Chapter Eighteen

Chapter Eighteen

Electrochemistry

1

Prentice Hall © 2005

General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry

Chapter Eighteen

Today…

• Turn in:

– Nothing

• Our Plan:

– Test Results

– Notes – Redox Equations

– Worksheet #1

• Homework (Write in Planner):

– Redox Equation WS due Friday



Chapter Eighteen

2


6% Curve

A

B

C

D

4

1

2

5

F

Average

4

72.72%

High Score 90% (x2)


Chapter Eighteen

3

Oxidation –Reduction:

The Transfer of Electrons

4

Electrons from copper metal are transferred to silver ions.



Silver metal is formed, and the solution turns blue from copper(II) ions formed.

Chapter Eighteen

Half-Reactions (Review)

• In any oxidation–reduction reaction, there are two half-

reactions:

–

Oxidation : a species loses electrons to another species.

(LEO)

–

Reduction : a species gains electrons from another species. (GER)

• Both oxidation and reduction must occur simultaneously.

– A species that loses electrons must lose them to something else (something that gains them).

– A species that gains electrons must gain them from something else (something that loses them).

Prentice Hall © 2005 Chapter Eighteen


5

Oxidation Numbers

• An oxidation number is the charge on an ion, or a hypothetical charge assigned to an atom in a molecule or polyatomic ion.

• Examples: in NaCl, the oxidation number of Na is

+1, that of Cl is –1 (the actual charge).

• In CO

2

(a molecular compound, no ions) the oxidation number of oxygen is –2, because oxygen as an ion would be expected to have a 2– charge.

• What is the oxidation number on carbon in CO

2

?



6

Rules for Assigning Oxidation Numbers

1. For the atoms in a neutral species —an isolated atom, a molecule, or a formula unit —the sum of all the oxidation numbers is 0 .

–

This includes elements in their standard state (Cu (s), Cl

2

(g), etc.)

2. For the atoms in an ion, the sum of the oxidation numbers is equal to the charge on the ion .

3. In compounds, the group 1A metals all have an oxidation number of +1 and the group 2A metals all have an oxidation number of +2.

4. In compounds, the oxidation number of fluorine is –1.

5. In compounds, hydrogen has an oxidation number of +1.

6. In most compounds, oxygen has an oxidation number of –2.

7. In binary compounds with metals, group 7A elements have an oxidation number of –1, group 6A elements have an oxidation number of –2, and group 5A elements have an oxidation number of –3.

7



Let’s Try It…

• Assign oxidation numbers to the elements in each compound:

1. NO

3

-1

2. SO

2

3. Fe

2

O

3

4. Cu (s)



8

Identifying Oxidation –Reduction

Reactions

• In a redox reaction, the oxidation number of a species changes during the reaction.

•

Oxidation occurs when the oxidation number increases (species loses electrons). LEO

•

Reduction occurs when the oxidation number decreases (species gains electrons). GER

•

Another mnemonic is OIL RIG!

• If any species is oxidized or reduced in a reaction, that reaction is a redox reaction.

• Examples of redox reactions: displacement of an element by another element; combustion; incorporation of an element into a compound, etc.



9

Review

• In the reactions below, write oxidation numbers for each substance and identify which substance is oxidized and which is reduced?

• N

2

O + H

2

--> H

2

O + NH

3

• K + KNO

3

--> N

2

+ K

2

O

• Fe

2

O

3

+ S --> Fe + SO

2



10

11

The Half-Reaction Method of

Balancing Redox Equations

1.

Separate a redox equation into two half-equations, one for oxidation and one for reduction.

2.

Balance the number of atoms of each element in each half-equation. Usually we balance O and H atoms last.

3.

Balance each half-reaction for charge by adding electrons to the left in the reduction half-equation and to the right in the oxidation half-equation.

4.

Adjust the coefficients in the half-equations so that the same number of electrons appears in each half-equation.

5.

Add together the two adjusted half-equations to obtain an overall redox equation.

6.

Simplify the overall redox equation as necessary.



Redox Reactions in Acidic and in

Basic Solution

• Redox reactions in acidic solution and in basic solution may be very different from one another.

• If acidic solution is specified, we must add H

2

O and/or H + as needed when we balance the number of atoms.

• If basic solution is specified, the final equation may have

OH

– and/or water molecules in it.

• A simple way to balance an equation in basic solution:

– Balance the equation as though it were in acidic solution.

