Download Unit 8: Oxidation and Reduction

Unit 8: Oxidation and Reduction
Lesson 1: Introduction to Oxidation and Reduction Reactions
Redox (reduction and oxidation) reactions form one of the largest groups of chemical reactions, and
include the reactions involved in rusting, combustion, batteries, respiration, photosynthesis and
ageing of the human body.
Check out the podcast on how oxidation reactions in apples causes them to brown
http://chemicalparadigms.wikispaces.com/Chemistry+Podcast
Magnesium burns brightly in oxygen to produce the white powder magnesium oxide.
2Mg(s) + O2(g) → 2MgO(s)
For some time, chemists classified reactions like this, where metal elements like magnesium gained
oxygen atoms, as oxidation. The opposite, removal of oxygen from a compound, was called
reduction.
The term ‘reduction’ was in use long before the term ‘oxidation’. Reduction was used to describe the
process of extracting a metal from its compound, such as the copper from copper oxide.
CuO(s) + H2(g) → Cu(s) + H2O(g)
The process by which the copper oxide loses oxygen atoms is called reduction. The definitions of
oxidation and reduction were later extended to include the removal and addition of hydrogen.
Check your Understanding
Deduce which reactant is reduced and which oxidized in the reaction:
CuO(s) + H2(g) → Cu(s)+ H2O(l)
These definitions, however, did not explain how the reactions occurred. Consider again the reaction
of magnesium with oxygen in terms of electrons. When the magnesium atom loses two electrons it
becomes a magnesium ion.
Mg → Mg2+ + 2e–
Each oxygen atom in the oxygen molecule becomes an oxide ion when it gains electrons.
O2 + 4e– → 2O2–
When written in this form, known as a half–equation, the reaction can be seen clearly to involve the
transfer of electrons. The loss of electrons by the magnesium atom is oxidation. The gain of electrons
by the oxygen atoms is reduction.
The two half equations can be combined to write an overall ionic equation for the reaction. The final
equation does not show the electrons being transferred and so the equations need to balance to cancel
them out. In this reaction the magnesium half equation is multiplied by two to make four electrons.
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Oxidation ½ equation
(Mg → Mg2+ + 2e–) x 2
Reduction ½ equation
O2 + 4e– → 2O2–
________________
2 Mg + O2 + 4e– → 2 Mg2+ + 2O2–+ 4e–
Ionic Equation
2 Mg + O2 → 2 Mg2+ + 2O2–
Oxidation is defined as the loss of electrons and reduction as the gain of electrons
A useful way to remember them is by using a mnemonic.
O
I
L
oxidation
is
Loss (of electrons)
R
I
G
Reduction
is
Gain (of electrons)
L
E
O
Loss
electrons
oxidation
G
E
R
Gain
electrons
reduction
or
Check your Understanding
For the following reactions, write the oxidation and reduction half equations, and overall ionic
equation.
1. F2(g) + Ni(s) → NiF2(aq)
2. Cu2+(aq) + Fe2+(aq) → Cu(s) + Fe3+(aq)
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Oxidation Numbers (States)
In the redox reaction between magnesium and oxygen, the transfer of electrons was easily seen
because ions were formed. However, electron transfers are not always obvious. Consider the
following redox reaction:
C(s) + O2(g) → CO2(g)
The element carbon has gained oxygen and so has been oxidised. The oxygen molecule has been
reduced, but this is not so obvious. To help identify which reactants have been oxidised or reduced
chemists have developed a system of oxidation numbers (ON).
Oxidation numbers are not real but are numbers assigned to atoms to show the transfer of electrons.
The table below shows the rules used for assigning oxidation numbers.
1
2
3
4
Rule for assigning oxidation numbers
The oxidation number of each atom in a pure
element is zero.
Examples of application of the rule
Zn, O in O2, and P in P4 all have an oxidation
number of zero.
The oxidation number of an atom in a monatomic
ion is equal to the charge on the ion.
Na+ has an oxidation number of +1.
S2– has an oxidation number of –2.
In compounds containing oxygen, each oxygen
atom has an oxidation number of –2 (exceptions
are OF2, where the oxidation number of oxygen is
+2, and peroxides (O22–), where it is –1).
In compounds containing hydrogen, each
hydrogen atom has an oxidation number of +1
(exceptions are metal hydrides like NaH, where
the oxidation number of hydrogen is –1).
In H2O and CO2 each oxygen atom has an
oxidation number of –2.
In H2O2 each oxygen atom has an oxidation
number of –1.
In NH3 and H2O each hydrogen atom has an
oxidation number of +1.
In KH the hydrogen atom has an oxidation
number of –1.
