CHAPTER SIX19 Electrochemistry Chapter 6 Electrochemistry Chapter Six

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CHAPTER SIX(19) Electrochemistry

CHAPTER SIX(19) Electrochemistry

Chapter 6 / Electrochemistry Chapter Six outline Chapter Six Contains: 6. 1 Redox Reactions

Chapter 6 / Electrochemistry Chapter Six outline Chapter Six Contains: 6. 1 Redox Reactions 6. 2 Galvanic Cells 6. 3 Standard Reduction Potentials 6. 4 Thermodynamics of Redox Reactions 6. 5 The Effect of Concentration on Cell Emf 6. 6 Batteries 6. 7 Corrosion 6. 8 Electrolysis 2

Chapter 6 / Electrochemistry 6. 1 Redox Reactions Electrochemistry is the branch of chemistry

Chapter 6 / Electrochemistry 6. 1 Redox Reactions Electrochemistry is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy. Electrochemical processes are redox (oxidation-reduction) reactions in which: • the energy released by a spontaneous reaction is converted to electricity or • electrical energy is used to cause a nonspontaneous reaction to occur 0 0 2+ 2 - 2 Mg (s) + O 2 (g) 2 Mg. O (s) 2 Mg 2+ + 4 e- Oxidation half-reaction (lose e-) O 2 + 4 e- 2 O 2 - Reduction half-reaction (gain e-) 3

Chapter 6 / Electrochemistry 6. 1 Redox Reactions Oxidation number The charge the atom

Chapter 6 / Electrochemistry 6. 1 Redox Reactions Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1. Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 = 0 2. In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe 3+, Fe = +3; O 2 -, O = -2 3. The oxidation number of oxygen is usually – 2. In H 2 O 2 and O 22 - it is – 1. 4

Chapter 6 / Electrochemistry 6. 1 Redox Reactions 4. The oxidation number of hydrogen

Chapter 6 / Electrochemistry 6. 1 Redox Reactions 4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is – 1. 5. Group IA metals are +1, IIA metals are +2 and fluorine is always – 1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. HCO 3 Oxidation numbers of all the atoms in HCO 3 - ? O = -2 H = +1 3 x(-2) + 1 + ? = -1 C = +4 5

Chapter 6 / Electrochemistry 6. 1 Redox Reactions Balancing Redox Equations The oxidation of

Chapter 6 / Electrochemistry 6. 1 Redox Reactions Balancing Redox Equations The oxidation of Fe 2+ to Fe 3+ by Cr 2 O 72 - in acid solution? 1. Write the unbalanced equation for the reaction ionic form. Fe 2+ + Cr 2 O 72 - Fe 3+ + Cr 3+ 2. Separate the equation into two half-reactions. +2 Oxidation: Reduction: 3. +3 Fe 2+ Fe 3+ +6 Cr 2 O 7 +3 2 - Cr 3+ Balance the atoms other than O and H in each half-reaction. Cr 2 O 72 - 2 Cr 3+ 6

Chapter 6 / Electrochemistry 6. 1 Redox Reactions Balancing Redox Equations 4. For reactions

Chapter 6 / Electrochemistry 6. 1 Redox Reactions Balancing Redox Equations 4. For reactions in acid, add H 2 O to balance O atoms and H+ to balance H atoms. Cr 2 O 72 - 2 Cr 3+ + 7 H 2 O 14 H+ + Cr 2 O 72 - 2 Cr 3+ + 7 H 2 O 5. Add electrons to one side of each half-reaction to balance the charges on the halfreaction. Fe 2+ Fe 3+ + 1 e 6 e- + 14 H+ + Cr 2 O 72 - 2 Cr 3+ + 7 H 2 O 6. If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 6 Fe 2+ 6 Fe 3+ + 6 e 6 e- + 14 H+ + Cr 2 O 72 - 2 Cr 3+ + 7 H 2 O 7

Chapter 6 / Electrochemistry 6. 1 Redox Reactions Balancing Redox Equations 7. Add the

