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

Size: px

Start display at page:

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

Ashlie Quinn
5 years ago
Views:

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

2 19.1 Redox Chemistry Revisited

another species loses e (oxidation) Oxidizing agents vs.

3 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)

4 A Voltaic Cell

5 19.2 Electrochemical Cells

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

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

8 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.

9 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).

10 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

11 19.3 Standard Potentials

12 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

13 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

14 19.3 Standard Reduction Potentials

15 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

16 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

17 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- Cd + 2 OH- Cd(OH)2 Eºreductiion = V Eºoxidation = V Eºcell = Eºoxidation + Eºreduction E o cell = ( V) + ( V) = V Eºcell = Eºcathode Eºanode

18 19.4 Chemical Energy and Electrical Work

19 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

20 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

21 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

22 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

23 19.5 A Reference Point: The Standard Hydrogen Electrode

24 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)

25 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.

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

27 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

28 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

29 19.6 The Effect of Concentration on E cell The Nernst Equation E and K

30 Nernst Equation G = ΔGº + RT ln Q ΔGº = nfeº cell for R = J/(K mol) Q = reaction quotient

31 5) Consider the cell reaction below, starting at standard conditions: Cu 2+ + Zn Zn 2+ + Cu Eºcell = Eºcathode Eºanode Eºcell = (0.342 V) ( V) = V Determine the cell potential after the reaction has proceeded enough for the [Zn 2+ ] to have changed by 0.35 mol/l at 25 o C. = J/mol (2) log Q = J/mol log 1.35 (2) 0.65 = V

32 K and ΔGº At equilibrium: Ecell = zero Q = K Rearranging the Nernst equation:

33 6) Calculate K from the standard cell potential for the reaction below at 25 o C. Fe(s) + Pb 2+ (aq) Pb(s) + Fe 2+ (aq) Fe(s) Fe 2+ (aq) Pb 2+ (aq) Pb(s) Eº = V Eº = V log K = (2 mol)(0.31 V) J/mol Eºcell = Eºcathode Eºanode Eºcell = (-0.45) Eºcell = log K = 10.4 K = 10.4 = 2.5 x 10 10

34 E cell, ΔG and K

35 E cell, ΔG and K Reaction K Gº Eºcell Spontaneous >1 <0 >0 At Equilibrium Nonspontaneous <1 >0 <0

36 Concentration Cell When the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow. If cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)

37 Concentration Cell = J/mol (2) log = V Cu(s) Cu 2+ (aq) (0.010 M) Cu2+ (aq) (2.0 M) Cu(s)

38 19.7 Relating Battery Capacity to Quantities of Reactants

39 7) Assuming no interfering reductions, what mass of Au can be plated in 25 min using 5.5 A for the half reaction: Au 3+ (aq) + 3 e- Au (s) Given: 1 A= 1 C/sec current A 1mol e-/96,485 C time = 25 min 3 mol e-/1 mol Au 1 molau = g Au 5 min x 60 sec x 5.5 C x 1 mol e- x 1 mol Au x 1 min 1 sec 96,485 C 3 mol e g Au 1 mol Au = 5.6 g Au

40 19.8 Electrolytic Cells Uses electrical energy to overcome the energy barrier and cause a non-spontaneous reaction to occur

41 anode cathode cathode anode DC source, sometimes over voltage required

42 Electrolysis The process of using electricity to break a compound apart Performed in an electrolytic cell Can be used to isolate elements from their compound H 2 and O 2 from water Naº and Cl 2 º from NaCl Various metals from ores

43 Electrolysis of Water

44 Electrolysis of Water

45 Electrolysis of NaCl (l)

46 Downs Cell for Sodium Production Cross-sectional view of a Downs cell for production of sodium metal by electrolysis of molten sodium chloride. The cell design keeps the sodium and chlorine apart so they can t react with teach other.

47 Mixtures of Oxidizable and Reducible Substances When more than one cation is present, the cation that is the easiest to reduce will be reduced first at the cathode. (The cation with the least negative or most positive Eº reduction ) When more than one anion is present, the anion that is the easiest to oxidize will be oxidized first at the anode. (The anion with the least negative or most positive Eº oxidation )

8) Electrolysis of NaI(aq) with Inert Electrodes NaI(aq) = Na+ (aq) + I-(aq) O2 + 4e - + 4H+ ----------> 2 H2O Eºred = +1.23 V I2 + 2e - ----------> 2 I - Eºred = +0.

49 8) Electrolysis of NaI(aq) with Inert Electrodes NaI(aq) = Na+ (aq) + I-(aq) O2 + 4e - + 4H > 2 H2O Eºred = V I2 + 2e > 2 I - Eºred = V 2 H2O + 2e > H2 + 2 OH - Eºred = V * Na + + e > Naº Eºred = V 2 H2O > O2 + 4e - + 4H+ Eºox = V 2 I > I2 + 2e - Eºox = V *

50 Electrolysis of NaI(aq) with Inert Electrodes Possible oxidations: 2 H2O > O2 + 4e - + 4H+ Eºox = V 2 I > I2 + 2e - Eºox = V Possible reductions: 2 H2O + 2e > H2 + 2 OH - Eºred = V Na + + e > Naº Eºred = V * *

51 Electrolysis of NaI(aq) with Inert Electrodes 2 I - (aq) + 2 H2O (l) I2 (aq) + H2 (g) + 2 OH -

Lecture #15. Chapter 18 - Electrochemistry

Lecture #15. Chapter 18 - Electrochemistry Lecture #15 Chapter 18 - Electrochemistry Chapter 18 - Electrochemistry the branch of chemistry that examines the transformations between chemical and electrical energy Redox Chemistry Revisited A Spontaneous