1. Applications of Redox Your last chapter! I know, …… kinda sad.
Electrochemistry
Applications of Redox
Your last chapter!
I know, …… kinda sad.

2. Redox Review
Oxidation reduction reactions involve a transfer of electrons.
OIL- RIG
Oxidation Involves Loss
Reduction Involves Gain
LEO-GER
Lose Electrons Oxidation
Gain Electrons Reduction

3. Applications Moving electrons is electric current.
8H++MnO4-+ 5Fe+2 +5e  Mn+2 + 5Fe+3 +4H2O
Helps to break the reactions into half reactions.
8H++MnO4-+5e-  Mn+2 +4H2O
5(Fe+2  Fe+3 + e- )
In the same mixture it happens without doing useful work. If separate, current can flow.

4. Connected this way the reaction starts
Stops immediately because charge builds up.
Fe2+
MnO4- H+

5. Galvanic Cell
Salt Bridge allows current to flow H+
MnO4-
Fe+2

6. Electricity travels in a complete circuit
Instead of a salt bridge e- H+
MnO4-
Fe+2

7. Porous Disk H+
MnO4-
Fe+2

8. e- e- e- e-
Anode
Cathode e- e-
Oxidation - supplies electrons
Reduction - uses electrons

9. 17.2 Cell Potential Oxidizing agent pulls the electron.
Reducing agent pushes the electron.
The push or pull (“driving force”) is called the cell potential Ecell
Also called the electromotive force (emf)
Unit is the volt(V) = 1 joule of work/coulomb of charge
Measured with a voltmeter (BDVD)half-reactions

11. 0.76
H2 in
Cathode
Anode
H+ Cl-
Zn+2 SO4-2
1 M ZnSO4
1 M HCl Pt

12. Standard Hydrogen Electrode
This is the reference all other oxidations are compared to
Eº = 0
º indicates standard states of 25ºC, 1 atm, 1 M solutions.
H2 in
H+ Cl-
1 M HCl

13. Cell Potential Zn(s) + Cu+2 (aq)  Zn+2(aq) + Cu(s)
The total cell potential is the sum of the potential at each electrode.
Eº cell = EºZn Zn+2 + Eº Cu+2  Cu
We can look up reduction potentials in a table.
One of the reactions must be reversed, so change its sign. Overall cell must be +
Potentials are not multiplies by the coefficients in the balanced equation. (BDVD) Galvanic cells

15. Cell Potential
Determine the cell potential for a galvanic cell based on the redox reaction.
Cu(s) + Fe+3(aq)  Cu+2(aq) + Fe+2(aq)
Fe+3(aq) + e-  Fe+2(aq) Eº = 0.77 V
Cu+2(aq)+2e-  Cu(s) Eº = 0.34 V
Cu(s)  Cu+2(aq)+2e Eº = V
2Fe+3(aq) + 2e-  2Fe+2(aq) Eº = 0.77 V

16. Cell Potential A galvanic cell based on the reaction:
Al3+(aq) + Mg(s) Al(s) + Mg2+(aq)
Give the balanced cell reaction and calculate E° for the cell. Ex.17.1
MnO4-(aq) + H+(aq) + ClO3-(aq) ClO4-(aq) + Mn2+(aq) + H2O (l)
(ClO4- + 2H+ + 2e- --> ClO3- + H2O = 1.19V)

17. Line Notation solidAqueousAqueoussolid
Anode on the leftCathode on the right
Single line different phases.
Double line porous disk or salt bridge.
If all the substances on one side are aqueous, a platinum electrode is indicated.
For the last reaction
Cu(s)Cu+2(aq)Fe+2(aq),Fe+3(aq)Pt(s)

18. Galvanic Cell Description
The reaction always runs spontaneously in the direction that produced a positive cell potential.
Four things for a complete description.
Cell Potential.
Direction of flow.
Designation of anode and cathode.
Nature of all the components- electrodes and ions.