The sum of the oxidation numbers of the
atoms in CH4 is zero. Since each hydrogen
atom has an oxidation number of +1, the
oxidation number of the carbon atom is –4 (–
4 + (4 × +1) = 0).
The sum of the oxidation numbers of the
atoms in PO43– is –3. Since each oxygen
atom has an oxidation number of –2, the
oxidation number of the phosphorus atom is
+5 (+5 + (4 × –2) = –3).
5
For a molecule, the sum of the oxidation numbers
of the atoms equals zero.
6
For a polyatomic ion, the sum of the oxidation
numbers of the atoms equals the charge on the ion.
7
When neither oxygen nor hydrogen is present in a
compound, the most electronegative atom is
assigned the negative oxidation number (equal to
the charge on the anion).
In SF6, the oxidation number of each fluorine
atom is –1. The oxidation number of the
sulfur atom is +6 (+6 + (6 × –1) = 0).
Notice that when oxidation numbers are written the number is preceded by a plus or minus
sign. Also note that in the examples given oxidation numbers are whole numbers. This is
usually, but not always, the case. Oxidation numbers are also based on one mole of the substance.
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Example 1:
(i) H2S
(ii) H2SO4
(iii) KClO4
Determine the oxidation number of each element in each of the following species:
Answer
(i) Rule 4: Each hydrogen atom has an oxidation number of +1 (which gives a total of 2 x +2 =
+2).
Rule 5: The sum of the oxidation numbers is 0 for a molecule.
Therefore the oxidation number of S is –2.
(ii) Rule 4: Each hydrogen atom has an oxidation number of +1 (a total of 2 x +1 = +2).
Rule 3: Each oxygen atom has an oxidation number of –2 (a total of 4 x-2 = –8).
Rule 5: The sum of the oxidation numbers is 0 for a molecule.
Therefore the oxidation number of S is +6 (+2 + 6 – 8 = 0).
(iii) KClO4 is an ionic compound, with K+ and ClO4– ions.
Rule 2: The oxidation number of K is +1.
Rule 3: Each oxygen atom has an oxidation number of –2 (a total of 4 x-2 = –8).
Rule6: The sum of the oxidation numbers of the Cl atom and the four O atoms is –1, the charge
of the ion.
Therefore the oxidation number of Cl is +7 (+7 – 8 = –1).
Recall that oxidation numbers are not real. When we say that carbon has an oxidation number of +4
in CO2, we do not mean that the C+4 ion actually exists. The +4 is assigned to help us identify
whether oxidation or reduction has occured. Consider again the reaction of carbon with oxygen to
produce carbon dioxide. Notice that the oxidation numbers are written above each element.
0
0
+4 –2
C(s) + O2(g) → CO2(g)
The oxidation number of carbon has increased from 0 to +4. This increase in oxidation number
shows that oxidation has occurred. For oxygen, the oxidation number has decreased from 0 to –2.
This decrease in oxidation number shows that reduction has occurred.
We now have four definitions of oxidation:
1. Gain of oxygen
2. Loss of hydrogen
3. Loss of electrons
4. Increase in oxidation number
While any of these definitions may be used to recognise a redox reaction, in practice it is the one
based on oxidation numbers that is usually the most convenient, because it works each time.
Example 2:
Use oxidation numbers to determine which reactant has been oxidised and which has
been reduced.
SO2(g) + NO2(g) → SO3(g) + NO(g)
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Answer
Assign oxidation numbers to all atoms.
+4 (–2x2=-4)
SO2(g)
•
•
•
+
+4 (–2x2=-4)
NO2(g)
→
+6 (–2x3=-6)
SO3(g)
+
+2 –2
NO(s)
The oxidation number of sulfur in SO2 has changed from +4 to +6. An increase in oxidation
number shows that oxidation has occurred.
The oxidation number of nitrogen in NO2 has changed from +4 to +2. A decrease in
oxidation number shows that reduction has occurred.
Oxygen is not oxidised or reduced because in the reaction it is in it’s compound form and will
have an oxidation number of -2. Its oxidation number does not change.
Test your Understanding:
1. Determine the oxidation number of each element in each of the following species.
a) MgO
b) H2O
c) HPO4–
d) C2H4
e) N2O4
f) KMnO4
g) CuO
h) CuO2
i) MnO2
j) V2O5
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2. Deduce the change in oxidation number of chromium in the reaction below. State with a
reason whether the element chromium has been oxidized or reduced.
