, digital

voltmeter.

• The metals may not have been completely pure. This is why we cleaned the electrodes with

emery paper to remove oxide layers and other impurities.

• There may have been impurities in the electrolyte solutions.

Conclusion: We were successfully able to measure the difference in potential of different

combinations of metals in electrolyte solution.

HC Notes Talent 100

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