Electrochemistry. Half-Reactions 1. Balancing Oxidation Reduction Reactions in Acidic and Basic Solutions

Similar documents
Electrochemistry. Learning Objectives. Half-Reactions 1. Balancing Oxidation Reduction Reactions in Acidic and Basic Solutions

Chemistry 132 NT. Electrochemistry. Review

Electrochemistry 1 1

Electrochemistry. Reduction: the gaining of electrons. Reducing agent (reductant): species that donates electrons to reduce another reagent.

Chem 163 Section: Team Number: ALE 24. Voltaic Cells and Standard Cell Potentials. (Reference: 21.2 and 21.3 Silberberg 5 th edition)

Chapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia:

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

CHEM 2400/2480. Lecture 19

Thermodynamics and Equilibrium

In the half reaction I 2 2 I the iodine is (a) reduced (b) oxidized (c) neither of the above

CHAPTER 21 ELECTROCHEMISTRY: CHEMICAL CHANGE AND ELECTRICAL WORK

Chapter 8 Reduction and oxidation

Downloaded from

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

2004 AP CHEMISTRY FREE-RESPONSE QUESTIONS

2-July-2016 Chemsheets A Page 1

Electrochemical Reactions

CHEM 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions?

Thermodynamics Partial Outline of Topics

Strategy Write the two half-cell reactions and identify the oxidation and reduction reactions. Pt H2 (g) H + (aq)

CHAPTER 6 / HARVEY A. CHEMICAL EQUILIBRIUM B. THERMODYNAMICS AND EQUILIBRIUM C. MANUPULATING EQUILIBRIUM CONSTANTS

CHEMISTRY 16 HOUR EXAM IV KEY April 23, 1998 Dr. Finklea. 1. The anti-cancer drug cis-platin is the complex: cis-[pt(nh ) (Cl) ]. In this complex, the

CHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics

Chapter 17 Free Energy and Thermodynamics

Lecture 13: Electrochemical Equilibria

Chem 112, Fall 05 (Weis/Garman) Exam 4A, December 14, 2005 (Print Clearly) +2 points

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY

Making and Experimenting with Voltaic Cells. I. Basic Concepts and Definitions (some ideas discussed in class are omitted here)

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes.

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics

BIT Chapters = =

A Chemical Reaction occurs when the of a substance changes.

N 2 (g) + 3H 2 (g) 2NH 3 (g) o Three mole ratios can be derived from the balanced equation above: Example: Li(s) + O 2 (g) Li 2 O(s)

Ch 18 Electrochemistry 電化學

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C?

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review)

4 Fe + 3 O 2 2 Fe 2 O 3

CHM 152 Practice Final

NUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command

Unit 3. Electrochemistry

Edexcel GCSE Physics

Semester 2 AP Chemistry Unit 12

In the spaces provided, explain the meanings of the following terms. You may use an equation or diagram where appropriate.

Chapter 19 ElectroChemistry

What factors influence how far a reaction goes and how fast it gets there?

In other words, atoms are not created nor destroyed in chemical reaction.

Unit 9: The Mole- Guided Notes What is a Mole?

1.0 Fundamentals. Fig Schematic diagram of an electrochemical cell.

19 Applications of Standard Electrode Potentials

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change?

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25

Electrochemistry Pulling the Plug on the Power Grid

Chem 111 Summer 2013 Key III Whelan

Thermochemistry. Thermochemistry

Supporting information

Lecture 12: Chemical reaction equilibria

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Acids and Bases Lesson 3

Unit 11 Solutions- Guided Notes. What are alloys? What is the difference between heterogeneous and homogeneous mixtures?

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

How can standard heats of formation be used to calculate the heat of a reaction?

Chapter 5: Diffusion (2)

SCIENCE 10: CHEMISTRY,

Lecture 16 Thermodynamics II

Chapter 9 Chemical Reactions NOTES

Name: Period: Date: ATOMIC STRUCTURE NOTES ADVANCED CHEMISTRY

Materials Engineering 272-C Fall 2001, Lecture 7 & 8 Fundamentals of Diffusion

Chemistry: Electrochemistry-1

**DO NOT ONLY RELY ON THIS STUDY GUIDE!!!**

Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra

Oxidation-Reduction Review. Electrochemistry. Oxidation-Reduction Reactions. Oxidation-Reduction Reactions. Sample Problem.

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY

5 th grade Common Core Standards

Lecture 17: Free Energy of Multi-phase Solutions at Equilibrium

A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1

Unit 14 Thermochemistry Notes

15.0 g Cr = 21.9 g Cr O g Cr 4 mol Cr mol Cr O

NUMBERS, MATHEMATICS AND EQUATIONS

Chapter 17. Electrochemistry

SCH4U: End of Year Review

General Chemistry II, Unit II: Study Guide (part 1)

W V. (d) W. (3) Which one is used to determine the internal resistance of a cell

Session #22: Homework Solutions

Matter Content from State Frameworks and Other State Documents

How can standard heats of formation be used to calculate the heat of a reaction?

Physics 2B Chapter 23 Notes - Faraday s Law & Inductors Spring 2018

ChE 471: LECTURE 4 Fall 2003

Entropy, Free Energy, and Equilibrium

Energies of Phase Changes (Some review from Chapter 6, Some New)

Chapter 18. Electrochemistry

General Chemistry II, Unit I: Study Guide (part I)

GOAL... ability to predict

Chapter 4 Thermodynamics and Equilibrium

How to Assign Oxidation Numbers. Chapter 18. Principles of Reactivity: Electron Transfer Reactions. What is oxidation? What is reduction?