Chapter 6 / Electrochemistry 6. 1 Redox Reactions Balancing Redox Equations 7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. Oxidation: Reduction: 6 Fe 2+ 6 Fe 3+ + 6 e 6 e- + 14 H+ + Cr 2 O 72 - 2 Cr 3+ + 7 H 2 O 14 H+ + Cr 2 O 72 - + 6 Fe 2+ 6 Fe 3+ + 2 Cr 3+ + 7 H 2 O 8. Verify that the number of atoms and the charges are balanced. 14 x 1 – 2 + 6 x 2 = 24 = 6 x 3 + 2 x 3 9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. 8

Chapter 6 / Electrochemistry 6. 8 Electrolysis is the process in which electrical energy

Chapter 6 / Electrochemistry 6. 8 Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. • In electrolytic solution, the charge is carried by ions. • Ions must be free to move. • This is only occur at molten salts or aqueous solutions of electrolyte. 9

Chapter 6 / Electrochemistry 6. 8 Electrolysis of molten Na. Cl Battery pump electrons

Chapter 6 / Electrochemistry 6. 8 Electrolysis of molten Na. Cl Battery pump electrons into the right hand electrode. (-ve) 10

Chapter 6 / Electrochemistry 6. 8 Electrolysis Anode reaction Electrolysis of molten Na. Cl

Chapter 6 / Electrochemistry 6. 8 Electrolysis Anode reaction Electrolysis of molten Na. Cl + 2 Cl- Cl 2 +2 e. Always oxidization occurs at the anode cathode Cathode reaction Na+ + e- Na Always reduction occurs at the cathode 11

Chapter 6 / Electrochemistry 6. 8 Electrolysis of Water Dose water conduct electricity? 12

Chapter 6 / Electrochemistry 6. 8 Electrolysis of Water Dose water conduct electricity? 12

Chapter 6 / Electrochemistry 6. 8 Electrolysis of Water + For oxidizes 4 H

Chapter 6 / Electrochemistry 6. 8 Electrolysis of Water + For oxidizes 4 H 2 O 4 H++ 4 OH- , (1) 4 OH- O 2+2 H 2 O + 4 e- , (2) 2 H 2 O ⇌ 2 OH- + 2 H+ , (1) 2 H+ + 2 e- → H 2 (g) (2) anode The summation of eqs. 1 and 2, we get 2 H 2 O For reduction , we have ; O 2 (g)+4 H++4 e- (4) cathode By summation eqs. 1 and 2 , we get 2 H 2 O+2 e- 2 OH-(aq) + H 2 (g) (3) 13

Chapter 6 / Electrochemistry 6. 8 Electrolysis of aqueous Na. Cl Anode reaction 2

Chapter 6 / Electrochemistry 6. 8 Electrolysis of aqueous Na. Cl Anode reaction 2 Cl- Cl 2 +2 e- X √ Cathode reaction Na+ + e- Na √ 2 H 2 O+2 e- 2 OH-(aq) + H 2 (g) 2 H 2 O O 2 (g)+4 H++4 e- X Na. Cl H 2 O 2 H 2 O + 2 Cl- + 2 Na+ H 2 (g) +Cl 2 (g) +2 Na+ OH- 14

Chapter 6 / Electrochemistry 6. 8 Electrolysis of aqueous Cu. SO 4 Anode reaction

Chapter 6 / Electrochemistry 6. 8 Electrolysis of aqueous Cu. SO 4 Anode reaction 2 SO 4 S 2 O 8 + -- -- 2 e- X 2 H 2 O O 2(g) + 4 H+ + e- √ √ Cathode reaction Cu++ + 2 e- Cu (s) X 2 H 2 O+2 e- 2 OH-(aq) + H 2 (g) 2 Cu++ + 2 H 2 O O 2(g) + 2 Cu(s) + 4 H+ 15

Chapter 6 / Electrochemistry 6. 8 Electrolysis of aqueous Cu. Cl 2 Anode reaction