Cr2O72– + 14H + + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O
3. Use oxidation numbers to identify which reactant has been oxidised in each of the following
redox reactions:
a) CuO(s) + H2(g) → Cu(s)+ H2O(l)
b) MnO2(s) + SO2(aq) → Mn2+(aq) + SO42–(aq)
4. Use oxidation numbers to show that glucose is oxidized in cellular respiration reaction.
C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O(l)
5. Which statement is a correct description of electron loss in a reaction?
2 Al + 3 S
A.
B.
C.
D.
→ Al2S3
Each aluminium atom loses two electrons
Each aluminium atom loses three electrons
Each sulphur atom loses two electrons
Each sulphur atoms loses three electrons
6. What are the oxidation numbers of the elements in the compounds phosphoric acid, H3PO4?
A.
B.
C.
D.
Hydrogen
+1
+1
+3
+3
Phosphorus
+1
+5
+1
+5
Oxygen
-2
-2
-4
-8
7. Which statement is correct?
A.
Oxidation involves loss of electrons and a decrease in oxidation state.
B.
Oxidation involves gain of electrons and an increase in oxidation state.
C.
Reduction involves loss of electrons and an increase in oxidation state.
D.
Reduction involves gain of electrons and a decrease in oxidation state.
8. Which are examples of reduction?
I.
Fe3+ becomes Fe2+
II.
Cl- becomes Cl2
III.
CrO3 becomes Cr3+
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III only
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9. The oxidation number of chromium is the same in the following compound except
A.
Cr(OH)3
B.
Cr2O3
C.
Cr2(SO4)3
D.
CrO3
Lesson 2:
Oxidizing and Reducing Agents
Consider the following reactions:
CuO(s) + Zn(s) → ZnO(s) + Cu(s)
Cu(s)
+
2 AgNO3(aq)
→
2Ag(s)
+
Cu(NO3)2(aq)
Using your current knowledge and the following statements dfor each reaction the oxidizing and
reducing agents.
A reducing agent causes reduction
An oxidizing agent causes oxidization
The reaction:
CuO(s) + Zn(s) → ZnO(s) + Cu(s)
can be written as two ionic half equations
Oxidation of zinc:
Zn → Zn2+ + 2e-
Reduction of copper ions:
Cu2+ + 2e- → Cu
The overall ionic equation is:
Zn + Cu2+ → Zn2+ + Cu
During this reaction, two electrons are transferred from the zinc to the copper ions. The oxidation
number of the element oxygen does not change. It is not oxidized or reduced, so it is not included in
the half reactions.
The element zinc donates its electrons to the copper ions, which allows the copper oxide to be
reduced. The element zinc is therefore called a reducing agent. Similarly, the copper ion is an
electron acceptor, it accepts two electrons from the zinc which allows the zinc to be oxidised. The
copper ion is therefore called an oxidising agent.
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A reducing agent causes another reactant to be reduced
An oxidizing agent causes another reactant to be oxidized
Zinc atoms donate electrons and are oxidised. Copper (II) ions accept electrons and are therefore
reduced. It is important to realize that in a redox reaction the reducing agent will always be oxidized
and the oxidizing will always be reduced.
When copper filings are placed in silver nitrate solution, AgNO3, the solution will gradually turn
blue and the solid will be a dark grey colour.
The equation for this reaction is
Cu(s)
+
Red-brown solid
2 AgNO3(aq)
colorless solution
→
2Ag(s)
silver-grey solid
+
Cu(NO3)2(aq)
blue solution
Examination of the oxidation numbers shows that the oxidation number of copper metal increases
from 0 to +2, while that of the silver ion decreases from +1 to 0. The copper is being oxidised and
the silver ion is being reduced.
The reactant that must be allowing the copper metal to lose electrons (be oxidized) is the silver ion,
Ag+, so it is the oxidizing agent. It is itself being reduced and so gains electrons to form Ag metal.
The reactant that is allowing silver ions to gain electrons (be reduced) is the element Cu, so it is the
reducing agent. It is itself being oxidized and so loses electrons to form copper ions, Cu2+.
Check your Understanding
For this reaction between Copper and Silver Nitrate write the:
Oxidation ½ reaction: ………………………………………………………………………………..
Reduction ½ reaction: ………………………………………………………………………………..
Ionic equation: ………………………………………………………………………………………...
Give a reason why the nitrate ions, NO3- are spectators?
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Colors of Oxidizing and Reducing Agents
The substances listed below are available for your observation. Deduce the colors of the ions using
the following reasoning:
• KCl is colorless
• so K+ (aq) and Cl- (aq) must also be colorless.
• K2Cr2O7 is orange
• Since K+ (aq) is colorless Cr2O72- (aq) must be orange and therefore the ion responsible for
the color.
