chapter 17 electrochemistry why does leo say ger?

Chapter 17

Electrochemistry

Why does LEO say GER?

Chapter 17

Table of Contents

Copyright © Cengage Learning. All rights reserved 2

17.0 Balancing Oxidation–Reduction Equations I

17.1 Galvanic Cells

17.2 Standard Reduction Potentials

17.3 Cell Potential, Electrical Work, and Free Energy

17.4 Dependence of Cell Potential on Concentration

17.5 Batteries

17.6 Corrosion

17.7 Electrolysis

17.8 Commercial Eletrolytic Processes

Section 17.0

Balancing Oxidation–Reduction Equations

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• To understand what oxidation – reduction reactions (REDOX)

• To assign the oxidation state to elements and ions

• To identify the oxidizing agent and the reducing agent

• To write balanced REDOX half-reactions

Objectives

Section 17.0


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• Reactions in which one or more electrons are transferred.

Redox Reactions

Section 17.0


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Reaction of Sodium and Chlorine

Section 17.0


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1. Oxidation state of an atom in an element = 0

2. Oxidation state of monatomic ion = charge of the ion

3. Oxygen = 2 in covalent compounds (except in peroxides where it = 1)

4. Hydrogen = +1 in covalent compounds

5. Fluorine = 1 in compounds

6. Sum of oxidation states = 0 in compounds

7. Sum of oxidation states = charge of the ion in ions

Rules for Assigning Oxidation States (Gangsta Switch)

Section 17.0


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Exercise

Find the oxidation states for each of the elements in each of the following compounds: (Start with what you know)

• K2Cr2O7

• CO32-

• MnO2

• PCl5• SF4

K = +1; Cr = +6; O = –2

C = +4; O = –2

Mn = +4; O = –2

P = +5; Cl = –1

S = +4; F = –1

Section 17.0


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• Transfer of electrons• Transfer may occur to form ions• Oxidation – increase in oxidation state

(loss of electrons); reducing agent• Reduction – decrease in oxidation state

(gain of electrons); oxidizing agent• Leo says Ger!

Redox Characteristics

Section 17.0


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Concept CheckUse the oxidation states to determine which of the following are oxidation-reduction reactions? Identify the oxidizing agent and the reducing agent.

a)Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g)

Section 17.0


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Concept CheckUse the oxidation states to determine which of the following are oxidation-reduction reactions? Identify the oxidizing agent and the reducing agent.

a)2CuCl(aq) CuCl2(aq) + Cu(s) (1/2 Rxs)

Section 17.0


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• To understand what oxidation – reduction reactions (REDOX)

• To assign the oxidation state to elements and ions

• To identify the oxidizing agent and the reducing agent

• Page 830 #13, 14(every other), • 15a-d also write the half reactions

Objectives Review

Section 17.0


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• To balance half reactions that occur in acid solutions

Objectives Continued

Section 17.0


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Review of Terms

• Oxidation–reduction (redox) reaction – involves a transfer of electrons from the reducing agent to the oxidizing agent

• Oxidation – loss of electrons (LEO)• Reduction – gain of electrons (GER)• Reducing agent – electron donor (loser)• Oxidizing agent – electron acceptor (gainer)

Section 17.0


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Half–Reactions

• Write the half reactions for the following:•

8H+ + MnO4- + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O

Oxidation: 5Fe2+ 5Fe3+ + 5e-

Reduction: 8H+ + MnO4- + 5e- Mn2+ + 4H2O

Section 17.0


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Write the half reactions for the following:

Cr2O72-(aq) + SO3

2-(aq) Cr3+(aq) + SO42-(aq)

• 6e- + Cr2O72-(aq) 2Cr3+(aq)

• SO32-(aq) + SO4

2-(aq) + 2e-

• Is this balanced? Rules:

Section 17.0


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1. Write separate equations for the oxidation and reduction half–reactions.

2. For each half–reaction:A. Balance all the elements except H and O.

B. Balance O using H2O.

C. Balance H using H+.

D. Balance the charge using electrons.

The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution

Section 17.0


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3. If necessary, multiply one or both balanced half–reactions by an integer to equalize the number of electrons transferred in the two half–reactions.

4. Add the half–reactions, and cancel identical species.

5. Check that the elements and charges are balanced.


Section 17.0


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Cr2O72-(aq) + SO3

2-(aq) Cr3+(aq) + SO42-(aq)

• 6e- + Cr2O72-(aq) 2Cr3+(aq)

• SO32-(aq) + SO4

2-(aq) + 2e-

Separate into half-reactions. Balance elements except H and O. Balance O with H2O and H with H+

Section 17.0


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• 6e- + Cr2O72-(aq) 2Cr3+(aq)

• SO32-(aq) + SO4

2-(aq) + 2e-

• How can we balance the oxygen atoms?