Chemistry 102 Chapter 19 OXIDATION-REDUCTION REACTIONS

Transcription:

Electrchemistry Half-Reactins 1. Balancing Oxidatin Reductin Reactins in Acidic and Basic Slutins Vltaic Cells 2. Cnstructin f Vltaic Cells 3. Ntatin fr Vltaic Cells 4. Cell Ptential 5. Standard Cell Ptentials and Standard Electrde Ptentials 6. Equilibrium Cnstants frm Cell Ptentials 7. Dependence f Cell Ptentials n Cncentratin 8. Sme Cmmercial Vltaic Cells Electrlytic Cells 9. Electrlysis f Mlten Salts 10. Aqueus Electrlysis 11. Stichimetry f Electrlysis 1

Learning Objectives Electrchemistry Balancing Oxidatin Reductin Reactins in Acidic and Basic Slutins a. Learn the steps fr balancing xidatin reductin reactins using the half-reactin methd. b. Balance equatins by the half-reactin methd (acidic slutins). c. Learn the additinal steps fr balancing xidatin reductin reactins in basic slutin using the half-reactin methd. d. Balance equatins using the half-reactin methd (basic slutin). 2. Cnstructin f Vltaic Cells Define electrchemical cell, vltaic (galvanic cell), electrlytic cell, and half-cell. Describe the functin f the salt bridge in a vltaic cell. State the reactin that ccurs at the ande and the cathde in an electrchemical cell. Define cell reactin. Sketch and label a vltaic cell. 3. Ntatin fr Vltaic Cells Write the cell reactin frm the cell ntatin. 4. Cell Ptential Define cell ptential and vlt. Calculate the quantity f wrk frm a given amunt f cell reactant. 2

5. Standard Cell Ptentials and Standard Electrde Ptentials Explain hw the electrde ptential f a cell is an intensive prperty. Define standard cell ptential and standard electrde ptential. Interpret the table f standard reductin ptentials. Determine the relative strengths f xidizing and reducing agents. Determine the directin f spntaneity frm electrde ptentials. Calculate cell ptentials frm standard ptentials. 6. Equilibrium Cnstants frm Cell Ptentials Calculate the free-energy change frm electrde ptentials. Calculate the cell ptential frm free-energy change. Calculate the equilibrium cnstant frm cell ptential. 7. Dependence f Cell Ptential n Cncentratin Calculate the cell ptential fr nnstandard cnditins. Describe hw ph can be determined using a glass electrde. 8. Sme Cmmercial Vltaic Cells Describe the cnstructin and reactins f a zinc carbn dry cell, a lithium idine battery, a lead strage cell, and a nickel cadmium cell. Explain the peratin f a prtn-exchange membrane fuel cell. Explain the electrchemical prcess f the rusting f irn. Define cathdic prtectin. 9. Electrlysis f Mlten Salts Define electrlysis. 3

10. Aqueus Electrlysis Learn the half-reactins fr water underging xidatin and reductin. Predict the half-reactins in an aqueus electrlysis. 11. Stichimetry f Electrlysis Calculate the amunt f charge frm the amunt f prduct in an electrlysis. Calculate the amunt f prduct frm the amunt f charge in an electrlysis. Our first step in studying electrchemical cell is t balance its xidatin reductin reactin. We will use the half-reactin methd frm Sectin 4.6 and extend it t acidic r basic slutins. In this chapter we will fcus n electrn transfer rather than prtn transfer s the hydrnium in, H 3 O (aq), will be represented by its simpler ntatin, H (aq). Only the ntatin, nt the chemistry, is different. Oxidatin-Reductin Reactins In Chapter 4 we intrduced the half-reactin methd fr balancing simple xidatin-reductin reactins. Oxidatin-reductin reactins always invlve a transfer f electrns frm ne species t anther. Recall that the species lsing electrns is xidized, while the species gaining electrns is reduced. 4

Oxidatin-Reductin Reactins Describing Oxidatin-Reductin Reactins An xidizing agent is a species that xidizes anther species; it is itself reduced. A reducing agent is a species that reduces anther species; it is itself xidized. Lss f 2 e -1 xidatin reducing agent 2 2 Fe ( s) Cu ( aq) Fe ( aq) Cu( s) xidizing agent Gain f 2 e -1 reductin What reactin is ging n and what is the equatin? A vltaic cell emplys a spntaneus xidatin reductin reactin as a surce f energy. It separates the reactin int tw halfreactins, physically separating ne half-reactin frm the ther. We will be studying mre cmplex situatins, s ur initial analysis is key. First, we need t identify what is being xidized and what is being reduced. Then, we determine if the reactin is in acidic r basic cnditins. 5

Balancing Oxidatin-Reductin Equatins in Acid Slutin 1. Assign xidatin numbers 2. Determine half reactins 3. Cmplete and balance each half-reactin a. Balance all atms except O and H b. Balance O atms by adding H 2 O s t ne side c. Balance H atms by adding H t ne side d. Balance electric charges by adding e - t the mre psitive side 4. Cmbine each half reactin t btain the final balanced xidatin reductin equatin (Multiply by apprpriate factrs and cancel species appearing n bth sides f equatin Balancing Oxidatin-Reductin Equatins in Basic Slutin 1. Assign xidatin numbers 2. Determine half reactins 3. Cmplete and balance each half-reactin a. Balance all atms except O and H b. Balance O atms by adding H 2 O s t ne side c. Balance H atms by adding H t ne side d. Balance electric charges by adding e - t the mre psitive side 4. Nte the number f H ins in the equatin. Add this number f OH - ins t bth sides f the equatin. 5. Simplify the equatin by nting that H and OH- react t frm H2O. Cancel waters that ccur n bth sides f the equatin and reduce the equatin t simplest terms. 6. Cmbine each half reactin t btain the final balanced xidatin reductin equatin (Multiply by apprpriate factrs and cancel species appearing n bth sides f equatin Balance in acid slutin: Mn 2 BiO 3 - MnO 4 - Bi 3 Balance in Basic slutin: S 2- I 2 SO 4 2- I - 6

