Download - Oxidation and Reduction
Oxidation and Reduction
Mr Field
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Using this slide show The slide show is here to provide structure to the lessons, but not to
limit them….go off-piste when you need to!
Slide shows should be shared with students (preferable electronic to save paper) and they should add their own notes as they go along.
A good tip for students to improve understanding of the calculations is to get them to highlight numbers in the question and through the maths in different colours so they can see where numbers are coming from and going to.
The slide show is designed for my teaching style, and contains only the bare minimum of explanation, which I will elaborate on as I present it. Please adapt it to your teaching style, and add any notes that you feel necessary.
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Menu: Lesson 1 – Oxidation and Reduction Lesson 2 – Redox Equations Lesson 3 – Reactivity Series Lesson 4 – Voltaic Cells Lesson 5-6 – Electrolytic Cells Lesson 7 – HL – Standard Electrode Potentials Lesson 8 – HL – Ecell and Non-Standard Conditions Lesson 9 – HL – Advanced Electrolysis Lesson 10 – HL – Quantitative Electrolysis Lesson 11–12 – Internal Assessment
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Lesson 1
Oxidation and Reduction
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Overview Copy this onto an A4 page. You should add to
it as a regular review throughout the unit.
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Assessment
This unit will be assessed by:
An internal assessment at the end of the topic
A joint test at the end of the Acids and Bases topic
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Lesson 1: Oxidation and Reduction
Objectives:
Reflect on prior knowledge of oxidation and reduction
Understand oxidation and reduction in terms of electron transfer
Calculate oxidation numbers
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Reflecting on Redox
Write down everything you know on oxidation and reduction
You have 60 seconds
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Defining Oxidation and Reduction O - oxidation I - is L - loss of electrons
R - reduction I - is G - gain of electrons
Often (but far from always) in practice:
Oxidation is gain of oxygen or loss of hydrogen This results in the loss of electrons
Reduction is loss of oxygen or gain of hydrogen This results in the gain of electrons
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Oxidation States / Oxidation Numbers Oxidation state is the charge an atom would have if all it’s bonds were ionic
It is the number of electrons an atom has gained or lost by forming bonds You even talk about oxidation state of covalent compounds!
It is important as the oxidation state of an atom has a significant impact on its chemistry
Fe(II) Fe(III)
Cr(III)Cr(VI)
Mn(II) Mn(VII)
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Calculating Oxidation States The oxidation state of an
element is zero
The oxidation states of a neutral compound sum to zero, and of an ion sum to the charge on the ion
The more electronegative atom in an ion assumes a negative oxidation state, the less electronegative one a positive oxidation state
Element Oxidation State
Notes
Fluorine -1 AlwaysOxygen -2 Except in
peroxides where -1, and F2O where +2
Chlorine -1 Except with O or F where +1
Gp I Metal
+1 Always
Gp II Metal
+2 Always
Hyrdogen +1 Except metal hydride where -1
Some rules of thumb:
Start with these, work the others out.
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For example Determine the oxidation states of each atom in the following:
CO2 O, -2 -2 except with F or a peroxide C, +4 to balance out the 2 lots of ‘-2’
H2SO4 O, -2 -2 except with F or peroxide H, +1 +1 except in metal hydrides S, +6 +6 since four lots of ‘-2’ and two of ‘-1’ sum to -6
BaO2 (barium peroxide) Ba, +2 +2 since Gp II metal O, -1 -1 since peroxide
CO32-
O, -2 -2 except with F or peroxide C, +4 +4 since 3 x -2 and +4 sums to the charge, -2
Note: oxidation states must be written with the sign in front: +2 NOT 2+
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Determine oxidation states for each atom in:
H2O
S8
CH4
H3PO4
CCl4
HClO
KMnO4
IO3-
Cr2O72-
Cr(H2O)63+
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Oxidation States and Names Oxidation states are used in the names of compounds
‘ate’ means an element is in a positive oxidation state Usually because it is bonded with oxygen
‘ide’ means an element is in a negative oxidation state
Oxidation state of transition metals is given in Roman numerals
FeCl2 – iron (II) chloride Iron (II) means Fe in the +2 ox. State Chloride means the chlorine is in a negative oxidation state
FeCl3 – iron (III) chloride Iron (III) mean Fe in the +3 ox. State
KClO3 – potassium chlorate Chlorate tells you the chlorine is in a positive oxidation state
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Name the following:
MnO2
CuO
Cu2O
KMnO4
K2Cr2O7
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Redox Reactions Whenever an oxidation occurs, a reduction also occurs, hence REDOX
For example:
Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq)0 +2 0 +2
Zinc is oxidised – it loses two electrons Zinc is the reducing agent because it reduces copper
Copper is reduced – it gains two electrons Cu2+ is the oxidising agent because it oxidises the zinc
The number of electrons gained by species is always equal to the number of electrons lost by species.