– Add as many OH – ions to each side as there are H + ions in the equation.

– Combine the H + and OH

– ions to give water molecules on one side, and simplify the equation as necessary.

12



Example 18.1

• Balance the equation in acidic solution:

MnO

4

-1 + S

2

O

3

-2 --> Mn +2 + SO

4

-2

13



Chapter Eighteen

Example 18.1 B

• Write a balanced equation for the oxidation of phosphorus by nitric acid, which is described by

P

4

(s) + H +1 (aq) + NO

3

-1 (aq) --> H

2

PO

4

-1 (aq) + NO (g)

14



Chapter Eighteen

Example 18.2

• Balance the following in basic solution:

Br

2

(l) --> Br -1 (aq) + BrO

3

-1 (aq)

15



Chapter Eighteen

Example 30a

Balance the following in basic solution:

CrO

4

-2 (aq) + AsH

3

(g) --> Cr(OH)

3

(s) + As (s)

16



Chapter Eighteen

WS Example 1

HCl + K

2

Cr

2

O

7

--> KCl + CrCl

3

+ H

2

O + Cl

2

17



Chapter Eighteen

WS Example 2

FeCl

2

+ KMnO

4

+ HCl --> FeCl

3

+ KCl + MnCl

2

+ H

2

O

18



Chapter Eighteen

WS Example 3

S -2 + MnO

4

-1 --> S + MnO

2

(basic solution)

19



Chapter Eighteen

WS Example 4

CuS + NO

3

-1 --> Cu +2 + S + NO (acidic)

20



Chapter Eighteen

Stop!

• Do the Redox Equations Worksheet.

21



Chapter Eighteen

Today…

• Turn in:

– Get out Redox WS

• Our Plan:

– Questions on Redox WS

– Quiz

– Notes

– Galvanic Cells WS


– Worksheet Due Wednesday



Chapter Eighteen

22

A Qualitative Description of

Voltaic Cells

• A voltaic cell uses a spontaneous redox reaction to produce electricity.

• A half-cell consists of an electrode (strip of metal or other conductor) immersed in a solution of ions.

This Zn 2+ becomes a Zn atom.

23


Both oxidation and reduction occur at the electrode surface, and equilibrium is reached.

This Zn atom leaves the surface to become a Zn 2+ ion.


Chapter Eighteen

24

Important Electrochemical Terms

• An electrochemical cell consists of two half-cells with the appropriate connections between electrodes and solutions.

• Two half-cells may be joined by a salt bridge (U shaped tube suspended in gel) that permits migration of ions, without completely mixing the solutions.

• The anode is the electrode at which oxidation occurs.

• The cathode is the electrode at which reduction occurs.



25

Helpful Mnemonic

• AUTO (anode = oxidation)

• CAR (cathode = reduction)

OR:

• To remember the charge: Ca+ions are attracted to the Ca+hode (the t is a plus sign)

• To remember which reaction occurs at which terminal: An Ox and Red Cat -

Anode Oxidation, Reduction Cathode



Chapter Eighteen

26

Important Electrochemical Terms

• In a voltaic cell , a spontaneous redox reaction occurs and current (electricity) is generated.

•

Cell potential ( E cell

) is the potential difference in volts between anode and cathode.

•

E cell is the driving force that moves electrons and ions.



A Zinc –Copper

Voltaic Cell

… the electrons produced move through the wire …

Positive and negative ions move through the salt bridge to equalize the charge.

27

… to the Cu(s) electrode, where they are accepted by Cu 2+ ions to form more Cu(s).

Zn(s) is oxidized to

Zn 2+ ions, and …


Reaction: Zn(s) + Cu 2+ (aq) --> Cu(s) + Zn 2+ (aq)

Chapter Eighteen


Cell Diagrams

• A cell diagram is “shorthand” for an electrochemical cell.

• The anode is placed on the left side of the diagram.

• The cathode is placed on the right side.

• A single vertical line ( | ) represents a boundary between phases, such as between an electrode and a solution.

• A double vertical line ( || ) represents a salt bridge or porous barrier separating two half-cells.