Balance in acid slutin: Mn 2 BiO - 3 MnO - 4 Bi 3 2(Mn 2 4 H 2 O MnO - 4 8 H 5e - ) 5(6 H BiO 3-2e- Bi 3 3 H 2 O) 2Mn 2 8 H 2 O 2MnO - 4 16 H 10e - ) 30 H 5 BiO 3-10e- 5Bi 3 15 H 2 O) 2Mn 2 14 H 5 BiO 3-2 MnO 4-5 Bi 3 7 H 2 O Balance in Basic slutin: S 2- I 2 SO 4 2- I - 8 OH - 4H 2 O S 2- SO 4 2-8 H 8 OH - 8e- I 2 2 e - 2I - 8 OH - 4H 2 O S 2- SO 2-4 8 H 8 OH - 8e- 4 (I 2 2 e - 2I - ) 8 OH - 4H 2 O S 2- SO 2-4 8 H 8 OH - 8e- 4 I 2 8 e - 8I - 8 OH - S 2-4 I 2 SO 4 2-8 I - 4 H 2 O Dichrmate in in acidic slutin is an xidizing agent. When it reacts with zinc, the metal is xidized t Zn 2, and nitrate is reduced. Assume that dichrmate in is reduced t Cr 3. Write the balanced inic equatin fr this reactin using the half-reactin methd. 7

First we determine the xidatin numbers f N and Zn: 6 3 0 2 Cr 2 O 7 2- (aq) Cr 3 (aq) and Zn(s) Zn 2 (aq) Cr was reduced frm 6 t 3. Zn was xidized frm 0 t 2. Nw we balance the half-reactins. Oxidatin half-reactin Zn(s) Zn 2 (aq) 2e - Reductin half-reactin First we balance Cr and O: Cr 2 O 7 2- (aq) 2Cr 3 (aq) 7H 2 O(l) Next we balance H: 14H (aq) Cr 2 O 7 2- (aq) 2Cr 3 (aq) 7H 2 O(l) Then we balance e - : 6e - 14H (aq) Cr 2 O 7 2- (aq) 2Cr 3 (aq) 7H 2 O(l) Nw we cmbine the tw half-reactins by multiplying the xidatin half reactin by 3. 3Zn(s) 3Zn 2 (aq) 6e - 6e - 14H (aq) Cr 2 O 7 2- (aq) 2Cr 3 (aq) 7H 2 O(l) 3Zn(s) 14H (aq) Cr 2 O 7 2- (aq) 3Zn 2 (aq) 2Cr 3 (aq) 7H 2 O(l) Check that atms and charge are balanced. 3Zn; 14H; 2Cr; 7O; charge 12 8

T balance a reactin in basic cnditins, first fllw the same prcedure as fr acidic slutin. Then 1. Add ne OH - t bth sides fr each H. 2. When H and OH - ccur n the same side, cmbine them t frm H 2 O. 3. Cancel water mlecules that ccur n bth sides. Balance in Basic slutin: S 2- I 2 SO 4 2- I - Did This Earlier Anther example Lead(II) in, Pb 2, yields the plumbite in, Pb(OH) 3-, in basic slutin. In turn, this in is xidized in basic hypchlrite slutin, ClO -, t lead(iv) xide, PbO 2. Balance the equatin fr this reactin using the halfreactin methd. The skeletn equatin is Pb(OH) 3- ClO - PbO 2 Cl - First we determine the xidatin number f Pb and Cl in each species. 2 1 4-1 Pb(OH) 3- ClO - PbO 2 Cl - Pb is xidized frm 2 t 4. Cl is reduced frm 1 t -1. Nw we balance the half-reactins. 9

Oxidatin half-reactin First we balance Pb and O: Pb(OH) 3- (aq) PbO 2 (s) H 2 O(l) Next we balance H: Pb(OH) 3- (aq) PbO 2 (s) H 2 O(l) H (aq) Then we balance e - : Pb(OH) 3- (aq) PbO 2 (s) H 2 O(l) H (aq) 2e - Reductin half-reactin First we balance Cl and O: ClO - (aq) Cl - (aq) H 2 O(l) Next we balance H: 2H (aq) ClO - (aq) Cl - (aq) H 2 O(l) Finally we balance e - : 2e - 2H (aq) ClO - (aq) Cl - (aq) H 2 O(l) Nw we cmbine the half-reactins. Pb(OH) 3- (aq) PbO 2 (s) H 2 O(l) H (aq) 2e - 2e - 2H (aq) ClO - (aq) Cl - (aq) H 2 O(l) H (aq) Pb(OH) 3- (aq) ClO - (aq) PbO 2 (s) Cl - (aq) 2H 2 O(l) 10