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Disproportionation* When some atoms of an element are oxidised and others
are reduced
For example:
2 H2O2 2H2O + O2
+1 -1 +1 -2 0
The oxygen ending in the H2O loses an electron and is oxidised
The oxygen ending in the O2 gains an electron and is reduced
*This is not on the syllabus but is useful and interesting
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Identify the species that are oxidised and reduced in each reaction, stating the number of electrons gained or lost and stating whether disproportionation takes place
Fe2O3 + 2 Al Al2O3 + 2 Fe
AgNO3 + NaCl AgCl + NaNO3
3 Cl2 + 6 OH− → 5 Cl− + ClO3− + 3 H2O
H2SO4 + 2HBr Br2 + SO2 + 2H2O
Cu + 4 HNO3 Cu(NO3)2 + 2 NO2 + 2 H2O
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Key Points
Oxidation state tells us the number of electorns an atom has gained or lost
The most electronegative atom in a bond gains electrons, to form a negative oxidation state and vice versa
Oxidation reactions are always accompanied by reductions
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Lesson 2
Redox Equations
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Fertilizers may cause health problems for babies because nitrates can change into nitrites in water used for drinking.
a) Define oxidation in terms of oxidation numbers.
b) Deduce the oxidation states of nitrogen in the nitrate, NO3
–, and nitrite, NO2–, ions.
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Lesson 2: Redox Equations
Objectives:
Deduce simple half-equations
Combine half-equations to form full equations
Conduct a series of redox reactions
Use H+ and H2O to balance redox equations
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Half-equations
Half equations show the changes to individual species in a redox reaction.
Fe2O3 + 2 Al 2 Fe + Al2O3
Fe3+ + 3 e- Fe ….this is the reduction
Al Al3+ + 3 e- ….this is the oxidation
A wide variety of half equations can be found in the data booklet
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Combining Half Equations and Balancing Redox Reactions – Example 1
Step Example 1: Reaction of iodine with copper
Write out half-equations side by side and the number of electrons lost or gained…can be found in the data booklet
½ I2 + e- I- … 1 e- gainedCu Cu2+ + 2 e- …2 e- lost
Combine them to form a single reaction, multiplying each half-equation in order to balance the electrons gained/lost
I2 + 2 e- + Cu 2 I- + Cu2+ + 2 e-
- the iodine ½-equation is doubled to make . 2 e- gained, the other remains unchanged
Ensure all atoms other than O and H balance
I2 + 2 e- + Cu 2 I- + Cu2+ + 2 e-
- no change neededCancel out any electrons that are duplicated on both sides
I2 + Cu 2 I- + Cu2+
- 2 e- appear on each side so are cancelled out
Balance oxygen atoms by adding H2O to the side that needs extra
I2 + Cu 2 I- + Cu2+
- no change neededBalance the hydrogens by adding H+ to the side that needs extra
I2 + Cu 2 I- + Cu2+
- no change neededCount the total charge on each side and add enough electrons to the more positive side to balance it
I2 + Cu 2 I- + Cu2+
- no change needed
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Combining Half Equations and Balancing Redox Reactions – Example 2
Step Example 2: Reaction of nickel with manganate ions
Write out half-equations side by side and the number of electrons lost or gained
MnO4- +5e- Mn2+ …5 e- gained by Mn
Ni Ni2+ + 2 e- … 2 e- lost by Ni
Combine them to form a single reaction
2MnO4- + 5Ni + 10e- 2Mn2+ + 5Ni2+ + 10e-
• Ni ½-equation multiplied by 5• MnO4- ½-equation multiplied by 2
Ensure all atoms other than O and H balance
2MnO4- + 5Ni + 10e- 2Mn2+ + 5Ni2+ + 10e-
• No change neededCancel out any electrons that are duplicated on both sides
2MnO4- + 5Ni 2Mn2+ + 5Ni2+
• 10 e- cancelledBalance oxygen atoms by adding H2O to the side that needs extra
2MnO4- + 5Ni 2Mn2+ + 5Ni2+ + 8H2O
• 8H2O on the right balance the 8 O on the left
Balance the hydrogens by adding H+ to the side that needs extra
2MnO4- + 5Ni + 16H+ 2Mn2+ + 5Ni2+ + 8H2O
• 16H+ added on left to balance 16 H on the right
Count the total charge on each side and add electrons to the more positive side to balance it
2MnO4- + 5Ni + 16H+ 2Mn2+ + 5Ni2+ + 8H2O
• Total charge balances so no change needed’• This step only needed on half-equations
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Microscale Redox Reactions
