1. ELECTROCHEMISTRY REDOX REVISITED!
Electrochemistry (Ch.17)

2. ELECTROCHEMISTRY redox reactions electrochemical cells
electrode processes
construction
notation
cell potential and Go
standard reduction potentials (Eo)
non-equilibrium conditions (Q)
batteries
corrosion
Electric
automobile

3. Oxidation Numbers
Is this a redox reaction?
Zn + HCl ZnCl2 + H2

4. Recall a redox reaction is a rxn where one or more electrons
are transferred betw. reactants.
One element is oxidized ( the reducing agent) while the
other is reduced (the oxidizing agent)

5. The Half-Rxn Method of Balancing Redox Eqns
A half-reaction occurs when a redox rxn is separated into
halves, i.e. an oxidation half and a reduction half.

6. Follow these steps to use the half-rxn method;
1. Write the oxidation half and then the reduction half from the redox equation.
2. Balance each half-reaction.
Balance elements other than H and O.
Balance O by adding H2O.
Balance H by adding H+.
Balance charge by adding e-

7. Half-Reaction Method 3. Add the half-reactions
4. Check the final equation for mass and charge balance.

8. Balance an Ionic Redox Equation
Cr2O72- + Cl1-  Cr3+ + Cl2 (acidic solution)
RED Cr6+2O72-  Cr3+
Cr2O72-  2Cr3+
Cr2O72-  2Cr3+ + 7H2O
14H1+ + Cr2O72-  2Cr H2O
6e H1+ + Cr2O72-  2Cr H2O
and OX Cl1-  Cl20
2Cl1-  Cl20
2Cl1-  Cl20 + 2e1-

9. Balance an Ionic Redox Equation
6e H1+ + Cr2O72-  2Cr H2O
3 x ( 2Cl1-  Cl20 + 2e1- )
6e H1+ + Cr2O Cl1-
 2Cr3+ + 7H2O + 3Cl20 + 6e1-
If solution is basic, add 14OH1- to both sides of the reaction to neutralize the H1+

10. The Half-Reaction Method
In basic solution hydrogen peroxide oxidizes chromite ions, Cr(OH)4-, to chromate ions, CrO42-. The hydrogen peroxide is reduced to hydroxide ions. Write and balance the net ionic equation for this reaction. 47 48 50 50 49

11. Electrochemical Cells
Batteries

12. Electrochemistry There are two kinds electrochemical cells.
Electrochemical cells containing nonspontaneous chemical reactions are called electrolytic cells.
Electrochemical cells containing spontaneous chemical reactions are called voltaic or galvanic cells.
Galvanic cell (voltaic cell)— energy released during a spontaneous reaction (G < 0) generates electricity
Electrolytic cell—consumes electrical energy from an external source to cause a nonspontaneous redox reaction to occur (G > 0)
The cathode (Reduction)is negative in electrolytic cells and positive in voltaic cells.
The anode (oxidation) is positive in electrolytic cells and negative in voltaic cells.

13. Electrochemistry
– Both types of cells contain two electrodes connected to an external circuit that provides an electrical connection between systems. Metallic Conduction.
– When circuit is closed, electrons flow from the anode to the cathode; electrodes are connected by an electrolyte, which is an ionic substance or solution that allows ions to transfer between the electrodes, thereby maintaining the system’s electrical neutrality. Ionic Conduction.

14. Electrochemistry
In all voltaic cells, electrons flow spontaneously from the negative electrode (anode) to the positive electrode (cathode).
Cell halves are physically separated so that electrons (from redox reaction) are forced to travel through wires; creating a potential difference.
A simple half-cell consists of:
A piece of metal immersed in a
solution of its ions.
A wire to connect the two half-cells.
And a salt bridge to
complete the circuit,
maintain neutrality, and
prevent solution mixing.

15. An Ox Ate a Fat Red Cat Anode – Oxidation- Lose Electrons
The anode = location for the oxidation half-reaction.
Reduction – Cathode- Gain electrons
The cathode = location for the reduction half-reaction.