Method of Half Reactions (continued)

Section 17.0


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• 6e- + Cr2O72-(aq) 2Cr3+(aq) + 7H2O

• H2O +SO32-(aq) + SO4

2-(aq) + 2e-

• How can we balance the hydrogen atoms?


Section 17.0


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• This reaction occurs in an acidic solution.

• 14H+ + 6e- + Cr2O72- 2Cr3+ + 7H2O

• H2O +SO32- SO4

2- + 2e- + 2H+

• How can we balance the electrons?


Section 17.0


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• 14H+ + 6e- + Cr2O72- 2Cr3+ + 7H2O

• 3[H2O +SO32- SO4

2- + 2e- + 2H+]

• Cancel to get the Final Balanced Equation:

Cr2O72- + 3SO3

2- + 8H+ 2Cr3+ + 3SO42- + 4H2O

WOW!!


Section 17.0


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Exercise

Balance the following oxidation–

reduction reaction that occurs in acidic solution.

Br–(aq) + MnO4–(aq) Br2(l)+ Mn2+(aq)

10Br–(aq) + 16H+(aq) + 2MnO4–(aq) 5Br2(l)+ 2Mn2+(aq) + 8H2O(l)

Section 17.0


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1. Use the half–reaction method as specified for acidic solutions to obtain the final balanced equation as if H+ ions were present.

2. To both sides of the equation, add a number of OH– ions that is equal to the number of H+ ions. (We want to eliminate H+ by forming H2O.)

The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution

Section 17.0


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3. Form H2O on the side containing both H+ and OH– ions, and eliminate the number of H2O molecules that appear on both sides of the equation.

4. Check that elements and charges are balanced.


Section 17.0


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• To balance half reactions that occur in acid solutions

• Work Session: page 830 # 16 a-c acid

Objectives Review

Section 17.1

Atomic MassesGalvanic Cells

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• To understand what a galvanic cell is and how it works

• To calculate cell potential for a given reaction

Objectives

cellE

Section 17.1


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Galvanic Cell

• Device in which chemical energy is changed to electrical energy.

• Uses a spontaneous redox reaction to produce a current that can be used to do work.

Section 17.1


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A Galvanic Cell

Section 17.1


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Galvanic Cell

• Oxidation occurs at the anode. (O and A vowels)• Reduction occurs at the cathode. (consonants)• Salt bridge or porous disk – devices that allow ions

to flow without extensive mixing of the solutions. Salt bridge – contains a strong electrolyte held

in a Jello–like matrix. Porous disk – contains tiny passages that allow

hindered flow of ions.

Section 17.1


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Voltaic Cell: Anode Reaction (losing e-s LEO)

Section 17.1


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Voltaic Cell: Cathode Reaction (gaining e-s GER)

Section 17.1


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Cell Potential

• A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment.

• The “pull”, or driving force, on the electrons is called the cell potential ( ), or the electromotive force (emf) of the cell. Unit of electrical potential is the volt (V).

1 joule of work per coulomb of charge transferred.

cellE

Section 17.2

The Mole Standard Reduction Potentials

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Electrochemical Half-Reactions in a Galvanic Cell

Section 17.2


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Galvanic Cell• All half-reactions are given as reduction

processes in standard tables. Table 17.1, 1 M, 1atm, 25°C

• When a half-reaction is reversed, the sign of E ° is reversed. Won’t be a factor in the calculation*

• When a half-reaction is multiplied by an integer, E ° remains the same.

• A galvanic cell runs spontaneously in the direction that gives a positive value for E °cell.

• E °cell = E °(cathode) - E °(anode) *Use numbers directly from Table 17.1**

Section 17.2


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Example: Fe3+(aq) + Cu(s) → Cu2+(aq) + Fe2+(aq)

• Write the Half-Reactions and list the E from 17.1 Fe3+ + e– → Fe2+ E ° = 0.77 V

Cu2+ + 2e– → Cu E ° = 0.34 V

• Balance the overall reaction to determine Anode/Cathode• Cu → Cu2+ + 2e– – E ° = – 0.34 V (LEO Anode)

2Fe3+ + 2e– → 2Fe2+ E ° = 0.77 V (GER Cathode) Cu + 2Fe3+ → Cu2+ + 2Fe2+

Calculate E °cell = E °(cathode) - E °(anode)

E °cell = 0.77 – 0.34 = 0.43 V (+ is spontaneous)

Section 17.2


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Example: Al 3+(aq) + Mg(s) → Al(s) + Mg2+(aq)