H (aq) Pb(OH) 3- (aq) ClO - (aq) PbO 2 (s) Cl - (aq) 2H 2 O(l) T cnvert t basic slutin, we add OH - t each side, cnverting H t H 2 O. H 2 O(l) Pb(OH) 3- (aq) ClO - (aq) PbO 2 (s) Cl - (aq) 2H 2 O(l) OH - (aq) Finally, we cancel the H 2 O that is n bth sides. Pb(OH) 3- (aq) ClO - (aq) PbO 2 (s) Cl - (aq) H 2 O(l) OH - (aq) D exercise 19.1 & 2 and see prblems 19.35-37 The next several tpics describe battery cells r vltaic cells (galvanic cells). An electrchemical cell is a system cnsisting f electrdes that dip int an electrlyte and in which a chemical reactin either uses r generates an electric current. A vltaic r galvanic cell is an electrchemical cell in which a spntaneus reactin generates an electric current. An electrlytic cell is an electrchemical cell in which an electric current drives an therwise nnspntaneus reactin. Oxidatin-Reductin Reactins In this chapter we will shw hw a cell is cnstructed t physically separate an xidatin-reductin reactin int tw halfreactins. The frce with which electrns travel frm the xidatin half-reactin t the reductin half-reactin is measured as vltage. 11

Electrchemistry An electrchemical cell is a system cnsisting f electrdes that dip int an electrlyte in which a chemical reactin either uses r generates an electric current. A vltaic, r galvanic, cell is an electrchemical cell in which a spntaneus reactin generates an electric current. An electrlytic cell is an electrchemical cell in which an electric current drives an therwise nnspntaneus reactin. In this chapter we will discuss the basic principles behind these cells and explre sme f their cmmercial uses. Vltaic Cells A vltaic cell cnsists f tw half-cells that are electrically cnnected. Each half-cell is a prtin f the electrchemical cell in which a half-reactin takes place. A simple half-cell can be made frm a metal strip dipped int a slutin f its metal in. Fr example, the zinc-zinc in half cell cnsists cnsists f a zinc strip dipped int a slutin f a zinc salt. Vltaic Cells A vltaic cell cnsists f tw half-cells that are electrically cnnected. Anther simple half-cell cnsists f a cpper strip dipped int a slutin f a cpper salt. In a vltaic cell, tw half-cells are cnnected in such a way that electrns flw frm ne metal electrde t the ther thrugh an external circuit. Figure 19.2 illustrates an atmic view f a zinc/cpper vltaic cell. 12

Figure 19.2: Atmic view f vltaic cell. The zinc metal atm lses tw electrns, frming Zn 2 ins. The Cu 2 ins gain tw electrns, frming slid cpper. The electrns flw thrugh the external circuit frm the zinc electrde t the cpper electrde. x Ins flw thrugh the salt bridge t maintain charge balance. Vltaic Cells As lng as there is an external circuit, electrns can flw thrugh it frm ne electrde t the ther. Because zinc has a greater tendency t lse electrns than cpper, zinc atms in the zinc electrde lse electrns t frm zinc ins. The electrns flw thrugh the external circuit t the cpper electrde where cpper ins gain the electrns t becme cpper metal. 13

Vltaic Cells The tw half-cells must als be cnnected internally t allw ins t flw between them. Withut this internal cnnectin, t much psitive charge builds up in the zinc half-cell (and t much negative charge in the cpper half-cell) causing the reactin t stp. Figure 19.3A and 19.3B shw the tw half-cells f a vltaic cell cnnected by salt bridge. Figure 19.3: Tw electrdes are cnnected by an external circuit. Vltaic Cells A salt bridge is a tube f an electrlyte in a gel that is cnnected t the tw half-cells f a vltaic cell. The salt bridge allws the flw f ins but prevents the mixing f the different slutins that wuld allw direct reactin f the cell reactants. Figure 19.3C shws an actual setup f the zinc-cpper cell. 14

Vltaic Cells The tw half-cell reactins, as nted earlier, are: Zn(s) Zn 2 (aq) Cu 2 (aq) 2e 2e The first reactin, in which electrns are lst, is the xidatin half-reactin. The electrde at which xidatin ccurs is the ande. Cu(s) (xidatin half-reactin) (reductin half-reactin) The secnd reactin, in which electrns are gained, is the reductin half-reactin. The electrde at which reductin ccurs is the cathde. Vltaic Cells Nte that the sum f the tw half-reactins Zn(s) Cu (aq) Zn 2 2 (aq) Cu(s) is the net reactin that ccurs in the vltaic cell; it is called the cell reactin Nte that electrns are given up at the ande and thus flw frm it t the cathde where reductin ccurs. The ande in a vltaic cell has a negative sign because electrns flw frm it. (See Figure 20.4 and Animatin: Ande Reactin) The cathde in a vltaic cell has a psitive sign (See Animatin: Cathde Reactin) 15

Figure 19.4: Vltaic Cell D exercise 19.3 And lk at Prblems 19.43-44 Ntatin fr Vltaic Cells It is cnvenient t have a shrthand way f designating particular vltaic cells. The cell cnsisting f the zinc-zinc in half-cell and the cpper-cpper in half-cell, is written Zn(s) Zn ande 2 2 (aq) Cu (aq) Cu(s) cathde The ande (xidatin half-cell) is written n the left. The cathde (reductin half-cell) is written n the right. The cell cnsisting f the zinc-zinc in half-cell and the cpper-cpper in half-cell, is written Zn(s) Zn ande 2 2 (aq) Cu salt bridge (aq) Cu(s) cathde The cell terminals are at the extreme ends in the cell ntatin. Zn(s) Zn ande 2 2 (aq) Cu salt bridge (aq) Cu(s) cathde A single vertical bar indicates a phase bundary, such as between a slid terminal and the electrde slutin. 16

Ntatin fr Vltaic Cells When the half-reactin invlves a gas, an inert material such as platinum serves as a terminal and an electrde surface n which the reactin ccurs. Figure 19.5 shws a hydrgen electrde; hydrgen bubbles ver a platinum plate immersed in an acidic slutin. The cathde half-reactin is 2H (aq) 2e H2(g) The ntatin fr the hydrgen electrde, written as a cathde, is H 2 (aq) H (g) Pt T write such an electrde as an ande, yu simply reverse the ntatin. Pt H2(g) H (aq) In the cell ntatin, these are written in parentheses. Fr example, 2 Zn(s) Zn (1.0 M) H (aq) H2(1.0 atm) Pt Figure 19.5: Hydrgen Electrde 17