Complete the experiment here, on microscale redox reactions Rather than doing it on a plastic sheet, use a
dropping tile For each change you observe, you should write a
balanced redox equation to describe it
Once you finish, you should practice balancing the redox equations on the following slide
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Produce balanced redox equations for the reactions of: Bromine with sodium
½ Br2 + e- Br-
Na Na+ + e-
Copper (II) oxide with hydrogen Cu2+ + 2e- Cu ½ H2 H+ + e-
Aluminium reacting with chromate ions Al Al3+ + 3e-
Cr2O72- + 6e- 2Cr3+
Iron (II) chloride reacting with manganate ions Fe2+ Fe3+ + e-
MnO4- + 5 e- Mn2+
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Key Points
Half-equations show the changes to each species in a redox reaction
To combine half-equations into a full equation Multiply each such that the electron-transfers
balance Add H2O to balance O Add H+ to balance H Add e- to balance charge
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Lesson 3
Reactivity Series
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Nitric acid reacts with silver in a redox reaction:
__ Ag(s) + __ NO3–(aq) + ____ → __ Ag+(aq) + __ NO(g) + ____
Using oxidation numbers, deduce the complete balanced equation for the reaction showing all the reactants and products.
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Lesson 3: Reactivity Series
Objectives:
Deduce reactivity series from chemical observations
Use reactivity series to predict the feasibility of reactions
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Redox and Reactivity
Redox behaviour is closely linked to reactivity
The most reactive metals are the best reducing agents
The most reactive non-metals are the best oxidising agents
The least reactive elements are neither good oxidising or reducing agents
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Reactivity and Displacement Reactions Reactive metals are better reducing agents
than un-reactive metals.
As such, a reactive metal can displace less reactive metals from their compounds.
Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
As a good reducing agent, the zinc reduces the Cu2+, causing it to gain two electrons.
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Constructing a Reactivity Series
Complete the experiment here in which you have to construct a reactivity series.
Analysis Use your reactivity series to predict the feasibility
of the reactions of: I- with Fe3+
Zn with Sn2+
Fe2+ with Cl Compare your series with the order found on Table
14 of the data booklet.
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Key Points
More reactive metals are better reducing agents
More reactive non-metals with oxidising agents
A more reactive metal will reduce (displace) ions of a less reactive metal
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Lesson 4
Voltaic Cells
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Refresh Consider the following three redox reactions.
Cd(s) + Ni2+(aq) → Cd2+(aq) + Ni(s)
Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s)
Zn(s) + Cd2+(aq) → Zn2+(aq) + Cd(s)
a) Deduce the order of reactivity of the four metals, cadmium, nickel, silver and zinc and list in order of decreasing reactivity.
b) Identify the best oxidizing agent and the best reducing agent.
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Lesson 4: Voltaic Cells
Objectives:
Explain in simple terms how voltaic cells use redox reactions to produce electricity
Understand that oxidation occurs at the anode and reduction at the cathode
Make a series of voltaic cells in order to better understand the how they work
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Voltaic Cells The reaction of Mg with Cu2+ ions:
Mg(s) + Cu2+(aq) Mg2+(aq) + Cu(s)
This reaction involves two electrons being transferred from the Mg to the Cu: Mg Mg2+ + 2e-
Cu2+ + 2e- Cu
The Mg reduces the copper ions as it is more reactive This is an exothermic reaction, and the energy is normally released as heat
A voltaic cell forces each half of the reaction to take place in a separate container, with the electrons moving through a circuit to get from one side to the next This is an exothermic reaction, where the energy is released as electrical rather
than thermal energy
The reactions in Voltaic cells usually involve only metals but do not have to.