16. Zn is below Cu, Zn is anode

17. Anode / Cathode
How do you know which electrode is which if there is no diagram?
Use 17.1 (pg 796) to predict which electrode is the anode and which electrode is the cathode
Anode = Oxidation = Electron Donor
The anode (oxidized) is the metal that has a lower value in table 17.1.
Cathode = Reduction = Electron Acceptor
The cathode is the metal that has a higher value Table 17.1.

18. Notation for Cells
ZnZn+2Cu+2Cu

19. Example Homework #26 5 e + 6 H+ + IO3 → 1/2 I2 + 3 H2O
(Fe2+ → Fe3+ + e) × 5
6 H+ + IO3 + 5 Fe2+ → 5 Fe3+ + 1/2 I2 + 3 H2O
Or 12 H+(aq) + 2 IO3-(aq) + 10 Fe2+(aq) → 10 Fe3+(aq) + I2(aq) + 6 H2O(l)
Cathode(Reduction): Pt electrode; IO3, I2 and H2SO4 (H+ source) in solution.
 Anode(oxidation): Pt electrode; Fe2+ and Fe3+ in solution

20. 26 b. b. (Ag+ + e → Ag) × 2 Zn → Zn2+ + 2 e-
______________________________
Zn(s) + 2 Ag+(aq) → 2 Ag(s) + Zn2+(aq)
Cathode(Reduction): Ag electrode; Ag+ in solution;
Anode(oxidation): Zn electrode; Zn2+ in solution

21. Calculating Voltage A.K.A. Cell Potentials

22. This waterfall is an analogy for cell potential.
Electron Rich
Electron Poor

23. Electrochemistry
Because the half-reactions in the table are arranged in order of their Eº values, the table can be used to predict the relative strengths of various oxidants and reductants
Any species on the left side of a half-reaction will spontaneously oxidize any species on the right side of another half-reaction that lies below it
Any species on the right side of one half-reaction will spontaneously reduce any species on the left side of another half-reaction that lies above it

24. Calculating Standard Cell Potentials
Electrochemistry
Calculating Standard Cell Potentials
• The standard cell potential for a redox reaction, Eºcell, is a measure of the tendency of the reactants in their standard states to form the products in their standard states—it is a measure of the driving force for the reaction (voltage)
The standard cell potential is the reduction potential of the reductive half-reaction minus the reduction potential of the oxidative half-reaction (Eºcell = Eºcathode – Eºanode).

25. Calculate the Cell Potential for Zn reacting with Cu 2+

27. Cell Potential for Zn reacting with Cu 2+
Ecell = Ecathode –Eanode
= V − (−0.76 V)
= V

28. Concept Check Question I
Determine the standard emf for this reaction.
Fe (s) + 2Fe 3+ (aq)
3Fe2+(aq)

29. Homework #28
26a. = 1.20 V – (0.77 V) = 0.43 V
26b. = 0.80 V – (-0.76 V) = 1.56 V

30. 17.3 Cell potential, electrical work and free energy
The work accomplished when electrons are transferred through a wire depends on the “push” (thermodynamic driving force) behind the electrons
The driving force (emf) is defined in terms of potential difference (in volts) between two points in the circuit
emf = potential difference (V) = (work)
(charge)

31. -w = qE q = nF = moles of e- x charge/mole e- Thus, DGo = -nFEo
emf = potential (V) = work (J) / Charge(C)
E = work done by system / charge
Charge is measured in coulombs. Thus,
-w = qE
q = nF = moles of e- x charge/mole e-
Thus,
DGo = -nFEo
Faraday Constant(F) = 96,485 C/mol e-

32. if E º > 0, then DGº < 0 spontaneous
Potential, Work, DG and spontaneity Relating Gibbs Free Energy and Elctrochemistry
DGº = -nFE º
if E º > 0, then DGº < 0 spontaneous
if E º < 0, then DGº > 0 nonspontaneous

33. Ex. 17.3/p.802 Calculate DGº for the following reaction:
Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq)
Fe+2(aq) + e-® Fe(s) Eº = V
Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V
DGº = -nFE º
n=2 mol
F=96,485 C/mol
E= cathode-anode (.34 – (-.44))= .78 J/C
G= - 2 mol*96,485 C/mol* .78 J/C
G=-1.5 x 10-5 J
Spontaneous reaction since G is negative or E is positive