A Prblem T Cnsider Give the verall cell reactin fr the vltaic cell 2 Cd(s) Cd (1.0 M) H (aq) H2(1.0 atm) Pt The half-cell reactins are 2 Cd(s) Cd (aq) 2e 2H (aq) 2e H2(g) 2 Cd(s) 2H (aq) Cd (aq) H2(g) D Exercise 19.4 and see prblems 19.49-50 Electrmtive Frce (Cell Ptential) The mvement f electrns is analgus t the pumping f water frm ne pint t anther. Water mves frm a pint f high pressure t a pint f lwer pressure. Thus, a pressure difference is required. The wrk expended in mving the water thrugh a pipe depends n the vlume f water and the pressure difference. An electric charge mves frm a pint f high electrical ptential (high electrical pressure) t ne f lwer electrical ptential. The wrk expended in mving the electrical charge thrugh a cnductr depends n the amunt f charge and the ptential difference. Electrmtive Frce Ptential difference is the difference in electric ptential (electrical pressure) between tw pints. Yu measure this quantity in vlts. The vlt, V, is the SI unit f ptential difference equivalent t 1 jule f energy per culmb f charge. Electrical wrk = charge x ptential difference Jules = culmbs x vlts 1 vlt = 1 J C 18

Electrmtive Frce The Faraday cnstant, F, is the magnitude f charge n ne mle f electrns; it equals 96,500 culmbs (9.65 x 10 4 C). In mving 1 ml f electrns thrugh a circuit, the numerical value f the wrk dne by a vltaic cell is the prduct f the Faraday cnstant (F) times the ptential difference between the electrdes. wrk(j) = F(culmbs) vlts(j/culmb) wrk dne by the system Electrmtive Frce In the nrmal peratin f a vltaic cell, the ptential difference (vltage) acrss the electrdes is less than than the maximum pssible vltage f the cell. The actual flw f electrns reduces the electrical pressure. Thus, a cell vltage has its maximum value when n current flws. Electrmtive Frce The maximum ptential difference between the electrdes f a vltaic cell is referred t as the electrmtive frce (emf) f the cell, r E cell. It can be measured by an electrnic digital vltmeter (Figure 19.6), which draws negligible current. 19

Electrmtive Frce We can nw write an expressin fr the maximum wrk attainable by a vltaic cell. Let n be the number f (ml) electrns transferred in the verall cell reactin. The maximum wrk fr mlar amunts f reactants is w = max nfe cell Wrk Exercise 19.6 Lk at Prblems 19.57-60 A Prblem T Cnsider The emf f the electrchemical cell belw is 0.650 V. Calculate the maximum electrical wrk f this cell when 0.500 g H 2 is cnsumed. 2 Hg2 (aq) H2(g) The half-reactins are 2 2Hg(l) 2H Hg 2 (aq) 2e 2Hg(l) H 2 (g) 2H (aq) 2e n = 2, and the maximum wrk fr the reactin is written as w w = max nfe cell 4 = 2 max (9.65 10 C) (aq) (0.650 V) 20

4 w = 2 max (9.65 10 C) (0.650 V) w = max nfe cell w = max 1.25 10 The emf f the electrchemical cell belw is 0.650 V. Calculate the maximum electrical wrk f this cell when 0.500 g H 2 is cnsumed. 2 Hg (aq) 2 H2(g) Fr 0.500 g H 2, the maximum wrk is 5 1 ml H 1.25 10 J 0.500 g H 2 = 3.09 2 10 2.02 g H 1 ml H 2 5 J 2Hg(l) 2H 2 (aq) 4 J Standard Cell emf s and Standard Electrde Ptentials A cell emf is a measure f the driving frce f the cell reactin. The reactin at the ande has a definite xidatin ptential, while the reactin at the cathde has a definite reductin ptential. Thus, the verall cell emf is a cmbinatin f these tw ptentials. E cell = xidatin ptential reductin ptential A reductin ptential is a measure f the tendency t gain electrns in the reductin half-reactin. Yu can lk at the xidatin half-reactin as the reverse f a crrespnding reductin reactin. The xidatin ptential fr an xidatin half-reactin is the negative f the reductin ptential fr the reverse reactin. 21

Standard Cell emf s and Standard Electrde Ptentials By cnventin, the Table f Standard Electrde Ptentials (Table 19.1) are tabulated as reductin ptentials. Cnsider the zinc-cpper cell described earlier. Zn(s) Zn 2 2 (aq) Cu The tw half-reactins are 2 Zn(s) Zn (aq) 2e Cu 2 (aq) 2e Cu(s) (aq) Cu(s) The zinc half-reactin is an xidatin. If yu write E Zn fr the reductin ptential f zinc, then E Zn is the xidatin ptential f zinc. Zn 2 (aq) 2e Zn(s) Zn 2 Zn(s) (E Zn ) Reductin (aq) 2e -(E Zn ) Oxidat. The cpper half-reactin is a reductin.. Write E Cu fr the electrde ptential. Cu 2 (aq) 2e Cu(s) (E Cu ) Fr this cell, the cell emf is the sum f the reductin ptential fr the cpper half-cell and the xidatin ptential fr the zinc half-cell. E E cell cell = E = E Cu Cu ( E E Nte that the cell emf is the difference between the tw electrde ptentials. In general, E cell is btained by subtracting the ande ptential frm the cathde ptential. Zn Zn E = E E cell cathde ande The electrde ptential is an intensive prperty whse value is independent f the amunt f species in the reactin. ) 22