Most oxidizing agents are made up of two ions. One of these ions will be the oxidizing agent the
other a spectator that does not take part in oxidation and reduction. It can be very difficult to
determine which one is the spectator ion, but often they are colorless. A familiarity of known
oxidizing agents and knowledge of common spectator ions will help.
Formula
Color of the
solution
Formula of the
ions present in
the solution
Formula of the
ions responsible
for the color
Formula of the
spectator ion
KCl
K2Cr2O7
K2CrO4
Cr(NO3)3
MnCl2
KBr
KMnO4
FeCl3
H2SO4
FeSO4
Cu(NO3)2
CuSO4
CoCl2
What do you notice about the nature of oxidizing agents, reducing agents and spectator ions?
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Test your Understanding
1. In each of the following equations, identify the oxidizing agent and the reducing agent.
a) Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)
b) 2Mg(s) + O2(g) → 2MgO(s)
c) CuO(s) + H2(g) → Cu(s) + H2O(g)
d) Cl2(g) + 2HI(aq) → 2HCl(aq) + I2(s)
2. When zinc powder is placed in a solution of hydrochloric acid zinc chloride and hydrogen
gas are formed.
a) Write an equation for the reaction
b) Write half-equations and an overall ionic equation for this reaction.
c) Identify the oxidizing agent and the reducing agent.
3. Given the equation:
Mg(s) + CuSO4(aq) → Cu(s) + MgSO4(aq)
a) Determine the oxidation number of magnesium and copper in the reactants.
b) Identify the oxidising agent in the reaction given in a).
c) Name the spectator ion.
4. When sodium burns in air, sodium oxide is produced. The equation for the reaction is:
Na(s) + O2(g) → Na2O(s)
a) Use oxidation numbers to identify which reactant is oxidised and which is reduced
b) Write half equations to show the oxidation and reduction reactions
c) Identify the oxidizing agent and the reducing agent.
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Lesson 3: Another Look at Oxidation and Reduction
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Lesson 4:
Reactivity Series
Redox reactions, involving metals placed in a metal ion solution, are also known as single
displacement reactions. In the reaction below Cu metal has pushed or displaced by Ag+ ions. The
metal and ion switch places.
Cu (s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
Cu (s) + 2 Ag+(aq) → 2 Ag(s) + Cu2+ (aq)
During this displacement reaction the Cu metal acts as a reducing agent and is oxidised. Metals
generally act as reducing agents (electron donors) because they lose outer-shell electrons to become
stable.
Metals differ in their ability to act as reducing agents. Sodium (Na), is so readily oxidised by water.
Others, like iron (Fe), are oxidized (corrode) readily in moist air. Gold (Ag), is very hard to oxidise
and so does not form compounds readily. This is why gold is found in its pure element state in
nature.
Check your Understanding
Write an equation for sodium metal reacting with water to show how it is oxidized.
Why do you think sodium metal is stored under oil?
By observing how readily metals react with oxygen, water, dilute acids and their ions, it is possible
to determine an order of reactivity of metals. The table below shows such a ranking, or activity
(reactivity) series.
Look at the reactivity series below and describe what you see.
Reactivity Series of Metals
K (s) ↔ K+(aq) + eIncreasing
Reactivity
Increasing
Strength as
Reducing
Agent
Increasing ability to
lose electrons
(become
oxidised)
Na (s) ↔ Na+(aq) + eLi (s) ↔ Li+(aq) + eCa (s) ↔ Ca2+(aq) + 2eMg (s) ↔ Mg2+(aq) + 2eAl (s) ↔ Al3+(aq) + 3eZn (s) ↔ Zn2+(aq) + 2eFe (s) ↔ Fe2+(aq) + 2ePb (s) ↔ Pb2+(aq) + 2e½ H2 (g) ↔ H+(aq) + eCu (s) ↔ Cu2+(aq) + 2eAg (s) ↔ Ag+(aq) + e-
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Understanding the reactivity series
•
•
The reactions in a reactivity series are by convention all written as reduction reactions where
electrons are gained by the reactants.
At the top of the series are the most reactive metals, those in Group 1 and 2 because they lose
electrons (become oxidised) more easily. They are the strongest reducing agents.
•
Towards the bottom are Group 3, 4 and transition metals.
•
Moving up the series metals are more easily oxidized and become more powerful reducing
agents. This is because it easier for them to lose electrons.
•
Metals above hydrogen will be oxidized by hydrogen ion in acid to form a salt and hydrogen gas.
Metals below hydrogen cannot displace hydrogen ions, but they may still react with acids
because the negative ion of some acids can act as an oxidizing agent.
