Lecture #15. Chapter 18 - Electrochemistry

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1 Lecture #15 Chapter 18 - Electrochemistry

2 Chapter 18 - Electrochemistry the branch of chemistry that examines the transformations between chemical and electrical energy

3 Redox Chemistry Revisited

another species loses e (oxidation) Oxidizing agents vs.

4 A Spontaneous Redox Reaction Znº(s) + Cu 2+ (aq) Zn 2+ (aq) + Cuº(s) Sum of two half-reactions: One species gains e (reduction) while another species loses e (oxidation) Oxidizing agents vs. reducing agents Znº(s) Zn 2+ (aq) Zn = red. agent; Cu 2+ = oxid. agent Cu 2+ (aq) Cuº(s)

5 Zn 0 + Cu 2+ Zn 2+ + Cu 0 electrons K +? Zn 2+ NO3 - NO3 - Cu 2+

6 A Voltaic Cell

7 Electrochemical Cells

8 The anode is for oxidation!! Reduction takes place at the cathode!!

9 Voltaic Cell Spontaneous Reaction Chemical energy is transformed into electrical energy.

10 Electrolytic Cell External source of electrical energy required Electrical energy is transformed into chemical energy.

11 Cell Components Anode = electrode at which oxidation halfreaction (loss of electrons) takes place. Cathode = electrode at which reduction halfreaction (gain of electrons) takes place. A bridge connects the two solutions of the cell; balances flow of electrons, eliminates accumulation of charge in either compartment.

12 Writing Cell Diagrams Write chemical symbol of anode at the far left, symbol of cathode at the far right, and a double vertical for connecting bridge halfway between them. Work inward from electrodes toward the bridge, using vertical lines to indicate phase changes and symbols of ions or compounds to represent electrolytes surrounding the electrode that are changed by the cell reaction. Indicate concentrations of dissolved species and partial pressures of any gases (if known).

13 Cell Diagram Example Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) 1. Cu(s) Ag(s) 2. Cu(s) Cu 2+ (aq) Ag + (aq) Ag(s) 3. Cu(s) Cu 2+ (1.00 M) Ag + (1.00 M) Ag(s) } anode half-cell } cathode half-cell

14 Standard Potentials

15 Recollection: The Activity Series of Metals ACTIVITY SERIES of Metals Li Rb K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb [ H 2 ] Sb Bi Cu Hg Ag Pt Au Replace hydrogen from cold water Replace hydrogen from steam Replace hydrogen from acids React with oxygen to form oxides Atoms are Strong Reducing Agents Cations are Strong Oxidizing Agents

16 Recollection: The Activity Series of Metals ACTIVITY SERIES of Metals Li Rb K Ba Sr Ca Na Mg Al Mn Zn Cr Fe Cd Co Ni Sn Pb [ H 2 ] Sb Bi Cu Hg Ag Pt Au Replace hydrogen from cold water Replace hydrogen from steam Replace hydrogen from acids React with oxygen to form oxides Atoms are Strong Reducing Agents 2 Naº + 2 H2O(l) 2 Na OH - + H2 Δ 2 Znº + 4 H2O(g) 2 Zn OH H2 Snº + H2O + H + Sn 2+ + OH H2 2 Cuº + O2 + 2 CuO Cations are Strong Oxidizing Agents

17 Standard Potentials Standard reduction potential (Eº) the potential of a halfreaction in which all reactants and products are their standard states at 25ºC. Standard cell potential (Eº cell ) a measure of how forcefully an electrochemical cell (in standard state) can pump electrons through an external circuit. Eº cell = Eº cathode Eº anode

18 19.3 Standard Reduction Potentials

19 Standard Cell Potential (E cell ) Znº(s) Zn 2+ (aq) Cu 2+ (aq) Cuº(s) Eºcell = Eºcathode Eºanode Eºcathode (Cu 2+ Cu(s)) = V Eºanode (Zn 2+ Zn(s)) = V E o cell = (0.342 V) ( V) = V

20 1) Calculate the standard cell potential for the reaction: 2 Fe 3+ (aq) + 2 I (aq) 2 Fe 2+ (aq) + I2(s) Fe 3+ (aq) Fe 2+ (aq) 2 I - (aq) I2º (s) Eºcell = Eºcathode Eºanode Eºcathode (Fe 3+ (aq) Fe 2+ (aq) ) = V Eºanode ( I2º(s) 2I - (aq)) = V E o cell = ( V) ( V) = V

21 2) Calculate the standard cell potential for the reaction: 2 NiO(OH)(s) + 2 H2O (l) + Cd (s) 2 Ni(OH)2 (s) + Cd(OH)2 (s) 2 NiO(OH) + 2 H2O 2 Ni(OH)2 + 2 OH- (Ni 3+ Ni 2+ ) Eºreductiion = V Cd + 2 OH- Cd(OH)2 (Cd Cd 2+ ) Eºoxidation = V Eºcell = Eºoxidation + Eºreduction E o cell = ( V) + ( V) = V Eºcell = Eºcathode Eºanode

22 Znº(s) + Cu 2+ (aq) Zn 2+ (aq) + Cuº(s) Znº(s) Zn 2+ (aq) Cu 2+ (aq) Cuº(s) 2 Fe 3+ (aq) + 2 I (aq) 2 Fe 2+ (aq) + I2(s) 2 I - (aq) I2º (s) Fe 3+ (aq) Fe 2+ (aq) 2 NiO(OH)(s) + 2 H2O (l) + Cd (s) 2 Ni(OH)2 (s) + Cd(OH)2 (s) 2 NiO(OH) + 2 H2O 2 Ni(OH)2 + 2 OH- Cd + 2 OH- Cd(OH)2 (Ni 3+ Ni 2+ ) (Cd Cd 2+ )