Chapter 6 / Electrochemistry 6. 8 Electrolysis of aqueous Cu. Cl 2 Anode reaction 2 Cl- Cl 2 +2 e- √ 2 H 2 O O 2 (g)+4 H++4 e- X √ Cathode reaction Cu++ + 2 e- Cu (s) X 2 H 2 O+2 e- 2 OH-(aq) + H 2 (g) Cu++ + 2 Cl- Cl 2 (g) + Cu (s) 16

Chapter 6 / Electrochemistry 6. 8 Electrolysis Anode reaction 2 SO 4 S 2

Chapter 6 / Electrochemistry 6. 8 Electrolysis Anode reaction 2 SO 4 S 2 O 8 + -- -- 2 H 2 O O 2(g) + Electrolysis of aqueous Cu. SO 4 with Cu electrode Cathode reaction Cu electrode 2 e- 4 H+ Cu Cu++ + 2 e- X + e- √ Cu++ + 2 e- Cu (s) X 2 H 2 O+2 e- 2 OH-(aq) + H 2 (g) X √ Cu++ + Cu++ 17

Chapter 6 / Electrochemistry 6. 8 Electrolysis • Electric current is measured in amperes

Chapter 6 / Electrochemistry 6. 8 Electrolysis • Electric current is measured in amperes (A). • Quantity of electricity is measured in coulombs (C). C = A x time ( in second) = A x t(s) charge (C) = current (A) x time (s) • Electromotive force is measured in Volts, V. V = J/C Vx. C=J 1 mole e- = 96, 500 C = 1 Faraday 18

Chapter 6 / Electrochemistry 6. 8 Electrolysis Na+ + e- Na From this half

Chapter 6 / Electrochemistry 6. 8 Electrolysis Na+ + e- Na From this half cell reaction we have 1 mole of e- 1 mol of Na 1 mole e- mwt of Na 1 mole e- 23 g of Na ≡ 96500 C ½ mole of e- will ppt. ½ mole of Na , 23/2 g Na. 19

Chapter 6 / Electrochemistry 6. 8 Electrolysis Example: How much Ca will be produced

Chapter 6 / Electrochemistry 6. 8 Electrolysis Example: How much Ca will be produced in an electrolytic cell of molten Ca. Cl 2 if a current of 0. 452 A is passed through the cell for 1. 5 hours? Anode: Cathode: 2 Cl- (l) Cl 2 (g) + 2 e. Ca 2+ (l) + 2 e- Ca (s) Ca 2+ (l) + 2 Cl- (l) Ca (s) + Cl 2 (g) C = A x ts = 0. 452 x 1. 5 x 60 = 2440. 8 C Ca 2+ (l) + 2 e- Ca (s) From the equation we have: 2 F = 2 x 96500 C ≡ 1 mol Ca = 40 g Ca 2440. 8 C ===== ? g of Ca = 2440. 8 x 40 g / 2 x 96500 = 0. 5 g Ca 20

Chapter 6 / Electrochemistry 6. 8 Electrolysis Example: a) In the electrolysis of Cu.

Chapter 6 / Electrochemistry 6. 8 Electrolysis Example: a) In the electrolysis of Cu. SO 4 , How much copper is plated out on the cathode by a current of 0. 75 A in 10 min ? b) What the volume of O 2 (g) at STP is librated ? a) C = A x ts = 0. 75 x 10 x 60 = 450 C Cu++ + 2 e- Cu (s) From the equation we have: 2 F = 2 x 96500 C ≡ 1 mol Cu = 63. 5 g Cu 450 C ===== ? g of Cu = 450 x 63. 5 g / 2 x 96500 = 0. 148 g Cu 21

Chapter 6 / Electrochemistry 6. 8 Electrolysis b) 2 H 2 O O 2(g)