28



Chapter Eighteen

29


Reaction: Zn(s) + Cu 2+ (aq) --> Cu(s) + Zn 2+ (aq)

Chapter Eighteen


30



Chapter Eighteen

Example 18.3

Describe the half-reactions and the overall reaction that occur in the voltaic cell represented by the cell diagram:

Pt(s) | Fe 2+ (aq), Fe 3+ (aq) || Cl

–

(aq) | Cl

2

(g) | Pt(s)

31



Chapter Eighteen

32

Standard Electrode Potentials

• Since an electrode represents only a half-reaction, it is not possible to measure the absolute potential of an electrode.

• The standard hydrogen electrode ( SHE ) provides a reference for measurement of other electrode potentials.

• The SHE is arbitrarily assigned a potential of 0.000 V.



33


• The standard electrode potential, E° , is based on the tendency for reduction to occur at an electrode. (Sometimes it is referred to as standard reduction potential)

• All standard reduction potentials are for 25 ᵒC and

1 M solutions.

•

E° for the standard hydrogen electrode is arbitrarily assigned a value of 0.000 V.

• All other values of

E° are determined relative to the standard hydrogen electrode.




• The standard cell potential ( E°

cell

) is the difference between

E°

of the cathode and

E°

of the anode .

•

E° cell

= E°

(cathode) –

E°

• Remember AUTO CAR!

(anode)

34



Chapter Eighteen

Measuring the Standard Potential of the Cu 2+ /Cu Electrode

The voltmeter reading and the direction of electron flow tell us that …

… Cu 2+ is more easily

reduced than H + , by

0.340 volts.

35

Standard hydrogen electrode

Cu 2+ + 2e --> Cu

E° = + 0.340 V



Measuring the Standard Potential of the Zn 2+ /Zn Electrode

The voltmeter reading and the direction of electron flow tell us that …

… Zn 2+ is harder to

reduce than H + , by

0.763 volts.

Standard hydrogen electrode

Zn 2+ + 2e --> Zn

E° = – 0.763 V

36



F

2 is the strongest oxidizing agent

F

– is the weakest reducing agent

37

Li is the strongest reducing agent



Chapter Eighteen

38

Important Note

• The formula

E° cell

= E°

(cathode) –

E°

(anode) allows us to simply use the numbers in the table, but if you were asked for the oxidation value of a reaction, you would have to reverse the sign since the values are all reduction potentials.



Chapter Eighteen

39

Important Points about

Electrode and Cell Potentials

• Standard electrode potentials and cell voltages are intensive properties; they do not depend on the total amounts of the species present.

• A “flashlight battery” (D-cell) and a “penlight battery” (AA cell) produce the same potential—

1.5 volts.

•

E

° does depend on the particular species in the reaction (or half-reaction).

• As we shall learn later, cell and electrode potentials can depend on concentration of the species present.



Chapter Eighteen

Example 18.4

Determine E

° for the reduction half-reaction

Ce 4+ (aq) + e

–

--> Ce 3+ (aq), given that the cell voltage for the voltaic cell

Co(s) | Co 2+ (1 M) || Ce 4+ (1 M), Ce 3+ (1 M) | Pt(s) is E ° cell

= 1.887 V.

40



Chapter Eighteen

Example 18.5

Balance the following oxidation –reduction equation, and determine E ° cell

O

2 for the reaction.

(g) + H + (aq) + I – (aq) --> H

2

O(l) + I

2

(s)

41



Chapter Eighteen

Electrode Potentials, Spontaneous

Change, and Equilibrium

• An electrochemical cell does work.

w elec

= nFE cell n = number of electrons in the balanced equation

F = 96,485 coulombs per mole.

• The amount of electrical work is also equal to – D

G :

D

G = – nFE cell

• Under standard conditions:

D

G° = – nFE° cell

42



43

Criteria for Spontaneous

Change in Redox Reactions

• If

E cell is positive spontaneous.

, the forward reaction is

• If

E cell is negative , the forward reaction is nonspontaneous (the reverse reaction is spontaneous).

• If

E cell

= 0, the system is at equilibrium

• When a cell reaction is reversed , E cell

D

G change signs.

.

and



Example 18.6

Will copper metal displace silver ion from aqueous solution? That is, does the reaction

Cu(s) + 2 Ag + (1 M) --> Cu 2+ (1 M) + 2 Ag(s) occur spontaneously from left to right?

44



Chapter Eighteen

45

The Activity Series Revisited

• In the activity series of metals

(Section 4.4), any metal in the series will displace a metal below it from a solution of that metal’s ions.

• Theoretical basis: The activity series lists metals in order of their standard potentials .