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Voltaic Cells Continued
Anode:Where oxidation happens
Cathode:Where reduction happens
+-
Electron Flow
Electron
Flow
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Key Parts of a Voltaic Cell Anode
Electrode or ‘half-cell’ where oxidation happens Contains the more reactive metal The negative electrode: produces electrons
Cathode Electrode or ‘half-cell’ where reduction happens Contains the less reactive metal The positive electrode: accepts electrons
Salt Bridge Contains a neutral salt such as potassium nitrate Made of a tube of jelly or a filter paper soaked in salt solution Ions diffuse in and out to balance charge and complete circuit
Voltmeter Measures the difference in potential between half-cells Could be replaced with other circuitry to do useful work
REMEMBER AnOx
Anode-Oxidation CaRe
Cathode-Reduction
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Drawing a cell Draw and fully label a zinc/iron cell. Include:
Labels for cathode and anode Labels for positive and negative Each half-equation Arrow showing direction of electron flow
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Constructing Voltaic Cells
You will need to build and measure the potential of voltaic cells comprising various combinations of the following: Cu/Cu2+
Fe/Fe2+
Mg/Mg2+
Sn/Sn2+
Zn/Zn2+
Follow the instructions here
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Key Points
Voltaic cells extract electrical energy from redox reactions by separating each half
At the anode, the more reactive of two metals is oxidised
At the cathode, the less reactive of two metals is reduced
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Lesson 5-6
Electrolytic Cells
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RefreshA particular voltaic cell is made from magnesium and iron half-cells. The overall equation for the reaction occurring in the cell is
Mg(s) + Fe2+(aq) → Mg2+(aq) + Fe(s)
Which statement is correct when the cell produces electricity?A. Magnesium atoms lose electrons.B. The mass of the iron electrode decreases.C. Electrons flow from the iron half-cell to the magnesium half-cell.D. Negative ions flow through the salt bridge from the magnesium
half-cell to the iron half-cell.
For each incorrect statement, explain why it is wrong.
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Lesson 5-6: Electrolysis
Objectives:
Describe electrolytic cells
Identify at which electrode oxidation and reduction takes place
Understand how current is conducted in electrolytic cells
Deduce the products of electrolysis of a molten salt
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Electrolytic Cells Electrolytic cells use electricity to provide the energy for an endothermic redox
reaction
Electrolysis of lithium chloride:
Li+(l) + Cl-(l) Li(s) + ½Cl2(g)
Current is carried by moving ions Cations (+) move to the cathode (negative electrode) Anions (-) move to the anode (positive electrode)
As ions need to be able to move, the ionic compound must be either: Molten Dissolved in solution
The opposite of a voltaic cell: Voltaic cells turn stored chemical energy into electrical Electrolytic cells turn electrical energy into stored chemical energy
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An Electrolytic Cell:
M+
M+
M+
M
M
M
X-X-
X-
X X
X X
CATHODE(-)
ANODE (+)
Anions move to anode
Cations move to cathode
Layer of metal formed
Bubbles of gas formed
MOLTEN SALTor
SALT SOLUTION
e- e-
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Products of Electrolysis Assuming a molten simple metal salt involving only monatomic ions
Cathode – metal Deposited on the surface of the electrode
Anode – non-metal Typically as bubbles of gas
For example, electrolysis of molten magnesium bromide: Cathode: a layer of magnesium metal (Mg2+ + 2e- Mg) Anode: bubbles of bromine gas (2Br- Br2 + 2e-)
More complicated systems Aqueous solutions – a range of possibilities depending on the stability of the
ions relative to water Polyatomic ions – a mixture of various products
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By the end of the next lesson you should: Complete the electrolysis of molten zinc chloride in the fume
hood. Instructions here: http://www.nuffieldfoundation.org/practical-chemistry/electrolysis-zinc-chloride
Produce an animation showing how electrolysis happens. Could be PowerPoint, flicker book, smart phone animation app, stop-motion (see: http://www.wikihow.com/Create-a-Stop-Motion-Animation )...be creative.
Draw a Venn diagram comparing and contrasting electrolytic and voltaic cells
Write three potential internal-assessment research questions and on electrolytic or voltaic cells, select one and start producing a plan for it.