34. Problem #33
33. Locate the pertinent half-reactions in Table 17.1, and then figure which combination will give a positive standard cell potential. In all cases, the anode compartment contains the species with the smallest standard reduction potential. For part a, the copper compartment is the anode,
a Au e → Au E° = 1.50 V
(Cu+ → Cu2+ + e) × E° = -0.16V
= 1.34 V
Au3+(aq) + 3 Cu + (aq) → Au(s) + 3 Cu 2+(aq)

35. 33 b and in part b, the cadmium compartment is the anode. b.
(VO H+ + e → VO2+ + H2O) × 2 E° = 1.00 V
Cd → Cd2+ + 2e E° = 0.40 V 2 VO2+(aq) + 4 H+(aq) + Cd(s) →
2 VO2+(aq) + 2 H2O(l) + Cd2+(aq) = 1.40 V

36. Problem 39
Because the cells are at standard conditions, wmax = ΔG = ΔG° = .
See Exercise for the balanced overall equations and for .
33a.wmax = -(3 mol e)(96,485 C/mol e)(1.34 J/C) = × 105 J = -388 kJ
33b.wmax = - (2 mol e)(96,485 C/mol e)(1.40 J/C) = × 105 J = kJ

37. Homework #34 34. a. (H2O2 + 2 H+ + 2 e → 2 H2O) × 3 E° = 1.78 V
2 Cr H2O → Cr2O72 + 14 H+ + 6 e
E° = V
3 H2O2(aq) + 2 Cr3+(aq) + H2O(l) → Cr2O72(aq) + 8 H+(aq)
= 0.45 V
b. (2 H+ + 2 e → H2) × E° = 0.00 V (Al → Al e) × E° = 1.66 V
6 H+(aq) + 2 Al(s) → 3 H2(g) + 2 Al3+(aq)
= 1.66 V

38. Homework #40
Because the cells are at standard conditions, wmax = ΔG = ΔG° =
See Exercise for the balanced overall equations
34a. wmax = - (6 mol e)(96,485 C/mol e)(0.45 J/C) = -2.6 × 105 J = -260 kJ
34b.wmax = - (6 mol e)(96,485 C/mol e)(1.66 J/C) = × 105 J = -961 kJ

39. The Nernst Equation (Detm. of Ecell Under Nonstandard Conditions

40. The Nernst equation is an equation used to determine cell potential under nonstandard conditions or to determine the concentration of a reactant or product by evaluating Q or K. As reactions proceed concentrations of products increase and reactants decrease. When equilibrium is reached Q = K ; Ecell = 0 and G = 0 (the cell no longer has the ability to do the work)

41. Nernst Equation Recall….. G = G + RT ln Q Therefore,
−nFE = −nFE + RT ln Q

42. Nernst Equation E = E − RT nF ln Q or, using base-10 logarithms,
Dividing both sides by −nF, we get the Nernst equation:
E = E − RT nF
ln Q
or, using base-10 logarithms,
E = E −
2.303 RT nF
log Q

43. Nernst Equation
At room temperature (298 K), where most reactions take place.
2.303 RT F
= V
The equation becomes…..
E = E −
0.0592 n
log Q

44. Calculation of Equilibrium Constants
for redox reactions
At equilibrium, Ecell = and Q = K.
Then,
at 25 oC

45. How does concentration Effect Ecell?
Qualitatively: we can predict direction of change in E from LeChâtelier pinciple
2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)
Predict if Ecell will be greater or less than Eºcell 1M) for the following cases:
if [Al+3] = 1.5 M and [Mn+2] = 1.0 M
if [Al+3] = 1.0 M and [Mn+2] = 1.5M
An increase in conc. of reactants would favor forward reaction causing Ecell to increase.