• Write the Half-Reactions and list the E from 17.1 Al3+ + 3e– → Al E ° = -1.66 V

Mg2+ + 2e– → Mg E ° = -2.37 V

• Balance the overall reaction to determine Anode/Cathode• 3[Mg → Mg2+ + 2e– ] – E ° = 2.37 V (LEO Anode)

2[Al3+ + 3e– → Al] E ° = -1.66 V (GER Cathode) 3Mg + 2Al3+ → 3Mg2+ + 2Al


E °cell = -1.66 –(- 2.37) = 0.71 V (+ is spontaneous)

Section 17.2


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MnO4- (aq) + H+ (aq) + ClO3

- (aq) → ClO4- (aq) + Mn2+(aq) + H2O

• Write the Half-Reactions and list the E from 17.1 • ClO4

- + 2H+ + 2e- ClO3- + H2O E ° = 1.19 V

• MnO4- + 5e- + 8H+ Mn+2 + 4H2O E ° = 1.51 V

• Balance the overall cell reaction • E ° = V (LEO Anode)

E ° = V (GER Cath)


E °cell = - = V (spontaneous??)

Section 17.2


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Line Notation

• Used to describe electrochemical cells.• Anode components are listed on the left.• Cathode components are listed on the right.• Separated by double vertical lines.• The concentration of aqueous solutions should

be specified in the notation when known.• Example: Mg(s)|Mg2+(aq)||Al3+(aq)|Al(s)

Mg → Mg2+ + 2e– (anode) Al3+ + 3e– → Al (cathode) ABC Order…

Section 17.2


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Line Notation

• Page 831 # 25 and 27 also show the line notation

• Also 26 and 28 with line notation

• Which way does the Potato Clock work?• Cu(s)|Cu2+(aq)||Zn2+(aq)|Zn(s)• Or• Zn(s)|Zn+2(aq)||Cu2+(aq)|Cu(s)

Section 17.2


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Concept Check

Sketch a cell using the following solutions and electrodes. Include:

The potential of the cell The direction of electron flow Labels on the anode and the cathode

a) Ag electrode in 1.0 M Ag+(aq) and Cu electrode in 1.0 M Cu2+(aq)

Section 17.2


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Concept Check

Sketch a cell using the following solutions and electrodes. Include:

The potential of the cell The direction of electron flow Labels on the anode and the cathode

b) Zn electrode in 1.0 M Zn2+(aq) and Cu electrode in 1.0 M Cu2+(aq)

Section 17.2


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Concept Check

Consider the cell from part b.

What would happen to the potential if you increase the [Cu2+]?

Explain.

The cell potential should increase.

Section 17.2


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Concept Check

Page 831 # 29 sketch the 2 cells, pickup #27

Quiz #30??

end

Section 17.3

Cell Potential, Electrical Work, and Free Energy

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• To understand that the work of a cell is always less that the calculated potential

• To predict the spontaneity of a galvanic cell

Objectives

Section 17.3


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Work

• Work is never the maximum possible if any current is flowing.

• In any real, spontaneous process some energy is always wasted – the actual work realized is always less than the calculated maximum.

Section 17.3


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Spontaneity Prediction

• Use the data from Table 17.1 to determine if 1 M HNO3 will dissolve gold metal to form a 1 M Au+3

solution.• In other words, calculate the E ° for this galvanic cell.

Remember, + E ° is spontaneous!

• E ° = -0.54 V

Section 17.3


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• To understand that the work of a cell is always less that the calculated potential

• To predict the spontaneity of a galvanic cell

• Work Session: 37, just calculate E °, no G

Objectives Review

Section 17.4

Dependence of Cell Potential on Concentration

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• To understand that the Nernst Equation is used to correct E ° for nonstandard conditions- namely different concentrations

• To calculate the cell potential using the Nernst Equation

• **have same metal, diff conc—add diff metal, diff conc

• **specify cathode/anode**smaller M is product, forming ion until [] = []

Objectives (Nernst Eq Excluded from AP Test 2014)

Section 17.4


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Concentration CellRemember, E ° is reported for 1 M solutions at 25 ° C

e-s flow to the higher concentration, forming more Ag + on left, Ag metal on right until the concentrations are equal. Lower concentration is the product side.

Section 17.4


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Nernst Equation• At 25°C:

Q is the reaction Quotient: Q = [products]A

[reactants]B

n = number of e-s transferred

What is the value of log(Q) when the concentrations are both 1 M?

0.0591 = log E E Q

n

Section 17.4


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Nernst Equation• At 25°C:

At equilibrium, the Reaction Quotient = K.