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Key Points
Electrolysis uses electricity to drive endothermic redox reactions
Metal salts must be molten or dissolved so ions can move and carry charge
The negative cathode reduces positive metal ions
The positive anode oxidises negative non-metal ions
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Lesson 7HL Only
Standard Electrode Potentials
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RefreshWhich processes occur during the electrolysis of molten sodium chloride?
I. Sodium and chloride ions move through the electrolyte.II. Electrons move through the external circuit.III. Oxidation takes place at the anode.
A. I and II onlyB. I and III onlyC. II and III onlyD. I, II and III
Justify your answer.
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Lesson 7: Standard Electrode Potentials
Objectives:
Describe the standard hydrogen electrode
Define the term standard electrode potential
Use standard electrode potentials to calculate the potential of a cell
Use standard electrode potentials to determine the feasibility of a reaction
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Half-cell potential When you set up a half-cell, it has a certain ‘POTENTIAL’.
When two half-cells are joined, electrons flow away from the half-cell with more negative potential towards the half-cell with more positive potential
You can think of this a little like enthalpy There is no absolute measure of potential, only relative You can only measure potential differences (voltage) between
two half-cells
The potential of a half-cell is always measured relative to the standard hydrogen electrode.
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The Standard Hydrogen Electrode Standard electrode potential defined as 0.00 V
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Standard Electrode Temperature = 298K
Pressure = 1 atm
Aqueous solution of metal ions:[M+] = 1.0 mol dm-3
Metal Electrode
Connecting wire
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Standard Electrode Potential, Eo
This is the potential of a standard electrode relative to the standard hydrogen electrode.
Always measure the potential of the reduction
Measured in Volts, V
Full table in the data booklet
Half Cell Standard Electrode Potential, Eo / V
H+(aq) + e- ⇌ ½ H2(g)
0.00
Li+(aq) + e- ⇌ Li(s)
-3.04
Mn2+(aq) + 2e- ⇌ Mn(s)
-1.19
Cu2+(aq) + 2e- ⇌ Cu(s)
+0.34
½ Br2(l) + e- ⇌ Br-
(aq) +1.07
Look at the table in the data booklet: What trends do you notice? How do the values relate to your
ideas of reactivity? How do the values compare to the
reactivity series you constructed earlier?
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The Potential of a Cell, Eocell
To make a cell, two half-cells are connected by a salt-bridge and a volt-meter
The potential of a cell is easy to calculate: Subtract the potential of the more negative ½-
cell from the potential of the more positive This gives the potential difference:
What is the potential of a Lithium-Manganese cell? Lithium: -3.04 V, Manganese: -1.19V Eo
cell = -1.19 – -3.04 = 1.85 V
What is the potential of a Manganese-Bromine cell? Manganese: -1.19 V, Bromine: 1.07 V Eo
cell = 1.07 – -1.19 = 2.26 V
Half Cell Standard Electrode Potential, Eo / V
H+(aq) + e- ⇌ ½ H2(g)
0.00
Li+(aq) + e- ⇌ Li(s)
-3.04
Mn2+(aq) + 2e- ⇌ Mn(s)
-1.19
Cu2+(aq) + 2e- ⇌ Cu(s)
+0.34
½ Br2(l) + e- ⇌ Br-
(aq) +1.07
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Determining the Reaction in a Cell: A little more tricky, but still OK
The more negative half-cell moves to the left (gets oxidised)
The more positive half-cell moves to the right (gets reduced)
Determine the reaction for each half cell and then combine them (making sure they balance)
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Determining the Reaction in a Cell: Example 1
The Lithium-Manganese cell?
Lithium: -3.04 V, Manganese: -1.19V Lithium: Li(s) Li+(aq) + e- ….more negative so goes
left Mn2+(aq) + 2e- Mn(s) …more positive so goes right
2 Li(s) + Mn2+(aq) 2 Li+(aq) + Mn(s)
Half Cell Standard Electrode Potential, Eo / V
H+(aq) + e- ⇌ ½ H2(g)
0.00
Li+(aq) + e- ⇌ Li(s)
-3.04
Mn2+(aq) + 2e- ⇌ Mn(s)
-1.19
Cu2+(aq) + 2e- ⇌ Cu(s)
+0.34
½ Br2(l) + e- ⇌ Br-
(aq) +1.07
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Determining the Reaction in a Cell: Example 2
The Manganese-Copper cell?