Chapter 6 / Electrochemistry 6. 8 Electrolysis b) 2 H 2 O O 2(g) + 4 H+ + 4 e- PV=n. RT 4 F ≡ 4 x 96500 C= 1 mole O 2 ≡ 24. 5 L 450 C → ? L = 450 x 24. 5 / 4 x 96500 = 2. 83 x 10 -2 L O 2 (g) 22

Chapter 6 / Electrochemistry 6. 8 Electrolysis Number of F in Cu. SO 4

Chapter 6 / Electrochemistry 6. 8 Electrolysis Number of F in Cu. SO 4 cell = Number of F passed in Ag. No 3 23

Chapter 6 / Electrochemistry 6. 8 Electrolysis Example: a) What mass of copper is

Chapter 6 / Electrochemistry 6. 8 Electrolysis Example: a) What mass of copper is plated in the electrolysis of Cu. SO 4 in the same time 1. 0 g of Ag is plated in a silver coulometer that arranged in series with Cu. SO 4 cell ? b) I f 1. 0 A is passed , how many minutes are required to plate this quantity ? a) Ag+ + e- Ag(s) Cu++ + 2 e- Cu 1 mole of e =1 F ≡ 96500 C ≡ 107. 868 g Ag 2 mole of e = 2 F = 2 x 96500 C = 63. 5 g 894. 61 C = ? g ? C ≡ 1. 0 g C= 1 x 96500 / 107. 868 = 894. 61 C ? G of Cu = 894. 61 x 63. 5 / 2 x 96600 = 0. 2948 g 24

Chapter 6 / Electrochemistry 6. 8 Electrolysis b) C = A x ts 894.

Chapter 6 / Electrochemistry 6. 8 Electrolysis b) C = A x ts 894. 61 = 1. 0 x ts t = 894. 615 s = 894. 615 /60 = 14. 9 min 25

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Galvanic cell (Voltaic cell) – device

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Galvanic cell (Voltaic cell) – device in which transfer of electrons takes place through an external circuit rather than directly between reactants. A spontaneous redox reaction generates an electric current when a piece of zinc metal is placed in a Cu. SO 4 solution, Zn is oxidized to Zn 2+ ions while Cu 2+ ions are reduced to metallic copper Zn Electrons are transferred from Zn to Cu++ , but there is no useful electric current. The electrons are transferred directly from the reducing agent (Zn) to the oxidizing agent (Cu 2+) in solution. However, if we physically separate the oxidizing agent from the reducing agent, the transfer of electrons can take place via an external conducting medium (a metal wire). Cu 2+ ions 26

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells 27

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells 27

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells SO 42– Zn 2+ Zn. SO

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells SO 42– Zn 2+ Zn. SO 4 solution 28

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells SO 42– Cu 2+ Zn. SO

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells SO 42– Cu 2+ Zn. SO 4 solution SO 42– Cu. SO 4 solution 29

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Salt bridge Cotton plugs SO 42–

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Salt bridge Cotton plugs SO 42– Zn 2+ Zn. SO 4 solution Cu 2+ SO 42– Cu. SO 4 solution 30

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Copper

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Copper cathode Cl– K+ Salt bridge Cotton plugs SO 42– Zn 2+ Zn. SO 4 solution Cu 2+ SO 42– Cu. SO 4 solution 31

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Copper

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Copper cathode Cl– K+ Salt bridge Cotton plugs SO 42– Zn 2+ Zn. SO 4 solution Cu 2+ SO 42– Cu. SO 4 solution Zn 32

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Cotton

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Cotton plugs SO 42– Zn 2+ Zn. SO 4 solution 2 e– Zn Copper cathode Cl– K+ Salt bridge Cu 2+ SO 42– Cu. SO 4 solution Zn 2+ 33

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Cotton

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Cotton plugs SO 42– Zn 2+ Zn. SO 4 solution 2 e– Zn Cu 2+ SO 42– Cu. SO 4 solution Zn 2+ Zn is oxidized to Zn 2+ at anode. Zn(s) Copper cathode Cl– K+ Salt bridge Zn 2+(aq) + 2 e– 34