At equilibrium, the cell no longer has any chemical driving force to move electrons, making the

battery unable to do work. E = 0

0.0591 = logE K

n

0.0591 = log E E Q

n

Section 17.4


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Concept Check

Explain the difference between E and E °.

When is E equal to zero?

When the cell is in equilibrium ("dead" battery).

When is E ° equal to zero?

E is equal to zero for a concentration cell.

0.0591 = log E E Q

n

Section 17.4


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ExerciseA concentration cell is constructed using two nickel electrodes with Ni2+ concentrations of 1.0 M and 1.00 x 10-4 M in the two half-cells.

Calculate the potential of this cell at 25°C.

1)Calculate E °

2)Balance Eq to get n

3)Use Nernst Eq

0.118 V

Section 17.4


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Concept Check

You make a galvanic cell at 25°C containing: A nickel electrode in 1.0 M Ni2+(aq) A silver electrode in 1.0 M Ag+(aq)

Sketch this cell, labeling the anode and cathode, showing the direction of the electron flow, and calculate the cell potential.

1.03 V

Section 17.4


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• To understand that the Nernst Equation is used to correct E ° for nonstandard conditions- namely different concentrations

• To calculate the cell potential using the Nernst Equation

• Work Session: 53 a-d• Quiz 54 a-d??

Objectives

Section 17.5

Batteries

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One of the Six Cells in a 12–V Lead Storage Battery

Section 17.5

Batteries

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A Common Dry Cell Battery

Section 17.5

Batteries

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A Mercury Battery

Section 17.5

Batteries

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Schematic of the Hydrogen-Oxygen Fuel Cell

Section 17.6

Corrosion

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• Process of returning metals to their natural state – the ores from which they were originally obtained.

• Involves oxidation of the metal.

Section 17.6

Corrosion

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The Electrochemical Corrosion of Iron

Section 17.6

Corrosion

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Corrosion Prevention

• Application of a coating (like paint or metal plating) Galvanizing

• Alloying• Cathodic Protection

Protects steel in buried fuel tanks and pipelines.

Section 17.6

Corrosion

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Cathodic Protection

Section 17.7

Electrolysis

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• Forcing a current through a cell to produce a chemical change for which the cell potential is negative.

Section 17.7

Electrolysis

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Stoichiometry of Electrolysis

• How much chemical change occurs with the flow of a given current for a specified time?

current and time quantity of charge

moles of electrons moles of analyte

grams of analyte

Section 17.7

Electrolysis

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Stoichiometry of Electrolysis

• current and time quantity of chargeCoulombs of charge = amps (C/s) × seconds (s)

• quantity of charge moles of electrons1 mol e

mol e = Coulombs of charge 96,485 C

Section 17.7

Electrolysis

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Concept Check

An unknown metal (M) is electrolyzed. It took 52.8 sec for a current of 2.00 amp to plate 0.0719 g of the metal from a solution containing M(NO3)3.

What is the metal?

gold (Au)

Section 17.7

Electrolysis

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Concept Check

Consider a solution containing 0.10 M of each of the following: Pb2+, Cu2+, Sn2+, Ni2+, and Zn2+.

Predict the order in which the metals plate out as the voltage is turned up from zero.

Cu2+, Pb2+, Sn2+, Ni2+, Zn2+

Do the metals form on the cathode or the anode? Explain.

Section 17.8

Commercial Electrolytic Processes

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• Production of aluminum• Purification of metals• Metal plating• Electrolysis of sodium chloride• Production of chlorine and sodium hydroxide

Section 17.8


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Producing Aluminum by the Hall-Heroult Process

Section 17.8


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Electroplating a Spoon

Section 17.8


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The Downs Cell for the Electrolysis of Molten Sodium Chloride

Section 17.8


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The Mercury Cell for Production of Chlorine and Sodium Hydroxide

Section 17.0


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Section 17.2


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Concept Check

Order the following from strongest to weakest oxidizing agent and justify. Of those you cannot order, explain why.

Fe Na F- Na+ Cl2

Section 17.2


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Description of a Galvanic Cell

• The cell potential is always positive for a galvanic cell where E °cell = E °(cathode)–E °(anode) and the balanced cell reaction.

• The direction of electron flow, obtained by inspecting the half–reactions and using the direction that gives a positive E °cell.

Section 17.2


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Description of a Galvanic Cell

• Designation of the anode and cathode.• The nature of each electrode and the ions

present in each compartment. A chemically inert conductor is required if none of the substances participating in the half–reaction is a conducting solid.

Section 17.3


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Maximum Cell Potential

• Directly related to the free energy difference between the reactants and the products in the cell. ΔG° = –nFE °

F = 96,485 C/mol e–

chapter 17 electrochemistry why does leo say ger?

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