Copper: +0.34 V, Manganese: -1.19 V Manganese: Mn(s) Mn2+(aq) + e- ….more negative so
goes left Copper: Cu2+(aq) + 2e- Cu(s) …more positive so goes
right
Mn(s) + Cu2+(aq) Mn2+(aq) + Cu(s)
Half Cell Standard Electrode Potential, Eo / V
H+(aq) + e- ⇌ ½ H2(g)
0.00
Li+(aq) + e- ⇌ Li(s)
-3.04
Mn2+(aq) + 2e- ⇌ Mn(s)
-1.19
Cu2+(aq) + 2e- ⇌ Cu(s)
+0.34
½ Br2(l) + e- ⇌ Br-
(aq) +1.07
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Drawing and Labelling a Cell Example: The manganese-copper cell
V
1.00 mol dm-3
Mn2+(aq)
1.00 mol dm-3
Cu2+(aq)
Manganese CopperSaltBridge
ANODE, (-)oxidation occurs:
Mn(s) Mn2+(aq) + e-
CATHODE, (+)reduction occurs
Cu2+(aq) + 2e- Cu(s)
electronflow
electronflow
Cu2+(aq)+ Mn(s) Cu(s) + Mn2+(aq)
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Write the equation for, and calculate the potential of the following cells. Draw and fully label two of them*.
1. Copper and Zinc
2. Silver and Lead
3. Hydrogen and Nickel
4. Iodine and Iron
5. Iron and Iron (II)
*Note: If you don’t have a solid metal, use platinum for your electrode, and make sure it is in contact with both phases of the ½-cell.
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Predicting The Feasibility of a Reaction Will iron (II) react with copper?
The only possible reaction between Cu and Fe2+ is:
Cu(s) + Fe2+(aq) Cu2+(aq) + Fe(s)
Select relevant ½-cells: Cu2+(aq) + 2e- ⇌ Cu(s) +0.34 V Fe2+(aq) + 2e- ⇌ Fe(s) -0.45 V
Compare the ½-cell potentials: The more negative ½-cell, (Fe2+(aq) + 2e- ⇌ Fe(s)), always gets
oxidised (moves to the left) This reaction shows it being reduced (moving to the right) Therefore this reaction can’t happen
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Predicting The Feasibility of a Reaction Will zinc ions react with magnesium?
The only possible reaction between Zn2+ and Mg is:
Mg(s) + Zn2+(aq) Mg2+
(aq) + Zn(s)
Select relevant ½-cells: Zn2+
(aq) + 2e- ⇌ Zn(s) -0.76 V Mg2+
(aq) + 2e- ⇌ Mg(s) -2.37 V
Compare the ½-cell potentials: The more negative ½-cell, (Mg2+
(aq) + 2e- ⇌ Mg(s)), always gets oxidised (moves to the left)
This reaction shows it being oxidised (moving to the left) Therefore this reaction will happen
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Determine whether each of the following reactions is feasible, stating why/why not in each case.
Pb2+ ions reacting with calcium
Zn2+ ions reacting with iron
Cl- ions reacting with fluorine
Silver reacting with chlorine
Silver reacting with iodine
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Key Points:
Eocell is the difference between the two Eo
values
The more negative half-cell gets oxidised
The more positive half-cell gets reduced
A reaction will only be feasible if it results in the more negative species being oxidised and the more positive being reduced
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Lesson 8HL Only
Eocell and non-standard conditions
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RefreshConsider the following standard electrode potentials.
Zn2+(aq) + 2e– Zn(s) Eo = –0.76 VCl2(g) + 2e– 2Cl–(aq) Eo = +1.36 VMg2+(aq) + 2e– Mg(s) Eo = –2.37 V
What will happen when zinc powder is added to an aqueous solution of magnesium chloride?A. No reaction will take place.B. Chlorine gas will be produced.C. Magnesium metal will form.D. Zinc chloride will form.
Justify your answer.
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Lesson 8: Deviations from Standard Conditions*
Objectives:
Understand the effects of non-standard conditions on the potential of a cell
Use the Nernst equation to calculate the effect of non-standard conditions on cell potential
*Note: this lesson does not lie within the syllabus but is useful to deepen your understanding and is a rich source of potential IA questions!