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Cotton

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Cotton plugs SO 42– Zn 2+ Zn. SO 4 solution 2 e– Zn Copper cathode Cl– K+ Salt bridge Cu 2+ SO 42– Cu. SO 4 solution Zn 2+ Cu Zn is oxidized to Zn 2+ at anode. Zn(s) Zn 2+(aq) + 2 e– 35

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Cotton

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Cotton plugs SO 42– Zn 2+ Zn. SO 4 solution 2 e– Copper cathode Cl– K+ Salt bridge Cu 2+ SO 42– Cu. SO 4 solution 2 e– Cu 2+ Zn Zn 2+ Cu Zn is oxidized to Zn 2+ at anode. Zn(s) Zn 2+(aq) + 2 e– 36

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Cotton

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Cotton plugs SO 42– Zn 2+ Zn. SO 4 solution 2 e– Copper cathode Cl– K+ Salt bridge Cu 2+ SO 42– Cu. SO 4 solution 2 e– Cu 2+ Zn Zn 2+ Cu Zn is oxidized to Zn 2+ at anode. Zn(s) Zn 2+(aq) + 2 e– Cu 2+ is reduced to Cu at cathode. 37 2 e– + Cu 2+(aq) Cu(s)

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Daniell

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Voltmeter e– Zinc anode e– Daniell cell Cotton plugs SO 42– Cu 2+ SO 42– Zn 2+ Zn. SO 4 solution 2 e– Copper cathode Cl– K+ Salt bridge Cu. SO 4 solution 2 e– Cu 2+ Zn Zn 2+ Cu Zn is oxidized to Zn 2+ at anode. Zn(s) Zn 2+(aq) + 2 e– Net reaction Zn(s) + Cu 2+(aq) Zn 2+(aq) + Cu(s) Cu 2+ is reduced to Cu at cathode. 38 2 e– + Cu 2+(aq) Cu(s)

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Anode – electrode at which oxidation

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Anode – electrode at which oxidation occurs Cathode – electrode at which reduction occurs Electrons always flow from anode to cathode Salt bridge – tube that contains an electrolytes , maintains charge neutrality for a voltaic cell 39

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Cell Diagram Zn (s) + Cu

Chapter 6 / Electrochemistry 6. 2 Galvanic Cells Cell Diagram Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+] = 1 M & [Zn 2+] = 1 M Zn (s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (s) anode cathode Cell Diagram Cl 2 (g) + 2 I – (aq) 2 Cl-(aq)+I 2 (s) Cell notation I- (aq)/ I(s) /Pt // Cl 2 (g)/ Cl- (aq)/ Pt anode cathode 40

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials An electric current flows from

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials An electric current flows from the anode to the cathode because there is a difference in electrical potential energy between the electrodes. The difference in electrical potential between the anode and cathode is called: • cell voltage • electromotive force (emf) • cell potential = electromotive force (EMF) = Ecell = Cell potential = Cell voltage ( V = J/C) = driving force that moves electrons from anode to cathode 41

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials Ecell= Erd. + Eox. (potential

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials Ecell= Erd. + Eox. (potential difference) E°cell= standard cell potential E°cell= E°re. + E°ox. Zn (s) + Cu 2+ (aq) Zn 2+ (aq) + Cu (s), E°cell= 1. 10 V Note: E˚cell must be +ve value for cell to operate, (spontaneous) It is impossible to measure the potential of just a single electrode, but if we arbitrarily set the potential value of a particular electrode at zero, we can use it to determine the relative potentials of other electrodes. 42

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials Standard reduction potential (E 0)

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials Standard reduction potential (E 0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Reduction Reaction 2 e- + 2 H+ (1 M) H 2 (1 atm) E 0 = 0 V Standard hydrogen electrode (SHE) 43

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials The standard hydrogen electrode (SHE)

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials The standard hydrogen electrode (SHE) is used as reference electrode. 44