Thus, the electrde ptential fr the half-reactin 2Cu 2 (aq) 4e 2Cu(s) is the same as fr Cu 2 (aq) 2e Cu(s) Tabulating Standard Electrde Ptentials The standard emf, E cell, is the emf f a cell perating under standard cnditins f cncentratin (1 M), pressure (1 atm), and temperature (25 C). Nte that individual electrde ptentials require that we chse a reference electrde. Yu arbitrarily assign this reference electrde a ptential f zer and btain the ptentials f the ther electrdes by measuring the emf s. By cnventin, the reference chsen fr cmparing electrde ptentials is the standard hydrgen electrde. (see Figure 19.5) Standard electrde ptentials (Table 19.1) are measured relative t this hydrgen reference. 23

Tabulating Standard Electrde Ptentials The standard electrde ptential, E, is the electrde ptential when cncentratins f slutes are 1 M, gas pressures are 1 atm, and the temperature is 25 C. (Table 19.1) Fr example, when yu measure the emf f a cell cmpsed f a zinc electrde cnnected t a hydrgen electrde, yu btain 0.76 V. Since zinc acts as the ande (xidatin) in this cell, its reductin ptential is listed as 0.76 V. 24

Strengths f Oxidizing and Reducing Agents Standard electrde ptentials are useful in determining the strengths f xidizing and reducing agents under standard-state cnditins. A reductin half-reactin has the general frm xidized species ne reduced species The xidized species acts as an xidizing agent. Cnsequently, the strngest xidizing agents in a table f standard electrde ptentials are the xidized species crrespnding t the half-reactins with the largest (mst psitive) E values. (Fr example F 2 (g)) An xidatin half-reactin has the general frm reduced species xidized species ne The reduced species acts as a reducing agent. Cnsequently, the strngest reducing agents in a table f standard electrde ptentials are the reduced species crrespnding t the half-reactins with the smallest (mst negative) E values. (fr example, Li(s)) 25

Which is the strnger reducing agent under standard cnditins: Sn 2 (t Sn 4 ) r Fe (t Fe 2 )? Which is the strnger xidizing agent under standard cnditins: Cl 2 r MnO 4-? The strnger reducing agent will be xidized and has the mre negative electrde ptential. The standard (reductin) ptentials are Sn 2 t Sn 4 E = 0.15 V Fe t Fe 2 E = 0.41 V The strnger reducing agent is Fe (t Fe 2 ). The strnger xidizing agent will be reduced. The standard (reductin) ptentials are Cl 2 t Cl - E = 1.36 V MnO 4- t Mn 2 E = 1.49 V The strnger xidizing agent is MnO 4-. Calculating Cell emf s frm Standard Ptentials The emf f a vltaic cell cnstructed frm standard electrdes is easily calculated using a table f electrde ptentials. Cnsider a cell cnstructed f the fllwing tw halfreactins. 2 = Cd (aq) 2e Ag (aq) 1e Cd Cd(s); E Ag(s); E = 0.40 V 0.80 V Yu will need t reverse ne f these reactins t btain the xidatin part f the cell reactin. 2 = (aq) 2e Ag (aq) 1e Cd(s); E Ag(s); E = 0.40 V 0.80 V 26

Cd This will be Cd, because has the mre negative electrde ptential. 2 = (aq) 2e Ag (aq) 1e Cd(s); E Ag(s); E = 0.40 V 0.80 V Therefre, yu reverse the half-reactin and change the sign f the half-cell ptential. Cd(s) Cd Ag (aq) 1e (aq) 2e ; E 2 = Ag(s); E = 0.40 V 0.80 V We must duble the silver half-reactin s that when the reactins are added, the electrns cancel. Cd(s) Cd Ag (aq) 1e 2 = (aq) 2e ; E Ag(s); E = 0.40 V 0.80 V This des nt affect the half-cell ptentials, which d nt depend n the amunt f substance. 2 = Cd(s) Cd Ag (aq) 2e (aq) 2e ; E Ag(s); E = 0.40 V 0.80 V Nw we can add the tw half-reactins t btain the verall cell reactin and cell emf. Cd(s) Cd 2Ag (aq) 2e (aq) 2e ; E 2 = 2Ag(s); E = 0.40 V 0.80 V Nw we can add the tw half-reactins t btain the verall cell reactin and cell emf. Cd(s) Cd 2Ag (aq) 2e 2Ag(s); Cd(s) 2Ag (aq) Cd 2 = (aq) 2e ; E E = 2 (aq) 2Ag(s); E = cell The crrespnding cell ntatin wuld be 0.40 V 0.80 V 1.20 V Cd(s) Cd 2 (1M) Ag (1M) Ag(s) D Exercise 19.7 See Prblems 19.61-64 27

Nte that the emf f the cell equals the standard electrde ptential f the cathde minus the standard electrde ptential f the ande. E cell = E cathde E ande Calculate the standard emf fr the fllwing vltaic cell at 25 C using standard electrde ptentials. What is the verall reactin? Al(s) Al 3 2 (aq) Fe (aq) Fe(s) The reductin half-reactins and standard ptentials are Al Fe 3 = 2 (aq) 3e (aq) 2e Al(s) Al Fe Al(s) Al 2 (aq) 2e Al(s); E 1.66 V Fe(s); E = 0.41V 3 2 (aq) Fe 3 = (aq) 3e ; E (aq) Fe(s) Yu reverse the first half-reactin and its half-cell ptential t btain 1.66 V Fe(s); E = 0.41 V T btain the verall reactin we must balance the electrns. 2Al(s) 2Al 3Fe Al(s) Al 2 (aq) 6e 3 2 (aq) Fe (aq) 6e ; E 3 = (aq) Fe(s) 1.66 V 3Fe(s); E = 0.41V Nw we add the reactins t get the verall cell reactin and cell emf. 2Al(s) 2Al 3Fe 2 2Al(s) 3Fe (aq) 6e 3 = (aq) 2Al (aq) 6e ; E 1.66 V 3Fe(s); E = 0.41V (aq) 3Fe(s); E 2 3 = 1.25 V 28