• Displacement of a metal from a solution of its ions by a metal higher in the series corresponds to a positive value of E cell spontaneous reaction. and a



Chapter Eighteen

Visual Example

• http://intro.chem.okstate.edu/1515F01/Labo ratory/ActivityofMetals/home.html

46



Chapter Eighteen

47

Equilibrium Constants in Redox

Reactions

• Whereas potential and free energy are related, and free energy and equilibrium are related, equilibrium and potential must be related to one another.

and

D

G ° = – nFE ° cell

D

G

° = –

RT ln K eq therefore

–

RT ln K eq

= – nFE o cell

E

° cell

RT ln K

= ––––––––– = –––– ln K nF eq

RT nF eq

R and F are constant, therefore at 298 K:

E ° cell

0.025693 V

= –––––––– ln K n eq



Example 18.8

Calculate the values of Δ G

° and

K eq at 25 °C for the reaction

Cu(s) + 2 Ag + (1 M) --> Cu 2+ (1 M) + 2 Ag(s)

48



Chapter Eighteen

Thermodynamics, Equilibrium, and

Electrochemistry: A Summary

From any one of the three quantities K eq

, ΔG°, E° cell

, we can determine the others.

49



Chapter Eighteen

Stop!

• Begin working on Worksheet #2!

50



Chapter Eighteen

Today…

• Turn in:

– Nothing

• Our Plan:

– Quick Review

– Notes – Concentration & Electrochemical

Cells

– Finish Electrochemical Cells WS


– Worksheet Due Wednesday



Chapter Eighteen

51

Effect of Concentrations on Cell

Voltage

• A nonstandard cell differs in potential from a standard cell

(1 M concentrations, 1 atm partial pressures).

52



Chapter Eighteen

53

Effect of Concentrations on Cell

Voltage

From the previous relationships we can show that:

E cell

= E

° cell

RT

– –––– ln Q nF

At 25 °C, and converting to common logarithms:

E cell

= E

° cell

0.0592 V

– ––––––– log

Q n

• This

Nernst equation relates a cell voltage for nonstandard conditions, E cell

, to a standard cell voltage, E

° cell

, and to the concentrations of reactants and products expressed through the reaction quotient, Q .

• We can use the Nernst equation to find cell potential from concentrations, or we can measure E cell concentration of a species in the cell.

and determine the



Example 18.9

Calculate the expected voltmeter reading for the voltaic cell pictured in Figure 18.13.

54



Chapter Eighteen

Another Nernst Equation Example

18.9 A

• Use the Nernst equation to determine E cell

25ᵒC for the following voltaic cells.

at a) Zn(s) | Zn 2+ (2.0 M) || Cu 2+ (0.050 M) | Cu (s) b) Zn(s) | Zn 2+ (0.050 M) || Cu 2+ (2.0 M) | Cu (s)

55



Chapter Eighteen

56

Batteries: Using Chemical

Reactions to Make Electricity

• We often call any device that stores chemical energy for later release as electricity a battery.

• Technically, a D, C, or AA “battery” is actually a single electrochemical cell .

• A battery consists of several cells joined together to produce higher current or higher voltage.

• A 9-volt “transistor” battery, an automobile battery, and a rechargable “battery pack” are all true batteries.



57

The Dry Cell

• A primary cell employs an irreversible chemical reaction.

• When the reactants inside the cell are largely used up, the cell is “dead.”

• The LeClanché cell or dry cell

(right) is the “ordinary” type of flashlight “battery.”

•

Alkaline cells cost more than the LeClanché cell but they have a longer shelf life and longer service life.



Chapter Eighteen

58

The Lead –Acid Storage Battery

• A lead–acid storage battery used in an automobile uses secondary cells ; they are rechargeable.

• By connecting the cell to an external electric energy source, the discharge reaction is reversed.

Cell reaction: Pb(s) + PbO

2

(s) + 2 H + (aq) + 2 HSO

4

–

(aq) --> 2 PbSO

4

(s) + 2 H

2

O(l)

Charging reaction: 2 PbSO

4

(s) + 2 H

2

O(l) --> Pb(s) + PbO

2

(s) + 2 H + (aq) + 2 HSO

4

–

(aq)



59

Other Secondary Cells

• The nickel –cadmium (NiCd) cell uses a cadmium anode and a cathode containing Ni(OH)

2

.

