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HSC Chemistry

Production of Materials

Week 6

Student name: ……………………………….

Class code: ……………………………………..

Teacher name: ……………………………….

T: (02) 8007 6824 E: [email protected]

W: www.duxcollege.com.au DUX

HSC Chemistry 1


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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

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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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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Measured potentials

Cell Measured potential Theoretical potential

Cu | Cu2+ || Ag+ | Ag 0.3v 0.46v

Zn | Zn2+ || Cu2+ | Cu 0.7v 1.1v

Mg | Mg2+ || Cu2+ | Cu 1.8v 2.71v

Mg | Mg2+ || Zn2+ | 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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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:

Anode: Pb(s) + H2SO4(aq) PbSO4(aq) + 2H+(aq) + 2e-

Cathode: 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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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:

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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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:

Anode: V2+ V3+ + e-

Cathode: 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:

Anode: vanadium (II) vanadium (III) + e-

Cathode: 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 Advantages of the lead acid 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

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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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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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 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

2Hg2+ + 2e- Hg22+ +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

Cu2+ + 2e- Cu +0.34

Sn4+ + 2e- Sn2+ +0.15

2H+ + 2e- H2 0.00 (by definition)

Pb2+ + 2e- Pb -0.13

Sn2+ + 2e- Sn -0.14

Ni2+ + 2e- Ni -0.26

Co2+ + 2e- Co -0.28

PbSO4 + 2e- Pb + SO42- -0.36

Cd2+ + 2e- Cd -0.40

Fe2+ + 2e- Fe -0.45

Zn2+ + 2e- Zn -0.76

Al3+ + 3e- Al -1.66

Mg2+ + 2e- Mg -2.37

Na+ + e- Na -2.71

Ca2+ + 2e- Ca -2.87

Ba2+ + 2e- Ba -2.91

K+ + e- K -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

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

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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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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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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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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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

2Hg2+ + 2e- Hg22+ +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

Cu2+ + 2e- Cu +0.34

Sn4+ + 2e- Sn2+ +0.15

2H+ + 2e- H2 0.00 (by definition)

Pb2+ + 2e- Pb -0.13

Sn2+ + 2e- Sn -0.14

Ni2+ + 2e- Ni -0.26

Co2+ + 2e- Co -0.28

PbSO4 + 2e- Pb + SO42- -0.36

Cd2+ + 2e- Cd -0.40

Fe2+ + 2e- Fe -0.45

Zn2+ + 2e- Zn -0.76

Al3+ + 3e- Al -1.66

Mg2+ + 2e- Mg -2.37

Na+ + e- Na -2.71

Ca2+ + 2e- Ca -2.87

Ba2+ + 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

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

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

b. Ag | Ag+ || Br-, Br2 | Pt

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

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c. Ni | Ni2+ || Cl- | Cl2, Pt

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

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

d. Ag | Ag+ || Cl- | Cl2, Pt

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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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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]

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4. Explain the need to define a particular cell voltage (2H+ + 2e- H2) to equal exactly zero. [1

mark]

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End of homework

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HSC Chemistry - Dux College · PDF fileask along the lines if ^In your study of HSC Chemistry,...

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