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials E 0 = 0. 76

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials E 0 = 0. 76 V cell Zn (s) | Zn 2+ (1 M) || H+ (1 M) | H 2 (1 atm) | Pt (s) Anode (oxidation): Zn (1 atm) Zn+2 (1 M) + 2 e. Cathode (reduction): 2 e- + H+ (1 M) H 2 (s) Zn (1 atm) + 2 H+ (1 M) H 2 (s) + Zn+2 (1 M) 45

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials E 0 = 0. 34

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials E 0 = 0. 34 V cell Pt (s) | H 2 (1 atm) | H+ (1 M) || Cu 2+ (1 M) | Cu (s) Anode (oxidation): H 2 (1 atm) 2 H+ (1 M) + 2 e. Cathode (reduction): 2 e- + Cu 2+ (1 M) Cu (s) H 2 (1 atm) + Cu 2+ (1 M) Cu (s) + 2 H+ (1 M) 46

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials • E 0 is for

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials • E 0 is for the reaction as written • The more positive E 0 the greater the tendency for the substance to be reduced (The more E is, the greater the driving force of the RXN) • The half-cell reactions are reversible • The sign of E 0 changes when the reaction is reversed Zn 2++ 2 e- Zn E°red= -0. 76 V Zn Zn 2++2 e E°oxid= +0. 76 V Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0 Zn 2++2 e- Zn E°red= -0. 76 V • 2 Zn 2++4 e- 2 Zn E°red= -0. 76 V 47

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials Example: Use electrode potential to

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials Example: Use electrode potential to determine whether the following proposed reactions are spontaneous with all substances present at unit activity: a) Cl 2 (g)+ 2 I- (aq) 2 Cl- (aq)+I 2 (s) b)2 Ag(s)+2 H+(aq) 2 Ag+(aq)+H 2(g) a) Fist we have to see who is reduced and who is oxidized. Cl 2 is reduced I- is oxidized Cathode Cl 2+2 e- 2 Cl. Cell reaction Anode reactor E =+1. 36 V 2 I- I 2 (s) + 2 e- E ox= -0. 536 V Cl 2+2 I- I 2(s)+2 Cl-(aq), E cell= +0. 824 V Since E cell is +ve the reaction is spontaneous 48

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials b: 2 Ag (s) +

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials b: 2 Ag (s) + 2 H+ (aq) → Ag+ (aq) +H 2 (g) H+ is reduced Ag is oxidized Anode reaction: 2 Ag(s) 2 Ag+(aq) + 2 e- E ox. = -0. 799 V Cathode reaction: 2 H++2 e- H 2 (g) E = 0. 00 V 2 Ag + 2 H+(aq) → 2 Ag+(aq) + H 2(g) , E cell = -0. 799 V The result is nonspontaneous, E cell = -ve 49

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials What is the standard emf

Chapter 6 / Electrochemistry 6. 3 Standard Reduction Potentials What is the standard emf of an electrochemical cell made of a Cd electrode in a 1. 0 M Cd(NO 3)2 solution and a Cr electrode in a 1. 0 M Cr(NO 3)3 solution? Cd 2+ (aq) + 2 e- Cd (s) E 0 = -0. 40 V Cd is the stronger oxidizer Cr 3+ (aq) + 3 e- Cr (s) E 0 = -0. 74 V Cd will oxidize Cr Anode (oxidation): Cathode (reduction): Cr (s) Cr 3+ (1 M) + 3 e- x 2 E 0 = 0. 74 V 2 e- + Cd 2+ (1 M) Cd (s) x 3 E 0 = -0. 40 V 2 Cr (s) + 3 Cd 2+ (1 M) 3 Cd (s) + 2 Cr 3+ (1 M) E°cell= E°re. + E°ox. E°cell= -0. 40 + 0. 74 = 0. 34 V. 50