• A NiCd cell can be recharged hundreds of times. It produces 1.2 V (a Leclanché cell produces 1.5 V).

• Nickel –metal hydride cells (NiMH) use a metal alloy anode that contains hydrogen.

• In use, the anode releases the hydrogen, forming water. Like the NiCd cell, a NiMH cell produces 1.2 V.

• Lithium-ion cells use a lithium –cobalt oxide or lithium– manganese oxide material as the anode. The electrolyte is an organic solvent containing a dissolved lithium salt.

• Many modern laptop computers and cellular phones use lithium-ion cells.



Chapter Eighteen

60

Fuel Cells

• In a fuel cell , the cell reaction is equivalent to a combustion reaction.

• The reactants are supplied externally; the cell does not “go dead” as long as the oxidizing and reducing agents are provided.

• Fuel cells are generally operated under nonstandard conditions and at temperatures considerably higher than 25 °C.

• H

2

–O

2 fuel cells are seeing use in some automobiles.



Chapter Eighteen

61

Corrosion: Metal Loss

Through Voltaic Cells

• In moist air, iron is oxidized to Fe 2+ , particularly at scratches, nicks, or dents. These areas are referred to as anodic areas.

• Other regions of the iron serve as cathodic areas, where the electrons from the anodic areas reduce O

2 to OH

–

.

• Iron(II) ions migrate from the anodic areas to the cathodic areas where they combine with the hydroxide ions.

• The iron(II) is then further oxidized to iron(III) by atmospheric oxygen.

• Common rust is Fe

2

O

3

· x H

2

O.



Corrosion of an Iron Piling

62

One way to minimize rusting is to provide a

different anode reaction.



Chapter Eighteen

Protecting Iron from Corrosion

• The simplest defense against corrosion of iron is to coat it

(with paint or metal) to exclude oxygen from the surface.

• An entirely different approach is to protect iron with a more active metal.

•

Galvanized iron has been coated with zinc.

• The zinc provides an alternative anode reaction; the zinc corrodes, protecting the iron.

63

Chapter Eighteen Prentice Hall © 2005


64

Cathodic Protection

• In cathodic protection , an iron object to be protected is connected to a chunk of an active metal.

• The iron serves as the reduction electrode and remains metallic. The active metal is oxidized.

• Water heaters often employ a magnesium anode for cathodic protection.



Chapter Eighteen

Stop!

• Finish Worksheet #2 –

Electrochemical Cells

Worksheet

65



Chapter Eighteen

Today…

• Turn in:

– Get out Electrochemical Cells WS

• Our Plan:

– Questions on WS

– Electrochemical Cell Review

– Notes

– Electrolysis WS


– Worksheet Due Friday

(LAST HOMEWORK)



Chapter Eighteen

66

67

Electrolytic Cells

• A voltaic cell corresponds to a spontaneous cell reaction.

• An electrolytic cell corresponds to a nonspontaneous cell reaction. The reaction is called electrolysis.

• An electrolytic cell is the opposite of a galvanic cell.

• The external source of electricity acts like an “electron pump.” It pulls electrons away from the anode , where oxidation takes place, and pushes them toward the cathode , where reduction takes place.

• The polarities of the electrodes are reversed from those in the voltaic cell, because now the external source controls the flow of electrons.




Electrolysis of Molten

Sodium Chloride

2 NaCl(l) --> 2 Na(l) + Cl

2

(g)

The nonspontaneous reaction is driven by external potential.

68

Molten NaCl

(around 1000 °C)

Chapter Eighteen


69

Predicting Electrolysis Reactions

• In an electrolytic cell, all combinations of cathode and anode half-reactions give negative values of E° cell

.

• The reaction most likely to occur is the one with the least negative value of E° cell

(requires the lowest applied voltage from the external electricity source). HOWEVER …

• In many half-reactions, particularly those involving gases, various interactions at electrode surfaces make the required voltage for electrolysis higher than the voltage calculated from E° data.

• Overvoltage is the excess voltage above the voltage calculated from E° values that is required in electrolysis.



70

Quantitative Electrolysis

• The unit of electric charge is the coulomb ( C ), and the charge carried by one electron is –1.6022 x 10

–19

C.

– The conversion factor we’ll use is

96,485 C/1 mole electrons

• Electric current, expressed in amperes ( A ), is the rate of flow of electric charge (C/s).