For example zinc will react with hydrochloric acid:
Overall equation:
Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2(g)
Oxidation ½ reaction
Zn (s) → Zn2+(aq) + 2e-
Reduction ½ reaction
H+(l) + e- → ½ H2(g)
Deduce the overall ionic equation for this reaction
The reactivity series gives information on which displacement reactions will occur, but not on how
fast they will occur. This will depend on the factors that affect the rate of a reaction like
concentration and temperature.
Deducing the Feasibility of a Redox Reaction
Any metal in its element form will be oxidised by the ion of a less-reactive metal.
Steps
1.
Write a balanced equation for the reaction.
2.
Write the oxidation half reaction and the reduction half reaction using oxidation numbers or
the gain and loss of electrons. Do not include spectator ions.
3.
Find the position of each reaction on the reactivity series. The reactant on top, is the most
reactive and will be oxidized and the one below reduced.
4.
If the oxidation reaction is below the reduction reaction then the reaction will not occur.
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Example 1
The Thermit reaction
According to the reactivity series, aluminium is more reactive than iron. So, in this reaction iron (III)
oxide, Fe2O3 is reduced by reacting its with aluminium, a metal higher in the reactivity series than
iron. When Thermit is ignited, the reaction is rapid and very exothermic, and the iron formed is
molten.
Fe2O3(s) + 2Al(s) → 2Fe(l) + Al2O3(s) + heat
Al (s) → Al3+(aq) + 3eFe3+(aq) + 3e- → Fe (s)
______________________________________
Al loses electrons - oxidised
Fe3+ gains electrons - reduced
Al (s) + Fe3+(aq) → Al3+(aq) + Fe(s)
Video of the thermit reaction can be found on YouTube.
Mythbusters http://www.youtube.com/watch?v=PPAYZMzGMwQ&feature=related
Steve Spangler Science http://www.youtube.com/watch?v=O5v3XxFfUOw
Test your Understanding
1. Predict whether the following reaction will occur:
a)
zinc and copper sulphate
b)
magnesium nitrate and lead
c)
copper and zinc chloride
d)
An iron nail and hydrochloric acid solution
e)
Silver and hydrochloric acid solution.
Explain your reasoning for each reaction. Write half equations and ionic equations for those
reactions that took place.
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2. Consider the following advertisement
Connoisseurs Silver Polish is formulated to gently clean and safely remove even the heaviest
tarnish from silver. The quick rinse formula lets you finish the job in less time. Your silver is
left with a sparkling shine and an anti-tarnish shield that controls tarnish build-up for
months.
When silver (Ag) reacts with naturally occurring hydrogen sulfide in the air it tarnishes. The
tarnish is a black coating of silver sulfide, Ag2S.
Ag + H2S → Ag2S
An easy way to clean silver at home without store bought products is to put the tarnished
silver in an aluminum dish and heat it in water. The chemistry of the reaction involves the
silver ion in silver sulfide being reduced by the more reactive aluminum.
a)
Using your knowledge of reactivity series prove using equations if this home cleaning
method will work.
b)
Give the name of another metal that can be cleaned this way.
c)
Advertisers often use the science to promote the effectiveness of their products and
make it claims more believable. How does the language in the advertisement for
Connoisseurs silver polish use the authority of science to sell silver polish.
f)
Are the claims made by the advertiser scientifically testable?
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General Chemistry April, 2010
Reactivity Series Simulation
Go to the reactivity series simulation at:
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redox/home.html
For activities 1-3:
a) Choose one metal and predict what you will observe when the four metal nitrates react
with your chosen metal. Explain the reason for your choice using your knowledge of the
reactivity series.
b) Write ionic ½ equations and an overall balanced equation for any reactions that occurred.
For activity 4:
c) Choose a metal that you predict will react with the hydrochloric acid. Explain the reason
for your choice using your knowledge of the reactivity series.
d) Carry out the reaction to confirm your prediction. Write an ionic ½ equations and overall
balanced equation for the reaction that occurred.
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Lessons 5: Create your own Reactivity Series
The reactivity series can be used to predict whether a reaction involving metals will occur.
Aim
To confirm the theoretical order of reactivity of zinc, copper, iron, magnesium and hydrogen.
Materials
• Pieces of zinc, copper, iron, magnesium and aluminum
• 0.1 M solutions of zinc (II), magnesium, iron (II), copper (II) and aluminium ions
• 0.5 M HCl solution
• Test tubes
• Thermometer
• Plastic pipettes
Safety
Wear safety glasses. Don’t contaminate the solutions.
The Task
Step 1
1. Using your theoretical knowledge of metal reactivity deduce with a reason the order of
reactivity of the metals and hydrogen.