20. Practice
Completely describe (sketch) the galvanic cell based on the following half-reactions under standard conditions. Ex.17.2
Ag+(aq) + e-  Ag(s)
Fe+3(aq) + e-  Fe2+(aq)
Describe the cell with this reaction:
IO3-(aq) + Fe+2(aq) Fe3+(aq) + I2(aq) #26a

21. Practice
Completely describe (sketch) the galvanic cell based on the following half-reactions under standard conditions. #30
H2O2 + 2H+ + 2e- 2H2O 1.78 V
O2 + 2H+ + 2e- H2O V
Give the standard line notation for the previous problems.

23. 17.3 Potential, Work and G emf = potential (V) = work (J)/Charge(C)
E = work done by system / charge
E = -w/q
Charge is measured in coulombs.
-w = qE (- sign means work is leaving the system)
Faraday (F) = 96,485 C/mol e-
q = nF = moles of e- x charge/mole e-
w = -qE = -nFE = G

24. Potential, Work and G Gº = -nFE º
if E º < 0, then Gº > 0 nonspontaneous
if E º > 0, then Gº < 0 spontaneous
Calculate Gº for the following reaction:
Cu+2(aq)+ Fe(s)  Cu(s) + Fe+2(aq)
Fe+2(aq) + 2e- Fe Eº = V
Cu+2(aq)+2e-  Cu(s) Eº = 0.34 V Ex.17.3

26. Practice
The amount of manganese in steel is determined by changing it to permanganate ion. The steel is first dissolved in nitric acid, producing Mn2+ ions. These ions are then oxidized to the deeply colored MnO4- ions by periodate ion (IO4-) in acid solution. #38
Complete and balance an equation describing each of the above reactions.
Calculate E° and G° at 25°C for each reaction.

27. Practice
The G° formula can also be used for half-reactions. Use standard reduction potentials to estimate G°f for Fe2+(aq) and Fe3+(aq). (G°f for e- = 0.) #42
Using data from Table 17.1 place the following in order of increasing strength as reducing agents. #44
Cr3+, H2, Zn, Li, F-, Fe2+

29. 17.4 Cell Potential and Concentration
Qualitatively - We can predict direction of change in E from LeChâtelier. Shift to the left reduces potential. To the right increases potential.
2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
Predict if Ecell will be greater or less than Eºcell if [Al+3] = 2.0 M and [Mn+2] = 1.0 M Ex. 17.5
if [Al+3] = 1.0 M and [Mn+2] = 3.0M
if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

30. Concentration Cells
A potential can be generated with the same components but in different concentrations.
The reaction proceeds in the direction that will equalize the ion concentration in the components.
These voltages are typically small.
By using the Nernst Equation, we can calculate the potential of a cell in which some or all of the components are not in their standard states.

31. Concentration Cells
This cell has a driving force that transfers electrons from left to right. Silver will be deposited on the right electrode, thus lowering the concentration of Ag+ in the right compartment. In the left, the silver electrode dissolves (producing Ag+ ions) to raise the concentration of Ag+ in solution.

32. Practice
Consider this cell. Predict the direction of electron flow, and designate the anode and cathode. If the Fe2+ concentration in the right compartment is changed from 0.1 M to 1 x 10-7 M Fe2+, what would change. #52

33. The Nernst Equation G = Gº + RTln(Q) -nFE = -nFEº + RTln(Q)
E = Eº - RTln(Q) nF
At 25°C, We use
Always have to figure out n by balancing.
If concentration can give voltage, then from the voltage we can tell the concentration.

34. Example
2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
Eº = 0.48 V under standard conditions.
Consider a cell [Mn2+] = 0.50 M and [Al3+] = 1.50 M
Q= [Al3+]2 = (1.50)2 = 18
[Mn2+]3 (0.50)3
n = 6 from the balanced equation = 6e-
Using Nernst Ecell = log(18) 6
= = 0.47V (p.844)

36. The Nernst Equation
As reactions proceed concentrations of products increase and reactants decrease.
Reach equilibrium where Q = K and Ecell = 0, this is a “dead” cell.
At equilibrium, the components in the two cell compartments have the same free energy and G = 0
0 = Eº log(K) n
log(K) = nEº