71

Quantitative Electrolysis

• To calculate the quantitative outcome of an electrolysis reaction:

1. Determine the amount of charge (C)—the product of current and time.

2. Convert the amount of charge to moles of electrons.

3. Use a half-equation to relate moles of electrons to moles of a reactant or a product.

4. Convert from moles of reactant or product to the final quantity desired.



Example 18.13

We can use electrolysis to determine the gold content of a sample.

The sample is dissolved, and all the gold is converted to Au 3+ (aq), which is then reduced back to Au(s) on an electrode of known mass. The reduction half-reaction is Au 3+ (aq) + 3e

– 

Au(s).

What mass of gold will be deposited at the cathode in 1.00 hour by a current of 1.50 A?

72



Chapter Eighteen

Try It Out!

• 18.13 A: For how many minutes must the electrolysis of a solution of CuSO

4

(aq) be carried out, at a current of 2.25 A, to deposit

1.00 g of Cu (s) at the cathode?

73



Chapter Eighteen

Example 18.14 An Estimation Example

Without doing detailed calculations , determine which of the following solutions will yield the greatest mass of metal at a platinum cathode during electrolysis by a 1.50-A electric current for 30.2 min: CuSO

4

(aq), AgNO

3

(aq), or AuCl

3

(aq).

74



Chapter Eighteen

75

Producing Chemicals by Electrolysis

Electrolysis plays an important role in the manufacture and purification of many substances, including chlorine, copper, silver, magnesium, aluminum, lead, zinc, sodium, fluorine, titanium, sodium hydroxide, hydrogen …

Electrolysis of NaCl(aq) is used to produce H

2

, Cl

2

, and

NaOH, all of which have important industrial uses.



Chapter Eighteen

76

Electroplating

• Electrolysis can be used to coat one metal onto another, a process called electroplating .

• Usually, the object to be electroplated, such as a spoon, is cast of an inexpensive metal. It is then coated with a thin layer of a more attractive, corrosionresistant, and expensive metal, such as silver or gold.



Chapter Eighteen

Stop

•Complete the

Electrolysis

Worksheet



Chapter Eighteen

77

Today…

• Turn in:

– Get out Electrolysis WS

• Our Plan:

– Questions on WS

– Electrochemical Cells Online Lab


– Take Home Test Due Monday OR prepare for the AP Exam!



Chapter Eighteen

78

A Little Redox Humor

79



Chapter Eighteen

Today…

• Turn in:

– Take Home Test

• Our Plan:

– Begin Qualitative Analysis Lab (read the pre-lab and procedure FIRST)


– Work on Lab Report



80

81

Lab Preparation

• Read the Packet, including the entire procedure.

• Answer the Pre-Lab Questions

• Begin working on your lab report

– I recommend recording data in the handout that

I gave you and then transferring it to your final copy.

– You could begin working on the title, purpose, pre-lab, and procedure, though.



82

Reminders

• Safety – Goggles, gloves, apron at ALL TIMES!

•

Clean all glassware and lab supplies before and after use.

•

Label things well (include your initials)!

•

Seal containers to store until next class period. Place them on the back bookshelf.

•

All solids go in the trash and liquids down the drain

EXCEPT barium (anion step 2 &3). Put them in the waste container in the hood!

• Run tests until you get clear and reproducible results

(you can repeat steps if you need to).

• Report is due 5/14 at the beginning of class!



Chapter Eighteen

Today…

• Turn in:

– Nothing

• Our Plan:

– Qualitative Analysis Lab


– Report Due on 8/14



Chapter Eighteen

83

Chapter Eighteen

Chapter Eighteen

Electrochemistry

6% Curve

• Assign oxidation numbers to the elements in each compound:

1. NO

2. SO

3. Fe

O

4. Cu (s)

• Fe

O

+ S --> Fe + SO

WS Example 1

WS Example 2

WS Example 3

WS Example 4

Stop!

• The standard cell potential ( E°

) is the difference between

of the cathode and

of the anode .

•

(cathode) –

• Remember AUTO CAR!

(anode)

• The formula

(cathode) –

(anode) allows us to simply use the numbers in the table, but if you were asked for the oxidation value of a reaction, you would have to reverse the sign since the values are all reduction potentials.

• Finish Worksheet #2 –

Electrochemical Cells

Worksheet

Stop

•Complete the

Electrolysis

Worksheet

• Homework (Write in Planner):

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Chapter Eighteen