Step 2
1. Design an experiment to confirm your prediction.
2. Present the data for those reactions that occurred and those that did not in an easily
interpretable table.
3. Record all the chemically relevant qualitative observations for those reactions that occurred.
Look at the reactants and products in the chemical equation. Describe the initial color and
state of the reactants and the final color and state of the products. Describe evidence of a
chemical reaction occurring. Examples include: color change in solution, solid reactant
disappears, temperature change, solid is precipitated.
4. Write balanced half equations and ionic equations for the reactions that occurred, including
those you were not expecting.
Questions
1.
A part of a reactivity series of metals, in order of decreasing reactivity is shown below.
Magnesium
Zinc
Iron
Lead
Copper
Silver
If a piece of copper metal were placed in separate solutions of silver nitrate and zinc nitrate
a)
Determine which solution would undergo reaction.
b)
State, giving a reason, what visible changes would take place in the solutions.
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2.
Both zinc and tin are used to coat cans to slow down rusting. Once the surface is scratched,
oxygen, water and acidic foods inside the can can react with the iron and the coating metal.
State and explain whether zinc or tin would be more effective in preventing iron from rusting
under these conditions.
3.
A group of students added small samples of metal powders to five different metal chloride
solutions. They repeated the experiment for each of the five different metals. Their results
were as follows:
Metal ion
Metal atom
Cu
CuCl2
(i)
(ii)
(iii)
(iv)
√ = metal in solution displaced X = no reaction
Mg
Zn
Fe
Ca
√
√
√
√
X
X
√
X
√
MgCl2
X
ZnCl2
X
√
FeCl2
X
√
√
Ca Cl2
X
X
X
√
X
Use the data to place the named metals in the correct order of reactivity, putting the most
reactive first
Metal Z is slightly more reactive than magnesium. Identify the metals chloride solutions
which metal Z would react. The formula for the metal chloride is ZCl3.
Write an ionic equation for the reaction of metal Z reacting with copper chloride
Write the overall equation for the reaction of metal Z with copper chloride.
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Lessons 6:
Electrochemical Cells and Batteries
Alessandro Volta (1745–1827) was the first to create electricity from chemicals. His
electrochemical cell consisted of a sheet of copper metal and a sheet of zinc metal in solutions of
their ions and separated by a sheet of paper soaked in concentrated salt solution. Several of these
cells together formed a battery able to produce a steady electric current.
Volta
is well known for his
contribution to the study of electrochemistry and he often carried out experiments on himself. Today,
ethics laws would prevent this from happening.
I introduced into my ears two metal rods with rounded ends and joined them to the
terminals of the apparatus. At the moment the circuit was completed I received a shock in
the head—and began to hear a noise—a crackling and boiling. This disagreeable sensation,
which I feared might be dangerous, has deterred me so that I have not repeated the
experiment.—Alessandro Volta
A single electrochemical cell is called a voltaic cell, after Volta. The other frequently used name,
galvanic cell, is in honour of an Italian physiologist, Luigi Galvani (1737–1798). Galvani discovered
that frog’s legs could be made to twitch by connecting the nerve and muscle tissues to different
metals such as copper and iron. He thought (incorrectly) that he had discovered a special form of
‘animal electricity’.
In 1836 the first battery manufactured for commercial use was used to power railway signals up until
the beginning of the twentieth century. Today the batteries used in flashlights, watches, calculators
and other electrical appliances, all use redox reactions to create electricity. As more and more
electronic devices are developed like iphones, laptops, and iPods new batteries need to be invented
with superior performance to those of the past. The implementation of electric vehicles and solar
energy are both dependent on the availability of efficient, low-cost electrochemical cells. For these
reasons, modern battery manufacturers are continuing to research and develop more efficient and
cheaper commercial cells.
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Producing Electricity in a Voltaic cell
Consider the reaction between iron and copper ions.
Fe (s) + Cu2+(aq) → Cu (s) + Fe2+(aq)
When the Cu2+ ions and Fe atoms are in contact with each other, electron transfer is rapid and a
significant amount of heat energy is released. To detect the transfer of electrons, the Cu2+ ions and Fe
atoms need to be separated and the electrons force to flow along a wire. The force required to move
the electrons is detected using a voltmeter (measured in Volts, V).
Look at the diagram of the voltaic cell below and describe what you see.
Fe (s) → Fe2+(aq) + 2e22 | P a g e
Cu2+(aq) + 2e- → Cu (s)
Constructing the Electrochemical Cell
Step
1
2
3
4
5
Explanation
The Fe and Cu2+ ions are separated and connected by a wire, which passes through a
detecting device (ammeter, A or voltmeter, V, light bulb, LED). No current can yet
be detected.