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Spontaneity of Redox RXNs

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Spontaneity of Redox RXNs E 0, spontaneous G = - n F E Faraday’s Constant mol electrons cell potential (96500 C/mol) G = - n F E° Eº and equilibrium constant G = - RT ln. K……. . . 1 G = - n F E ……… 2 From eqns. (1) and (2) we get E˚ = ( RT/n F ) ln. K 51

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions 0 G 0 =

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions 0 G 0 = -RT ln K = -n. FEcell 52

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions What is the equilibrium

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions What is the equilibrium constant for the following reaction at 250 C? Fe 2+ (aq) + 2 Ag (s) Fe (s) + 2 Ag+ (aq) Oxidation: Reduction: E˚ = ( RT/n F ) ln. K 2 Ag 2 Ag+ + 2 e 2 e- + Fe 2+ Fe E 0 = -0. 80 V E 0 = -0. 44 V n = 2, T=25 °C= 298 K, F=96500 E˚ = ( RT/n F ) ln. K E°cell= E°re. + E°ox. E 0 = -0. 44 + (-0. 80) E 0 = -1. 24 V ln. K = E˚/ ( RT/n F ) = -1. 24 /(8. 314 X 298/2 X 96500) = -96. 6 K =e-96. 6 = 1. 12 x 10 -42 53

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: a) Use electrochemical

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: a) Use electrochemical data to calculate the value of G for the reaction 2 Ag (s) + Cl 2 (g) → 2 Ag. Cl (s) b) If H = - 254. 0 KS, calculate S a) We have to calculate E cell from the two half cell reactions. Anode reaction: 2 Ag (s) + Cl – (aq) 2 Ag Cl(s) + 2 e- E ox = -0. 222 V Cathode reaction: 2 e - + Cl 2 (s) 2 Cl – (aq) , E = 1. 369 V 2 Ag (s) + Cl 2 (g) 2 Aq. Cl (s) , E cell = 1. 137 V 54

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions G = -nf E

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions G = -nf E cell = -2 x 96500 x 1. 137 J = -219. 4 k. J b) G = H T S - 219. 4 x 1000 J = - 254. 0 x 1000 J - 298 SJ/K S = - ( 254. 0 219. 4 ) 1000 J 298 K = - 116 J/K 55

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: Use electrochemical data

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: Use electrochemical data to calculate the equilibrium constant K for the following reaction at 25ºC Fe++ (aq) + Ag+ (aq) ⇌ Fe +++(aq) + Ag (s) Fe++(aq) Fe+++ (aq) + e E = -0. 771 V e + Ag+(aq) Ag(s) E = 0. 799 V n = 1, T=25 °C= 298 K, F=96500 E°cell= E°re. + E°ox. E 0 = -0. 771 + (0. 799) E 0 = 0. 028 V E cell = - (RT /n. F ) ln K 0. 028 = - ( 8. 314 x 298 /1 x 96500 ) ln K = 1. 091 K = 2. 98 56

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions G = G +

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions G = G + RT ln Q -n. FE = -n. FE + RTl. Q Nernst equation Nonstandard E = E - (RT / n F) ln Q standard At equilibrium, E=0, K=Q E = (RT /n F) ln K 57

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: What is the

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: What is the electrode potential of Zn++/Zn electrode in which the concentration of Zn++ = 0. 1 M . Zn+++2 e- Zn , E˚ = -0. 76 V E = E - (RT / n F) ln Q n = 2, T=25 °C= 298 K, F=96500, E˚ = -0. 76 V , Q = ( Zn)/ (Zn++) = 1 /0. 1 E= - 0. 76 –( 8. 314 x 298 / 2 x 96500 ln (1/0. 1) = -0. 76 – 0. 0128 (2. 303) = - 0. 76 – 0. 0257 = - 0. 79 V 58