46. The Nernst Equation
As reactions proceed concentrations of products increase and reactants decrease.
When equilibrium is reached
Q = K ; Ecell = 0
and G = 0 (the cell no longer has
the ability to do the work)

47. 2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)
Exercise- p. 805
Determine the cell potential at 25oC for the following cell, given that
2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)
[Mn2+] = 0.50 M; [Al3+]=1.50 M;
Q= (1.5)2/(.5)3
E0 cell = cathode (Al)+ anode (Mn)
1.66V V = 0.48V
Always we have to figure out n from the balanced equation n=6
2Al(s)+ ® 2Al+3(aq) + 6e-
3Mn+2(aq) + 6e- ® 3Mn(s)

48. Pg 805 continued .48V – (.059/6 * Log (1.5)2/(.5)3)
So,
.48V – (.059/6 * Log (1.5)2/(.5)3)
.48V - (.0591/6) * 1.26
.48V-.01V= .47V

49. Other types of Electrochem Problems

50. Electrolysis (animations)
Using electrical energy to drive a reaction in a non-spontaneous direction to force reactions to happen.

51. Electrolysis or Plating
Using electrical energy to drive a reaction in a nonspontaneous direction
Used for electroplating, electrolysis of water, separation of a mixture of ions, etc. (Most negative reduction potential is easiest to plate out of solution.)

52. Calculating plating Have to count charge.
Measure current I (in amperes)
1 amp = 1 coulomb of charge per second
q = I x t
q/nF = moles of metal
Mass of plated metal
Faraday Constant (F)
(96,485 C/mol e-) gives the amount of charge (in coulombs that exist in 1 mole of electrons passing through a circuit.
1volt = 1joule/coulomb

53. Calculating plating Current x time = charge
Charge ∕Faraday = mole of e-
Mol of e- to mole of element or compound
Mole to grams of compound
or the reverse these steps if you want to find the time to plate
How many grams of copper are deposited on the cathode of an electrolytic cell if an electric current of 2.00A is run through a solution of CuSO4 for a period of 20min?
How many hours would it take to produce 75.0g of
metallic chromium by the electrolytic reduction of Cr3+ with a current of 2.25 A?

54. How many grams of copper are deposited on the cathode of an electrolytic cell if an electric current of 2.00A is run through a solution of CuSO4 for a period of 20min?
Answer:
Cu2+(aq) + 2e- →Cu(s)
2.00A = 2.00C/s and 20min (60s/min) = 1200s
Coulombs of e- = (2.00C/s)(1200s) = 2400C
mol e- = (2400C)(1mol/96,480C) = .025mol
(.025mol e-)(1mol Cu/2mol e-) = .0125mol Cu
g Cu = (.0125mol Cu)(63.55g/mol) = .79g

55. How many hours would it take to produce 75.0g of
metallic chromium by the electrolytic reduction of Cr3+ with a current of 2.25 A?
Answer:
75.0g Cr/(52.0g/mol) = 1.44mol Cr
mol e- = (1.44mol Cr)(3mol e-/1mol Cr) = 4.32mol e-
Coulombs = (4.32mol e-)(96,480C/mol) = 416,793.6 C (4.17x105C)
Seconds = (4.17x105C)/(2.25C/s) = 1.85x105 s
Hours = (1.85x105 s)(1hr/3600 s) = 51.5 hours

56. AP Chem Practice
An external direct-current power supply is connected to two platinum electrodes immersed in a beaker containing 1.0 M CuSO4(aq) at 25˚C, as shown in the diagram above. As the cell operates, copper metal is deposited onto one electrode and O2(g) is produced at the other electrode. The two reduction half-reactions for the overall reaction that occurs in the cell are shown in the table below.

57. O2(g) + 4 H+(aq) + 4 e-  2 H2O(l)
Half-Reaction
E0(V)
O2(g) + 4 H+(aq) + 4 e-  2 H2O(l)
+1.23
Cu2+(aq) + 2 e-  Cu(s)
+0.34
On the diagram, indicate the direction of electron flow in the wire.
Ans: from the right to the left (this is nonspontaneous)
(b)Write a balanced net ionic equation for the electrolysis reaction that occurs in the cell.
2 Cu2+(aq) + 2 H2O(l)  2 Cu(s) + O2(g) + 4 H+(aq)
(c)Predict the algebraic sign of ∆G˚ for the reaction. Justify your prediction.
+, a non-spontaneous reaction that requires the input of energy to take place
(d)Calculate the value of ∆G˚ for the reaction.
E˚ = +0.34v + (–1.23v) = –0.89v;
∆G˚ = –nE˚ = –(4)(96500)(–0.89) = J = 340 kJ