Hw d yu determine the directin f spntaneity? G Lk at example 19.7 and example 19.8 See prblems 19.65 and 19.66 Lk at Cncept check 19.2 Page 790 Equilibrium Cnstants frm emf s Sme f the mst imprtant results frm electrchemistry are the relatinships amng E cell, free energy, and equilibrium cnstant. In Chapter 18 we saw that DG equals the maximum useful wrk f a reactin. Fr a vltaic cell, wrk = -nfe, s when reactants are in their standard states G = nfe The measurement f cell emf s gives yu yet anther way f calculating equilibrium cnstants. 29

Cmbining the previus equatin, G = -nfe cell, with the equatin G = -RTlnK, we get nfe cell = RTlnK Or, rearranging, we get 2.303RT E = cell lg K nf Substituting values fr the cnstants R and F at 25 C gives the equatin 0.0592 E = cell lg K n (values in vlts at 25 C) D Exercise 19.9 and see prblems 19.69 and 19.70 Lk at Examples 19.9 and 19.10 D Exercises 19.10 and 19.11 Lk at prblems 19.73 & 74, 81 & 82 Figure 19.7 summarizes the varius relatinships amng K, G, and E cell. 30

A Prblem T Cnsider The standard emf fr the fllwing cell is 1.10 V. Zn(s) Zn 2 2 (aq) Cu (aq) Cu(s) Calculate the equilibrium cnstant K c fr the reactin Zn(s) Cu 2 2 (aq) Zn (aq) Cu(s) Nte that n=2. Substituting int the equatin relating E cell and K gives 2.303 RT E = lg cell nf 0.0592 1.10 V = lgk 2 K Slving fr lg K c, yu find lg K = 37.2 Nw take the antilg f bth sides: K = c anti lg(37.2) = 1.6 10 The number f significant figures in the answer equals the number f decimal places in 37.2 (ne). Thus K = c 2 10 37 37 Dependence f emf n Cncentratin Recall that the free energy change, G, is related t the standard free energy change, G, by the fllwing equatin. G = G RTlnQ Here Q is the thermdynamic reactin qutient. Cmbining the previus equatin, G = -nfe cell, with the equatin G = -RTlnK, we get nfe = cell nfe cell RTlnQ 31

Dependence f emf n Cncentratin The result rearranges t give the Nernst equatin, an equatin relating the cell emf t its standard emf and the reactin qutient. E cell = E cell 2.303RT lgq nf Substituting values fr R and F at 25 C, we get E cell = E cell 0.0592 lgq n (values in vlts at 25 C) Dependence f emf n Cncentratin The result rearranges t give the Nernst equatin, an equatin relating the cell emf t its standard emf and the reactin qutient. The Nernst equatin illustrates why cell emf decreases as the cell reactin prceeds. As reactant cncentratins decrease and prduct cncentratins increase, Q increases, thus increasing lg Q which in turn decreases the cell emf. A Prblem T Cnsider What is the emf f the fllwing vltaic cell at 25 C? Zn(s) Zn (1 10 2 5 2 M) Cu The standard emf f the cell is 1.10 V. The cell reactin is Zn(s) Cu 2 2 (aq) Zn (0.100M) Cu(s) (aq) Cu(s) The number f electrns transferred is 2; hence n = 2. The reactin qutient is [Zn Q = [Cu 2 2 ] 1.00 10 = ] 0.100 5 = 1.00 10 4 32

E The standard emf is 1.10 V, s the Nernst equatin becmes cell E cell = E cell 0.0592 lgq n 0.0592 = 1.10 V lg(1.00 10 2 E cell = 1.10 V ( 0.12) = 1.22 V 4 ) The cell emf is 1.22 V. D Exercise 19.13 Lk at Prblems 19.85 &86 Cncept Check 19.3 Page 796 Sme Cmmercial Vltaic Cells The Leclanché dry cell, r zinc-carbn dry cell, is a vltaic cell with a zinc can as the ande and a graphite rd in the center surrunded by a paste f manganese dixide, ammnium and zinc chlrides, and carbn black, as the cathde. The electrde reactins are Zn(s) Zn 2 (aq) 2e ande 2NH (aq) 2MnO (s) 4 2 2e ) Mn O (s) H O(l) 2 3 2 2NH3(aq cathde (see Figure 19.9) 33

Figure 19.9: Leclanché Dry Cell Sme Cmmercial Vltaic Cells The alkaline dry cell, is similar t the Leclanché cell, but it has ptassium hydrxide in place f ammnium chlride. The electrde reactins are Zn(s) Zn 2 MnO (s) H O(l) 2 2e (aq) 2e ande 2 Mn O (s) 2 3 2OH (aq) cathde (see Figure 19.10) Figure 19.10: A Small Alkaline Dry Cell 34