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The Nernst Equation You have learnt to calculate the potential of a cell in standard conditions
Cells are rarely under standard conditions
We can still calculate the cell potential using the Nernst Equation
Where: E is the non-standard cell potential E is the standard cell potential R is the gas constant, 8.31 T is the temperature in Kelvins N is the number of electrons transferred in the reaction F is the Faraday constant, 96,500 Q is the reaction quotient: concentration of products divided by concentration
of reactants
QnFRTEE o ln
Walther Nernst
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What does the Nernst Equation mean?
From the maths of the Nernst equation we can draw three important conclusions
1. Increasing the temperature will decrease the cell potential
2. Increasing the concentration of the anode (the oxidised species, our ‘product’) will reduce cell potential
3. Increasing the concentration of the cathode (the reduced species, our reactant) will increase cell potential
Note: effects two and three will be relatively small as they are logarithmic
QnFRTEE o ln
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Testing the Nernst Equation
In this experiment you will design and plan experiments to confirm the findings of the Nernst equation.
You will be using a magnesium-copper cell as this gives about the largest potential we can feasibly investigate with our facilities
Follow the instructions here
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Key Points
The Nernst Equation describes how Ecell changes under non-standard conditions
Temperature reduces Ecell
Increasing the anode concentration reduces Ecell
Increasing the cathode concentration increases Ecell
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Lesson 9HL Only
Advanced Electrolysis
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RefreshConsider the following cell:
Pb(s) + 2 Ag+(aq) Pb2+(aq) + 2 Ag(s)
Describe and explain what happens to cell potential upon:
Increasing the temperature
Reducing the concentration of Pb2+ ions
Increasing the concentration of Ag+ ions
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Lesson 9: Advanced Electrolysis
Objectives:
Describe the products of electrolysis in aqueous solution
Explain the uses of electrolysis for electroplating
Complete an experiment to investigate electroplating
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Electrolysis of Water Water itself can be readily electrolysed
Since Kw of water is very low, H+ is added to help carry current
Anode (oxidation, moves to left): ½ O2(g) + 2 H+(aq) ⇌ H2O(l) Eo = +1.23V
Cathode (reduction, moves to right): 2 H2O(l) + e- ⇌ H2(g) + 2 OH-(aq) Eo = -0.83 V
Overall*: 2 H2O(l) 2 H2(g) + O2(g)
*Note: the reverse of this is the reaction that takes place in a hydrogen fuel cell and has an Eo
cell of 2.06 V.
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Electrolysis of Aqueous Solutions Electrolysis of molten salts is simple:
Cathode Metal Anode Non Metal
Electrolysis of aqueous solutions is less straightforward due to the fact that the water can compete with the salts to undergo electrolysis.
At the Anode: If an anion has an Eo more positive than +1.23 V, water will be oxidised instead of the anion Bubbles of O2 gas will be formed and H+ ions will enter solution This happens for anions including (but not only): fluoride and sulphate
Note: Although chlorine has a potential of +1.36 V, it WILL be discharged in aqueous solution., contrary to what you might expect. The reasons for this are complex and are to do with changes in Cl - concentrations at the anode once current starts to flow.
At the Cathode: If a cation has an Eo more negative than -0.83 V, water will be reduced instead of the cation Bubbles of H2 gas will be formed and OH- ions will enter solution This happens for cations including (but not only): lithium, potassium, sodium and
magnesium
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Example 1: What are the products of the electrolysis of aqueous copper sulphate? Consider the standard electrode potentials:
At the cathode: 2 H2O(l) + e- ⇌ H2(g) + 2 OH-(aq) -0.83 V Cu2+(aq) + 2e- ⇌ Cu(s) +0.34 V
Copper less negative than water so Cu2+ is reduced, cathode gains coating of metallic copper
At the anode: ½ O2(g) + 2 H+(aq) ⇌ H2O(l) +1.23 V S2O8
2–(aq) + 2e– ⇌ 2 SO42– +2.01 V
Sulphate is more positive than water so water is oxidised, bubbles of O2 gas produced and solution becomes acidic
Overall reaction:
Cu2+(aq) + SO42-(aq) + 2H2O(l) Cu(s) + SO4
2-(aq) + 2H+(aq) + O2(g)
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Example 2: Electrolysis of Sodium Chloride Consider the standard electrode potentials:
At the cathode: 2 H2O(l) + e- ⇌ H2(g) + 2 OH-(aq) -0.83 V Na+(aq) + e- ⇌ Na(s) -2.71 V
Sodium more negative than water so water is reduced, cathode produces bubbles of H2 gas and solution becomes more alkaline
At the anode: ½ O2(g) + 2 H+(aq) ⇌ H2O(l) +1.23 V ½ Cl2(g) + e– ⇌ Cl- +1.36 V
Chlorine is more positive than water, so you expect it water to be oxidised, but chloride is due to reasons stated earlier.