A solid conducting surface called an electrode is used as the site for each halfreaction. The electrode must not react the electrolyte, so a Cu strip is chosen for the
Cu2+ solution. Fe serves as the other electrode for the Fe2+ solution.
A conducting solution called an electrolyte is used to allow the movement of ions in
each half-reaction. The electrolyte must not react with the electrolyte, so a Fe(NO3)2
solution is chosen for the Fe half-cell. Cu(NO3)2 serves as the other electrolyte. No
current can yet be detected because the electrical circuit is not complete.
The beakers containing the Fe/Fe2+ and Cu2+/Cu must be connected by another
electrolyte that allows the flow of ions to and from each beaker to make a complete
circuit. A convenient form is a piece of filter paper soaked in a salt solution,
commonly NH4Cl or KNO3. This electrolytic conductor is called a salt bridge. A
current can now be detected.
Since iron is the more reactive metal oxidation occurs at the Fe electrode, giving this
electrode called the anode a negative charge.
Fe(s) → Fe2+(aq) + 2e–
The electrons flow along the wire towards the copper electrode where they are used
in reduction. This gives the copper electrode, called the cathode a more positive
charge.
Cu2+(aq) + 2e– → Cu(s)
Ions flow through the NH4Cl salt bridge. Cl– ion towards the Fe/Fe2+ beaker, NH4+
ions towards the Cu2+/Cu beaker) to complete the circuit.
More details on how the cell works
The electrodes serve as the site for the redox reactions. At the surface of each electrode, electrons
are either produced or accepted. The electrode where oxidation occurs is called the anode. It is the
negative electrode (has negative polarity), because electrons are produced here (the electron source).
The electrode where reduction occurs is called the cathode. It is the positive electrode (called the
electron sink), because electrons are gained by the oxidizing agent at this electrode.
Fe(s) → Fe2+(aq) + 2eAN OX: Oxidation occurs at the anode e–
Fe metal electrode produces electrons
Fe(s) is the reducing agent
Cu2+(aq) + 2e– → Cu(s)
RED CAT: Reduction occurs at the cathode
Cu metal electrode accepts electrons
Cu2+(aq) is the oxidizing agent
Electrodes conduct current by the flow of electrons. They may be made of metal or graphite. Where
a half-cell contains a metal reducing agent, the metal serves as the electrode. Where the reducing
agent is a non-metal, but a gas or aqueous solution the electrode is either graphite or a non-reactive
metal such as platinum.
23 | P a g e
The ammonium nitrate (NH4NO3) salt bridge completes the circuit between the anode and cathode.
It acts as an electrolyte conducting current by the flow of ions.
Consider the Fe/Fe2+ half-cell. As the Fe atoms lose electrons during oxidation, Fe2+ ions enter the
solution. There are now more moles of positive ions (Fe2+) than negative nitrate ions (NO3–) in the
solution. This build-up of extra positive Fe2+ ions would soon stop the flow of the negatively charged
electrons away from the half-cell and into the wire because they would be attracted to these extra
positive ions. Build-up is therefore prevented because the negative nitrate ions in the salt bridge
move into the half-cell.
Similarly, in the Cu2+/Cu half-cell, Cu2+ ions are used up in reduction, leading to an excess of
negative nitrate ions (NO3–) in the electrolyte. The positive NH4+ ions from the salt bridge move into
the half-cell to prevent build-up of negative ions, maintaining the electrical neutrality of each halfcell.
A voltmeter connected between the two electrodes measures the force required to push the electrons
from the anode and cathode. In other words it measures the ability of the cell to move electrons
through a circuit. This tendency is called the cell potential (or voltage) measured in Volts (V). The
rate of electron flow is measured using an ammeter measured in Amps (A).
Both metals have a tendency to lose electrons, however of the two iron is the strongest reducing
agent and therefore has a greater tendency to lose electrons than copper. The electrons produced at
the iron anode travel along the wire to the copper cathode where they are used for the reduction of
copper(II) ions.
Check your Understanding
Design a voltaic cell using silver and lead electrodes. Draw a diagram of the cell and fully label it.
Show the:
a)
relevant half-equations
b)
overall ionic equation
c)
polarity/charge on the electrodes
d)
direction of flow of both electrons through the wires and ions through the salt bridge.
e)
Label the anode and cathode
f)
Identify the chemical composition of the electrolytes and salt bridge
g)
Label the oxidizing and reducing agent.