58. using a 2:1 ratio from equation in part (b), this gives 0.00933 mol O2
An electric current of 1.50 amps passes through the cell for 40.0 minutes.
(e)Calculate the mass, in grams, of the Cu(s) that is deposited on the electrode.
(1.50 amps)(2400 sec) = 3600 coul.;
3600 coul.  = 1.19 g Cu
(f)Calculate the dry volume, in liters measured at 25˚C and 1.16 atm, of the O2(g) that is produced.
1.19 g Cu = mol Cu;
using a 2:1 ratio from equation in part (b), this gives mol O2
V = = L O2

59. Commercial Galvanic Cells
Electrochemistry
Commercial Galvanic Cells
• Galvanic cells can be self-contained and portable and can be used as batteries and fuel cells
A battery (storage cell) is a galvanic cell (or a series of galvanic cells) that contains all the reactants needed to produce electricity.
A fuel cell is a galvanic cell that requires a constant external supply of one or more reactants in order to generate electricity.

60. Electrochemistry • Two basic kinds of batteries
1. Disposable, or primary, batteries in which the electrode reactions are effectively irreversible and which cannot be recharged
2. Rechargeable, or secondary, batteries, which form an insoluble product that adheres to the electrodes; can be recharged by applying an electrical potential in the reverse direction, which temporarily converts a rechargeable battery from a galvanic cell to an electrolytic cell
• Major difference between batteries and galvanic cells is that commercial batteries use solids or pastes rather than solutions as reactants to maximize the electrical output per unit mass

61. The Dry Cell
One example of a dry cell is flashlight and radio batteries.
The cell’s container is made of zinc which acts as an electrode.
A graphite rod is in the center of the cell which acts as the other electrode.
The space between the electrodes is filled with a mixture of:
ammonium chloride, NH4Cl
manganese (IV) oxide, MnO2
zinc chloride, ZnCl2
and a porous inactive solid.

62. Secondary Voltaic Cells
Secondary cells are reversible, rechargeable.
The electrodes in a secondary cell can be regenerated by the addition of electricity.
These cells can be switched from voltaic to electrolytic cells.
One example of a secondary voltaic cell is the lead storage or car battery.
The Lead Storage Battery
In the lead storage battery the electrodes are two sets of lead alloy grids (plates).
Holes in one of the grids are filled with lead (IV) oxide, PbO2.
The other holes are filled with spongy lead.
The electrolyte is dilute sulfuric acid.

63. The Lead Storage Battery
Provides the starting power in automobiles and boats; can be discharged and recharged many times
The Lead Acid Storage Battery is an example of a very successful recylcing program.
The anodes in each cell of this rechargeable battery are plates or grids of lead containing spongy lead metal, while the cathodes are similar grids containing powdered lead dioxide, PbO2
The electrolyte is an aqueous solution of
sulfuric acid
The value of Eº for such a cell is 2 V;
connecting three cells in series produces
a 6-V battery, and a typical 12-V car
battery contains six of these cells
connected in series.

64. The Lead Storage Battery
Diagram of the lead storage battery.

65. Batteries • Lithium-iodine battery Water-free battery
Consists of two cells separated by a metallic nickel mesh that collects charge from the anodes
The anode is lithium metal, and the cathode is a solid complex of 2
Electrolyte is a layer of solid Li that allows Li+ ions to diffuse from the cathode to the anode
Highly reliable and long-lived
Used in cardiac pacemakers, medical implants, smoke alarms, and in computers
Disposable

66. The Nickel-Cadmium (Nicad) Cell
Nicad batteries are the rechargeable cells used in calculators, cameras, watches, etc.
A water-based cell with a cadmium anode and a highly oxidized nickel cathode
This design maximizes the surface area of the electrodes and minimizes the distance between them, which gives the battery both a high discharge current and a high capacity
Lightweight, rechargeable, and high capacity but tend to lose capacity quickly and do not store well; also presents disposal problems because of the toxicity of cadmium

67. Fuel Cells
• A galvanic cell that requires an external supply of reactants because the products of the reaction are continuously removed
• Does not store electrical energy but allows electrical energy to be extracted directly from a chemical reaction
• Have reliability problems and are costly
• Used in space vehicles
Hydrogen is oxidized at the anode.
Oxygen is reduced at the cathode.