23 Stronger Oxidizing Agents Standard Reduction Potentials Fe 3+ (aq) Fe 2+ (aq) Weaker Reducing Agents Ni 3+ Ni I2º(s) 2I- (aq) Cu 2+ (aq) Cuº(s) Zn 2+ (aq) Zn º (s) Cd 2+ Cdº Weaker Oxidizing Agents Stronger Reducing Agents

24 Alessandro Volta Luigi Galvani

25 Dry Cells

26 Lead Storage Battery

27 Fuel Cell

28 Chemical Energy and Electrical Work

29 Current and Voltage Current - the number of electrons that flow through the system per second 1 A = 1 Ampere = 1 Coulomb of charge/second = x electrons/second Potential difference - the difference in potential energy between reactants and products 1 V of force = 1 J of energy/coulomb of charge (The potential difference can also be thought of as the voltage needed to drive electrons through the external circuit.) Electromotive force (emf) - the amount of force pushing the electrons through the wire

30 Voltage and Electrical Work ΔGºcell = Welec = CEºcell Welec = work done by the cell C = charge (coulombs) Eºcell = electromotive force (emf); cell voltage,volts = J/C ΔGºcell = -nfeºcell Faraday constant (F) is C/(mol e ) n = number of moles of electrons x C/e x e mol

31 3) Calculate the value of ΔGº and the work done on the circuit for the reaction: Mg(s) + Cu 2+ (aq) Mg 2+ (aq) + Cu(s) taking place in a voltaic cell that produces 2.71V. Mg(s) Mg 2+ (aq) Cu 2+ (aq) Cu(s) [oxidation] [reduction] Eºcell = Eºcathode Eºanode E o cell = ( V) ( 2.37 V) = V ΔGºcell = Welec = -nfeºcell = (-2)(9.65 x 10 4 C/mol e)(2.71 J/C) = -523 J

32 4) Calculate the value of Eºcell for the following reaction by calculating ΔGº: Cu(s) + 2 Fe 3+ (aq) Cu 2+ (aq) + 2 Fe 2+ (aq) ΔGºcell =[ (-78.9)] - [ (-4.7)] ΔGºcell =[-92.3] - [ ] = kj Eºcell = x 10 3 J/mol -(2)(9.65 x 10 4 C) = J/C = V

33 A Reference Point: The Standard Hydrogen Electrode

34 A Reference Point: The Standard Hydrogen Electrode 2 H + (aq) + 2 e H2(g) ESHE = 0.00 V Pt H2(g), 1.0 atm H + (1.0 M) (Can serve as anode or cathode)

35 A Reference Point: The Standard Hydrogen Electrode The Standard Hydrogen Electrode (SHE) reduction potential is defined to be exactly 0.00 V. Half reactions with a stronger tendency toward reduction that the SHE have a positive value for Eº reduction. Half reactions with a strong tendency toward oxidation than the SHE have a negative value for Eº reduction. Eºcell = Eºcathode Eºanode Eºcell = Eºoxidation + Eºreduction Eºoxid = -Eºred When adding Eº values for the half-cells, do not multiply the half-cell Eº values.

36 Determination of E o!!h 2 "!!(1!atm)! V = ESHE EZn V = 0.00 V EZn V = ECu ESHE V = ECu 0.00 V

37 Selected Standard Electrode Potentials (298K) Half-Reaction E 0 (V) F 2 (g) + 2e - 2F - (aq) strength of oxidizing agent Cl 2 (g) + 2e - 2Cl - (aq) MnO 2 (g) + 4H + (aq) + 2e - Mn 2+ (aq) + 2H 2 O(l) NO 3- (aq) + 4H + (aq) + 3e - NO(g) + 2H 2 O(l) Ag + (aq) + e - Ag(s) Fe 3+ (g) + e - Fe 2+ (aq) O 2 (g) + 2H 2 O(l) + 4e - 4OH - (aq) Cu 2+ (aq) + 2e - Cu(s) 2H + (aq) + 2e - H 2 (g) N 2 (g) + 5H + (aq) + 4e - N 2 H 5+ (aq) Fe 2+ (aq) + 2e - Fe(s) 2H 2 O(l) + 2e - H 2 (g) + 2OH - (aq) Na + (aq) + e - Na(s) strength of reducing agent Li + (aq) + e - Li(s) -3.05

38 Overall Cell Potential Eº(V) Cu 2+ (aq) + 2e Cu (s) Zn 2+ (aq) + 2e Zn (s) Eºcell = Eºoxid + Eºred Eºcell = - (-0.76) =1.10 V Eºcell = Eºcathode - Eºanode Eºcell = (-0.76) = 1.10 V

Chapter 19 - Electrochemistry. the branch of chemistry that examines the transformations between chemical and electrical energy

Chapter 19 - Electrochemistry. the branch of chemistry that examines the transformations between chemical and electrical energy Chapter 19 - Electrochemistry the branch of chemistry that examines the transformations between chemical and electrical energy 19.1 Redox Chemistry Revisited A Spontaneous Redox Reaction Znº(s) + Cu 2+