24 | P a g e
Lesson 7:
More about Batteries
1. Research how one of the following kinds of batteries are made.
Acidic dry cell battery
Alkaline dry cell battery
Silver dry cell battery
Mercury dry cell battery
Nickel cadmium dry cell battery
2. a)
Using a labelled diagram describe the construction/parts of the battery. [cathode
charge and materials; anode charge and materials; electrolyte and salt bridge
composition]. (9 points)
b)
Describe common everyday uses of the battery. (1 point)
c)
Write ½ equations and an ionic equation for the overall reaction. (3 points
Lesson 8:
Creating your own Battery
Make your own battery that will produce 1.5V or more. Write a procedure for constructing
the battery that could be reproduced by someone else.
Lesson 9 and 10: Battery Investigation Lab
Design and carry out your own investigation into lab one factor affecting the voltage
produced by the battery you made in lesson 8. See the rubric attached.
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Voltaic Cells Lab Report Assessment RUBRIC
Scoring rubric: completely met = 3 ; partially met = 2 ; not met = 1
Title
Relates the dependent and independent variables in general terms. Can be written as a
question.
Research Question or Aim
Clearly stated in about two sentences. It clarifies the relationship between the factor being
investigated (independent variable) and what is being measured (dependent variable). It does
not describe the method.
A balanced half equation and ionic equation is written for the redox reaction under
investigation.
Hypothesis
The prediction or expected outcome is clearly stated and explained.
Variables
A list of all the important dependent, independent and controlled variables.
o Independent variable – the factor being investigated.
o Dependent variable(s) – what is measured or the type of numerical data collected
about the independent variable.
o Controlled / constant variables – what is kept constant or remains unchanged between
the different experiments into the factor being investigated.
Procedure
Written in sufficient detail that it is easy to follow, quantities are given
Clearly explains the method, and demonstrates a sound understanding of the principles of
voltaic cells.
There is evidence that the measurements have been repeated or data is collected over a
suitable time period.
Data Collection
26 | P a g e
Sufficient relevant qualitative observations and quantitative measurements are recorded.
All units of measurement and absolute uncertainties are recorded. Column headings in tables
are clearly labelled and include units and uncertainties in measurement.
/13
The quantitative data collected is presented in easily interpretable, organized and labeled
table(s). See the guidelines attached for the features of a good scientific table.
Data Processing
The processed data is presented in easily interpretable, organized graph(s) so that patterns in
the results are made obvious. See the guidelines attached for the features of a good scientific
graph.
/9
Complete and correct quantitative analysis of the data is carried out. This could include
combining, manipulating the collected data to determine the value of an answer, taking the
average of several measurements and transforming the data into a form suitable for graphing.
Conclusion
A statement is made which indicates if the aim was achieved. This is stated explicitly. It is
stated whether the results supported your expectations.
Describe and explain the relationship found between the independent (factor changed) and
dependent (measured) variables.
Evaluation
Compare what you expected and the actual results obtained in the experiment quantitatively.
Suggest and explain improvements and modifications that could be made to minimize the
differences between the expected and the actual results described above.
References
Bibliography of all sources cited is included. This includes information sourced when
planning the lab, diagrams, graphs, literature values and expected results.
Tables should have:

Clear column and row headings

Units of measurement included in the column heading

Borders/lines around text and numerical data

Don’t run over two pages

Columns to be compared are placed next to one another

Called tables and numbered consecutively
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
Concise descriptive title that relates the measured (dependent) and changed (independent)
variables on top of the table

Consistent and correct use of decimal places/significant figures for numerical data

Text and data is centered

Rows and columns are evenly distributed

11 – 12 point font size for electronic tables

Consistent font type for electronic tables

Correctly placed in data collection part of the lab report
Graphs should:
 have axes labelled with the correct variable on the x and y axes.
 include units of measurement for each variable in brackets.
 have an appropriate scale for the data – it should reflect the precision in the data and intervals
of measurement used.
 be an appropriate size (at least ½ page)
 have key/legend for two or more sets of data on the same graph is clear (or series deleted
from excel graphs if there is only one set of data)
 be 11 -12 point font and a consistent font type for electronic graphs
 have accurately plotted data points
 Correctly placed in document.
 Appropriate best fit straight or curved line connecting the points. Plotting scatter graphs with
data point to data point connecting lines without a line is not appropriate.
28 | P a g e
Teacher Notes:
Must have 1.5V or more from a voltmeter
Factors: width of electrode, concentration of electrolyte, number of cells connected
One cell with Cu and Zn about 0.7 V
Use vernier voltage meter or multimeters
Factors affecting voltage size:
Number cells connected in series
Concentration of electrolyte
Type of electrode material
Mass/moles of electrode
29 | P a g e