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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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.