Sme Cmmercial Vltaic Cells The lithium-idine battery is a slid state battery in which the ande is lithium metal and the cathde is an I 2 cmplex. The slid state electrdes are separated by a thin crystalline layer f lithium idide. Althugh it prduces a lw current, it is very reliable and is used t pwer pacemakers. (see Figure 19.11) Figure 19.11: Slid-State Lithium-Idine Battery Sme Cmmercial Vltaic Cells The lead strage cell (a rechargeable cell) cnsists f electrdes f lead ally grids; ne electrde is packed with a spngy lead t frm the ande, and the ther electrde is packed with lead dixide t frm the cathde. The electrde reactins are 4 Pb(s) HSO (aq) PbSO 4(s) H (aq) 2e PbO2(s) 3H (aq) HSO4 (aq) 2e PbSO (s) 2H2O(l) 4 ande cathde (see Figure 20.12) 35

Figure 19.12: Lead Strage Battery Sme Cmmercial Vltaic Cells The nickel-cadmium cell (nicad cell) cnsists f an ande f cadmium and a cathde f hydrated nickel xide n nickel; the electrlyte is ptassium hydrxide. The electrde reactins are Cd(s) 2OH (aq) Cd(OH) 2(s) 2e NiOOH(s) H O(l) e 2 Ni(OH) (s) 2 OH (aq) ande cathde (see Figure 19.14) Figure 19.14: Nicad Strage Batteries 36

Sme Cmmercial Vltaic Cells A fuel cell is essentially a battery, but differs by perating with a cntinuus supply f energetic reactants, r fuel. Fr a hydrgen-xygen fuel cell, the electrde reactins are 2H2(g) 4OH (aq) H2 O(l) 4e (g) 2H O(l) 4e O2 2 4OH (aq) ande cathde (see Figure 19.15) Figure 19.15: Hydrgen-Oxygen Fuel Cell T d: Research hw a fuel cell wrks and what different types exist. Figure 19.16: The Electrchemical Prcess Invlved in the Rusting f Irn 37

Figure 19.17: Cathdic Prtectin f Buried Steel Pipe Figure 19.18: A Demnstratin f Cathdic Prtectin Unprtected nail Fe 2 Ferricyanide 38

Electrlytic Cells An electrlytic cell is an electrchemical cell in which an electric current drives an therwise nnspntaneus reactin. (See Vide: Electrlysis f Water) The prcess f prducing a chemical change in an electrlytic cell is called electrlysis. Many imprtant substances, such as aluminum metal and chlrine gas are prduced cmmercially by electrlysis. Figure 19.19: Electrlysis f Mlten Sdium Chlride Electrlysis f Mlten Salts A Dwns cell is a cmmercial electrchemical cell used t btain sdium metal by electrlysis f mlten NaCl. (see Figure 19.20) A number f ther reactive metals are btained by the electrlysis f a mlten salt. Lithium, magnesium, and calcium metals are all btained by the electrlysis f the chlrides. D Exercise 19.15 See Prblems 19.91-92 39

Figure 19.20: Dwns Cell fr Preparatin f Sdium Metal Figure 19.22: Chlr-Alkali Membrane Cell Figure 19.23: Chlr-Alkali Mercury Cell 40

Figure 19.24: Purificatin f Cpper by Electrlysis Figure 19.24: Purificatin f Cpper by Electrlysis (cnt d) 41

Stichimetry f Electrlysis What is new in this type f stichimetric prblem is the measurement f numbers f electrns. Yu d nt weigh them as yu d substances. Rather, yu measure the quantity f electric charge that has passed thrugh a circuit. T determine this we must knw the current and the length f time it has been flwing. Electric current is measured in amperes. An ampere (A) is the base SI unit f current equivalent t 1 culmb/secnd. The quantity f electric charge passing thrugh a circuit in a given amunt f time is given by Electric charge(cul) = electric current (cul/sec) time lapse(sec) A Prblem T Cnsider When an aqueus slutin f ptassium idide is electrlyzed using platinum electrdes, the halfreactins are I (aq) I (aq) 2e 2 2 2H2O(l) 2e H2(g) 2OH (aq) Hw many grams f idine are prduced when a current f 8.52 ma flws thrugh the cell fr 10.0 min? When the current flws fr 6.00 x 10 2 s (10.0 min), the amunt f charge is (8.52 10 3 = 2 A) (6.00 10 s) 5.11 C 42

Figure 19.21: Electrlysis f Aqueus Ptassium Idide A Prblem T Cnsider When an aqueus slutin f ptassium idide is electrlyzed using platinum electrdes, the halfreactins are 2I (aq) I2 2 2 2 Hw many grams f idine are prduced when a current f 8.52 ma flws thrugh the cell fr 10.0 min? 5.11 C (aq) 2e H O(l) 2e H (g) 2OH (aq) Nte that tw mles f electrns are equivalent t ne mle f I 2. Hence, 1 ml e 9.65 10 1 ml I 2 254 g I 2 3 = 4 6.73 10 g I 2 C 2 ml e 1 ml I 2 D Exercise 19.16 See Prblems 19.93-94 D Exercise 19.17 See Prblems 19.95-96 D Exercise 19.18 See prblems 19.97-98 43

Operatinal Skills Balancing xidatin-reductin reactins Sketching and labeling a vltaic cell Writing the cell reactin frm the cell ntatin Calculating the quantity f wrk frm a given amunt f cell reactant Determining the relative strengths f xidizing and reducing agents Determining the directin f spntaneity frm electrde ptentials Calculating the emf frm standard ptentials Operatinal Skills Calculating the free-energy change frm electrde ptentials Calculating the cell emf frm free-energy change Calculating the equilibrium cnstant frm cell emf Calculating the cell emf fr nnstandard cnditins Predicting the half-reactins in an aqueus electrlysis Relating the amunts f prduct and charge in an electrlysis End f Chapter 19 44