37. Practice For the redox reaction: log(K) = nEº 0.0591
S4O62-(aq) + Cr2+(aq)  Cr3+(aq) + S2O32-(aq)
Half reactions are:
S4O e- S2O32- E°= 0.17 V
Cr3+ + e-  Cr2+ E°= V
Balance the redox reaction, and calculate E° and K at 25°C Ex. 17.8
log(K) = nEº

38. Practice The overall reaction in a lead storage battery is
Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq)  2PbSO4(s) + 2H2O(l)
Calculate E at 25°C for this battery when [H2SO4] = 4.5 M. At 25°C, E° = 2.04V for the lead storage battery. #56
Al|Al3+(1.00 M)||Pb2+(1.00M)|Pb
In the above cell, calculate the cell potential after the reaction has operated long enough for the [Al3+] to have changed by 0.60 mol/L. #58

39. 17.5 Batteries are Galvanic Cells
A battery is a galvanic cell or, more commonly, a group of galvanic cells connected in series.
Car batteries are lead storage batteries.
Pb + PbO2 + H2SO4 PbSO4(s) + H2O
Dry Cell Zn + NH4+ + MnO2  Zn+2 + NH3 + H2O
Alkaline Zn + MnO2  ZnO + Mn2O3 (in base)
NiCad
NiO2 + Cd + 2H2O  Cd(OH)2 + Ni(OH)2

40. 17.6 Corrosion Corrosion (rusting) - spontaneous oxidation.
Many metals develop a metal oxide coating that prevents further oxidation.
Most structural metals have reduction potentials that are less positive than O2 .
Fe  Fe+2 +2e- Eº= 0.44 V
O2 + 2H2O + 4e-  4OH- Eº= 0.40 V
Fe+2 + O2 + H2O  Fe2 O3 + H+
Reaction happens in two places.

41. Salt speeds up process by increasing conductivity
Water
Rust
Anodic area
Cathodic area e-
Iron Dissolves - Fe  Fe+2

42. Preventing Corrosion Coating to keep out air and water.
Galvanizing - Putting on a zinc coat.
Has a lower reduction potential, so it is more easily oxidized.
Alloying with metals that form oxide coats.
Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

43. 17.7 Electrolysis
Forcing a current through a cell to produce a chemical change for which the cell potential is negative.
Running a galvanic cell backwards.
Put a voltage bigger than the potential and reverse the direction of the redox reaction.
Used for electroplating.

44. 1.10 e- e- Zn Cu
1.0 M Zn+2
1.0 M Cu+2
Anode
Cathode

45. A battery >1.10V e- e- Zn Cu
1.0 M Zn+2
1.0 M Cu+2
Cathode
Anode

46. Stoichiometry of Electrolysis
How much chemical change occurs with the flow of a given current for a specified time?
Measure current I (in amperes)
1 amp = 1 coulomb of charge per second
1 Faraday (F) = 96,485 C/mol e-
Problems are essentially a unit conversion.
current and time  quantity of charge moles of electrons  moles of substance grams of substance
How long must 5.00 amp current be applied to produce 10.5 g of Ag from Ag+? Ex.17.9

47. Practice
How long would it take to produce 10.0 g of Bi by the electrolysis of a BiO+ solution using a current of 25.0A? #74
Aluminum is produced commercially by the electrolysis of Al2O3 in the presence of a molten salt. If a plant has a continuous capacity of 1.00 million amp, what mass of aluminum can be produced in 2.00 hours? #76
Electrolysis of an alkaline earth metal chloride using a current of 5.00A for 748 seconds deposit g of metal at the cathode. What is the identity of the alkaline earth metal? #78

48. Other uses Electrolysis of water. Separating mixtures of ions.
More positive reduction potential means the reaction proceeds forward.
A solution of Ag+, Cu2+, and Zn2+ were introduced to a low voltage that was slowly turned up. The most positive reduction potential is easiest to plate out of solution. Therefore Ag+> Cu2+ >Zn2+