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: What is the

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: What is the potential for the cell : Ni / Ni++ (0. 01 M) // Cl – (0. 2 M) / Cl 2 M) / Pt Ni++ + 2 e - E = + 0. 25 V 2 e - + Cl 2 (g) 2 Cl - E = 1. 36 V 2 Ni + Cl 2 (g) → 2 Cl- (aq) + Ni++ (aq) , E˚cell = 1. 61 V n = 2, T=25 °C= 298 K, F=96500, E˚ = 1. 61 V , Q = Cl - 2 Ni-++ / PCl 2 = ((0. 2)2 (0. 01)/1=0. 0004 E = E - (RT / n F) ln Q E= 1. 61 –( 8. 314 x 298 / 2 x 96500 ln (0. 0004) = 1. 71 V 59

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: What is the

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: What is the E for the cell : Sn / Sn++ (1. 0 M) // Pb++ (0. 001 M) / Pb. Sn++ + 2 e- E. = 0. 136 V Pb++ + 2 e- Pb (s) E = - 0. 126 V Sn (s) + Pb++ Sn ++ (aq) + Pb (s) E cell = 0. 010 V n = 2, T=25 °C= 298 K, F=96500, E˚ = 0. 01 V , Q = ( Sn+2)/ (Pb+2) = 1 /0. 001 E = E - (RT / n F) ln Q E = 0. 01 - ( 8. 314 x 298/ 2 x 96500) ln (1/0. 001) =- 0. 079 V 60

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Will the following reaction

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Will the following reaction occur spontaneously at 250 C if [Fe 2+] = 0. 60 M and [Cd 2+] = 0. 010 M? Fe 2+ (aq) + Cd (s) Fe (s) + Cd 2+ (aq) Cd 2+ + 2 e- E. = 0. 40 V 2 e- + Fe 2+ 2 Fe E. = -0. 44 V n = 2, T=25 °C= 298 K, F=96500, E˚ = -. 04 V , Q = ( Cd+2)/ (Fe+2) = 0. 01 /0. 6 E = E - (RT / n F) ln Q E = -0. 04 - ( 8. 314 x 298/ 2 x 96500) ln (0. 01/0. 6) =0. 013 V 61

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: Consider a cell

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Example: Consider a cell reaction Mg(s) + 2 H+ (aq) Mg++(aq)+ H 2 (g) , E cell=2. 363 V What is the concentration of H+ (aq) in a cell in which Mg++ = 1. 00 M and PH 2 =1. 0 atm, if the Ecell = 2. 099 V? n = 2, T=25 °C= 298 K, F=96500, E˚ = 2. 363 V , E= 2. 099 V Q = ( Mg+2)PH 2/ (H+)2 = (1)(1) / (H+)2 E = E - (RT / n F) ln Q 2. 099 = 2. 363 - ( 8. 314 x 298/ 2 x 96500) ln (1/ (H+)2 ) 2. 099 – 2. 363 = -0. 0128 ln [ H+]-2 = 20. 625 [ H+]-2 = 906407915. 01 [H+]= 3. 3 X 10 -5 M 62

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Concentration cell : two

Chapter 6 / Electrochemistry 6. 3 Thermodynamics of Redox Reactions Concentration cell : two half-cells composed of the same material but differing in ion concentrations High concentration = reduction , low concentration = oxidation Example: What is the E for the cell : Cu / Cu++ (0. 01 M) // Cu++ (0. 1 M) / Cu. Cu++ + 2 e- E. = 0. 34 V Cu++ + 2 e- Cu (s) E = - 0. 34 V Cu (s) + Cu++ Cu ++ (aq) + Cu (s) E cell = 0. 0 V n = 2, T=25 °C= 298 K, F=96500, E˚ = 0. 0 V , Q = ( Cu+2)/ (Cu+2) = 0. 01 /0. 1 E = E - (RT / n F) ln Q E = 0. 0 - ( 8. 314 x 298/ 2 x 96500) ln (0. 01/0. 1) =- 0. 0296 V 63

Chapter 6 / Electrochemistry Thank you for listening 64

Chapter 6 / Electrochemistry Thank you for listening 64