Overall reaction:
2Na+(aq) + 2Cl-(aq) + 2H2O(l) 2Na+(aq) + 2OH-(aq) + H2(g) + Cl2(g)
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Predict the products at the anode and cathode for electrolysis of the following aqueous solutions. For each one, write the overall equation for the reaction.
KCl
NiSO4
PbI2
ZnCl2
LiOH
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Electroplating
Electroplating uses electrolysis to coat metal objects in a fine layer of another metal, by setting it as the cathode.
Examples include: Coating things in nice shiny chrome Coating with hold to improve conductivity Coating with sacrificial metal for protection
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Investigating Electroplating
In this short experiment, you will electroplate some small metal objects with a variety of different metals
Follow the instructions here
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Key Points
In electrolysis of aqueous solutions, it is possible for water to be oxidised and reduced.
At the cathode: If a metal has a more negative Eo than water:
Water will be reduced Hydrogen gas and OH- ions will be produced
At the anode: If a species has a more positive Eo than water:
Water will be oxidised O2 gas and H+ ions will be produced
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Lesson 10HL Only
Quantitative Electrolysis
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Sodium metal can be obtained by the electrolysis of molten sodium chloride.
a. Explain why it is very difficult to obtain sodium from sodium chloride by any other method.
b. Explain why an aqueous solution of sodium chloride cannot be used to obtain sodium metal by electrolysis.
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We Are Here
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Lesson 10: Advanced Electrolysis
Objectives:
Describe the effect of time on the amount of amount of product formed during electrolysis
Describe the effect of ionic charge on the amount of product formed during electrolysis
Complete an experiment to determine Avogadro’s constant by electrolysis of copper sulphate solution.
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First some physics The ‘amount’ of electricity (charge, Q) is measured in
Coloumbs, C
Current (I), the flow of electricity, is measured in Amperes, A A current of 1.00 A means 1.00 C of charge flows every
second
So, the total amount of charge flowing through a circuit is given by:
Q = I.t Charge = Current x time
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And a bit more physics
The amount of charge on a single electron, e, is: e = 1.602x10-19
If you If you divide 1.00 C (charge) by the charge on a single electron (e), you can determine the number of electrons that make up a Coloumb:
Electrons per Coulomb = 1.00 / 1.602x10-19
= 6.24x1018
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The Effect of Time
The greater the time of electrolysis, the more products formed:
This stands to reason but is clear if you look at the equation for charge and currents:
Q = I x t The longer the time, the bigger Q The bigger Q, the bigger the number of electrons that have
flowed The bigger the number of electrons, the more
reduction/oxidation can take place
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The effect of ionic charge Of three solutions of equal concentration, electrolysed at equal current
for an equal time, which would form the greatest quantity (in mol) of metal at the cathode? PbCl2 PbCl4 CrCl3
Answer: PbCl2
Reason: The same current for the same time means the same number of electrons flows PbCl contains Pb2+ ions which requires the fewest electrons (2) to reduce them,
this means more ions can be reduced for the same number of electrons: PbCl4 contains Pb4+ ions – needs 4 electrons CrCl3 contains Cr3+ ions – needs 3 electrons
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Calculating Avogadro’s Constant
Avogadro’s constant can be determined by electrolysis
Follow the instructions here
This goes beyond the requirements of the syllabus but is interesting and will deepen your understanding
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Key Points
The longer the electrolysis, the more products will form
All else being equal, metal ions with a lower charge will produce more metal during electrolysis than those with a higher charge
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Lesson 11-12
Internal Assessment
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Internal Assessment
You should design and conduct and internal assessment on an aspect of oxidation and reduction.
Electrolysis and voltaic cells both provide rich hunting grounds.