chapter 17 electrochemistry why does leo say ger?
TRANSCRIPT
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
Return to TOC
Copyright © Cengage Learning. All rights reserved 3
• 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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 4
• Reactions in which one or more electrons are transferred.
Redox Reactions
Section 17.0
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 5
Reaction of Sodium and Chlorine
Section 17.0
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 6
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 7
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 8
• 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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 9
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 10
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 11
• 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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 12
• To balance half reactions that occur in acid solutions
Objectives Continued
Section 17.0
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 13
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 14
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 15
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 16
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 17
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.
The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
Section 17.0
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 18
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 19
• 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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 20
• 6e- + Cr2O72-(aq) 2Cr3+(aq) + 7H2O
• H2O +SO32-(aq) + SO4
2-(aq) + 2e-
• How can we balance the hydrogen atoms?
Method of Half Reactions (continued)
Section 17.0
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 21
• This reaction occurs in an acidic solution.
• 14H+ + 6e- + Cr2O72- 2Cr3+ + 7H2O
• H2O +SO32- SO4
2- + 2e- + 2H+
• How can we balance the electrons?
Method of Half Reactions (continued)
Section 17.0
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 22
• 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!!
Method of Half Reactions (continued)
Section 17.0
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 23
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 24
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
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 25
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.
The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
Section 17.0
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 26
• 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
Return to TOC
Copyright © Cengage Learning. All rights reserved 27
• To understand what a galvanic cell is and how it works
• To calculate cell potential for a given reaction
Objectives
cellE
Section 17.1
Atomic MassesGalvanic Cells
Return to TOC
Copyright © Cengage Learning. All rights reserved 28
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
Atomic MassesGalvanic Cells
Return to TOC
Copyright © Cengage Learning. All rights reserved 29
A Galvanic Cell
Section 17.1
Atomic MassesGalvanic Cells
Return to TOC
Copyright © Cengage Learning. All rights reserved 30
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
Atomic MassesGalvanic Cells
Return to TOC
Copyright © Cengage Learning. All rights reserved 31
Voltaic Cell: Anode Reaction (losing e-s LEO)
Section 17.1
Atomic MassesGalvanic Cells
Return to TOC
Copyright © Cengage Learning. All rights reserved 32
Voltaic Cell: Cathode Reaction (gaining e-s GER)
Section 17.1
Atomic MassesGalvanic Cells
Return to TOC
Copyright © Cengage Learning. All rights reserved 33
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
Return to TOC
Copyright © Cengage Learning. All rights reserved 34
Electrochemical Half-Reactions in a Galvanic Cell
Section 17.2
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 35
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
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 36
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
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 37
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
Calculate E °cell = E °(cathode) - E °(anode)
E °cell = -1.66 –(- 2.37) = 0.71 V (+ is spontaneous)
Section 17.2
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 38
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)
Calculate E °cell = E °(cathode) - E °(anode)
E °cell = - = V (spontaneous??)
Section 17.2
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 39
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
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 40
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
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 41
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
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 42
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
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 43
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
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 44
Concept Check
Page 831 # 29 sketch the 2 cells, pickup #27
Quiz #30??
end
Section 17.3
Cell Potential, Electrical Work, and Free Energy
Return to TOC
Copyright © Cengage Learning. All rights reserved 45
• 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
Cell Potential, Electrical Work, and Free Energy
Return to TOC
Copyright © Cengage Learning. All rights reserved 46
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
Cell Potential, Electrical Work, and Free Energy
Return to TOC
Copyright © Cengage Learning. All rights reserved 47
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
Cell Potential, Electrical Work, and Free Energy
Return to TOC
Copyright © Cengage Learning. All rights reserved 48
• 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
Return to TOC
Copyright © Cengage Learning. All rights reserved 49
• 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
Dependence of Cell Potential on Concentration
Return to TOC
Copyright © Cengage Learning. All rights reserved 50
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
Dependence of Cell Potential on Concentration
Return to TOC
Copyright © Cengage Learning. All rights reserved 51
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
Dependence of Cell Potential on Concentration
Return to TOC
Copyright © Cengage Learning. All rights reserved 52
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
Dependence of Cell Potential on Concentration
Return to TOC
Copyright © Cengage Learning. All rights reserved 53
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
Dependence of Cell Potential on Concentration
Return to TOC
Copyright © Cengage Learning. All rights reserved 54
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
Dependence of Cell Potential on Concentration
Return to TOC
Copyright © Cengage Learning. All rights reserved 55
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
Dependence of Cell Potential on Concentration
Return to TOC
Copyright © Cengage Learning. All rights reserved 56
• 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
Return to TOC
Copyright © Cengage Learning. All rights reserved 57
One of the Six Cells in a 12–V Lead Storage Battery
Section 17.5
Batteries
Return to TOC
Copyright © Cengage Learning. All rights reserved 58
A Common Dry Cell Battery
Section 17.5
Batteries
Return to TOC
Copyright © Cengage Learning. All rights reserved 59
A Mercury Battery
Section 17.5
Batteries
Return to TOC
Copyright © Cengage Learning. All rights reserved 60
Schematic of the Hydrogen-Oxygen Fuel Cell
Section 17.6
Corrosion
Return to TOC
Copyright © Cengage Learning. All rights reserved 61
• 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
Return to TOC
Copyright © Cengage Learning. All rights reserved 62
The Electrochemical Corrosion of Iron
Section 17.6
Corrosion
Return to TOC
Copyright © Cengage Learning. All rights reserved 63
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
Return to TOC
Copyright © Cengage Learning. All rights reserved 64
Cathodic Protection
Section 17.7
Electrolysis
Return to TOC
Copyright © Cengage Learning. All rights reserved 65
• Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
Section 17.7
Electrolysis
Return to TOC
Copyright © Cengage Learning. All rights reserved 66
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
Return to TOC
Copyright © Cengage Learning. All rights reserved 67
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
Return to TOC
Copyright © Cengage Learning. All rights reserved 68
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
Return to TOC
Copyright © Cengage Learning. All rights reserved 69
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
Return to TOC
Copyright © Cengage Learning. All rights reserved 70
• Production of aluminum• Purification of metals• Metal plating• Electrolysis of sodium chloride• Production of chlorine and sodium hydroxide
Section 17.8
Commercial Electrolytic Processes
Return to TOC
Copyright © Cengage Learning. All rights reserved 71
Producing Aluminum by the Hall-Heroult Process
Section 17.8
Commercial Electrolytic Processes
Return to TOC
Copyright © Cengage Learning. All rights reserved 72
Electroplating a Spoon
Section 17.8
Commercial Electrolytic Processes
Return to TOC
Copyright © Cengage Learning. All rights reserved 73
The Downs Cell for the Electrolysis of Molten Sodium Chloride
Section 17.8
Commercial Electrolytic Processes
Return to TOC
Copyright © Cengage Learning. All rights reserved 74
The Mercury Cell for Production of Chlorine and Sodium Hydroxide
Section 17.0
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 75
The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Acidic Solution
Section 17.0
Balancing Oxidation–Reduction Equations
Return to TOC
Copyright © Cengage Learning. All rights reserved 76
The Half–Reaction Method for Balancing Equations for Oxidation–Reduction Reactions Occurring in Basic Solution
Section 17.2
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 77
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
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 78
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
The Mole Standard Reduction Potentials
Return to TOC
Copyright © Cengage Learning. All rights reserved 79
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
Cell Potential, Electrical Work, and Free Energy
Return to TOC
Copyright © Cengage Learning. All rights reserved 80
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–