1. Electrochemistry
Chapter 20

2. Electrochemistry = the study of relationships between electricity and chemical reactions
Topics Include:
Batteries
Spontaneity of reacation
Corrosion of metals
Electroplating

3. Redox Reactions (Chapter 4.4)
REDOX reactions always involve a transfer of electrons
Oxidation = loss of electrons
Reduction = gain of electrons
2 ways to remember:
LEO says GER
Loss Electrons Oxidation; Gain Electrons Reduction
OIL RIG
Oxidation Involves Loss; Reduction Involves Gain

4. Redox Terminology
Oxidation Number  a positive or negative value assigned to each atom in a compound or ion for the purpose of keeping track of electrons
Oxidizing Agent  the substance that makes it possible for another substance to be oxidized
Oxidizing agents remove electrons. They pick them up and become reduced (become more negative)

5. Reducing Agent  the substance that gives up electrons
Reducing agents lose electrons, become oxidized, and become less negative/more positive

6. Rules for Assigning Ox. #’s
Summary
1. Ox. # of the atom of a free element is Zero
Element = 0
2. Ox # of a monatomic ion equals its charge
Calcium = +2
Sulfide = -2
3. In compounds, Oxygen is always (-2), except in peroxides (-1)
H2O; O=-2
H2O2; O = -1
4. In compounds, Hydrogen is always +1
Hydrogen = +1
5. In compounds, Fluorine is ALWAYS -1
Fluorine = -1
6. Sum of the oxidation states for an electrically neutral compound must be zero

7. Helpful Number Line +4 +3 +2 +1 oxidation 0 reduction 1 2 3 4
Reducing
Agent
Oxidizing
Agent

8. Example
The nickel-cadmium battery, a rechargeable “dry cell” used in battery-operated devices, uses the following redox reaction to generate electricity:
Cd(s) + NiO2(s) + H2O(l) Cd(OH)2(s) + Ni(OH)2(s)
Assign all the oxidation numbers
Identify the substances that are oxidized and reduced.
Identify which are oxidizing agents and which are reducing agents.

9. Redox Every redox reaction has both oxidation and reduction occurring.
What is being oxidized? reduced ?
Cd(s) + NiO2(s) + 2H2O(l) --> Cd(OH)2(s) + Ni(OH)2(s)
   2+1
Identify the oxidation numbers
Identify the substances which change in oxidation no.
Cd ---> Cd e Ni e --> Ni
Cd is oxidized Nickel is reduced
Ni is the oxidizing agent Cd is the reducing agent

10. Practice
Label the oxidation numbers for all elements on both sides of the reaction.
Determine the Oxidizing Agent and the Reducing Agent in each reaction.
Zn(s) + 2H+(aq)  Zn+2(aq) + H2(g)
2H2 (g) + O2 (g)  2H2O(l)

11. Balancing Redox Redox reactions are sometimes difficult to balance.
Must balance both atoms of each element AND charge.
We only show the substances involved in the reaction, and add H, OH and/or H2O
Requires following a series of steps and practice!

12. Half Reactions
Although REDOX takes place simultaneously, it is easier to look at it as two separate processes.
For Example:
Sn+2 + 2Fe+3  Sn+4 + 2Fe+2
Oxidation: Sn+2  Sn e-
Reduction: 2Fe e-  2Fe+2
Notice: In oxidation, electrons are products, while in reduction electrons are reactants

13. Balancing using Half Reactions
MnO4- + C2O4-2  Mn+2 + CO2
Lets look at the half reactions:
MnO4-  Mn+2
C2O4-2  CO2
First add coefficients to balance out atoms (this may need to be done in an acidic or basic solution)
Note: if the process is done in acidic solution, H+ and H2O can be added; in basic solution, OH - and H2O can be added.

14. MnO4- + C2O4-2  Mn+2 + CO2 MnO4-  Mn+2 + H2O (added water)
H+ + MnO4-  Mn+2 + H2O (added H)
Now we can balance:
8H+ + MnO4-  Mn+2 + 4H2O
Now equal types of atoms on both sides, but charges aren’t equal, so need to add electrons:
5e- + 8H+ + MnO4-  Mn+2 + 4H2O
Now to balance the oxalate ½ reaction…

15. C2O4-2  CO2 (add coefficients to balance)
C2O4-2  2CO2(add electrons to balance charge)
C2O4-2  2CO2 + 2e-
Now we have to put both balanced half reactions together to get the overall balanced equation:
5e- + 8H+ + MnO4-  Mn+2 + 4H2O
We need the same number of electrons to appear on both sides of the reaction so they will cancel out.

16. 10e- + 16H+ + 2MnO4-  2Mn+2 + 8H2O
5C2O4-2  10CO2 + 10e-
16H+ + 2MnO4- + 5C2O4-2 
2Mn+2 + 8H2O + 10CO2
The best way to do this is through practice and we will do various examples in class and for HW

17. AP Type question
For the following: 1) write a balanced equation and 2) answer the question about the reaction.
1)Calcium metal is heated strongly in nitrogen gas. 2) What is the change in oxidation number of the nitrogen in this RXN?
1)Acidified potassium dichromate solution is mixed with potassium iodide solution. 2) What is the reducing agent in the above reaction?
*Hint: Water is one of the products for the above reaction.

18. Redox Reactions - the reality: electricity
Moving electrons is electric current.
A voltaic cell (or galvanic cell) is a device which allows the transfer of e to take place.
Voltaic cells produce electric current.
A battery is a voltaic cell.
Batteries contain chemicals which undergo redox reactions.

19. Features of a Voltaic Cell
Two different substances connected by a conductor in 2 separate compartments containing an electrolytic solution which is allowed to flow between the 2 compartments
When assembled: a complete circuit forms which allows electrons to flow in a one-way direction. Direct Current (DC)

20. The one-way direction is the direction of the spontaneous reaction.
Each substance is called an electrode.
Anode - electrode where oxidation occurs
Cathode - electrode where reduction occurs
Electrons always flow from anode to the cathode.

21. Both metals in same mixture?
Metal ions of one metal is deposited on another
Reaction happens without doing useful work
Zn Cu
Cu+2 H+
Zn+2 SO4-2
but if separate, the electricity is available to do work

22.  Connected this way the reaction starts, but
Stops immediately because charge builds up. For a cell to work, the solutions in each half cell must remain electrically neutral. 
2e 
2e 
2e 
Zn (s) --> Zn+2 + 2e
Cu e --> Cu(s) Zn Cu
Cu+2
NO3-
Zn+2
NO3-
A half cell

23. A bridge is needed Salt Bridge allows current to flow Zn+2 Cu+2 NO3-

24. Now, Electricity travels in a complete circuit
Zn (s) --> Zn+2 + 2e
e 
Cu e --> Cu(s) Zn Cu
Zn+2
NO3-
Cu+2
NO3-

25. Instead of a salt bridge
Porous Disk or Barrier works too
Zn+2
NO3-
Cu+2
NO3-

26. e- e- e- e- e- e- Reducing Agent Zn Oxidizing Agent Cu+2  +
Reduction occurs at the cathode, Oxidation at the anode e- e- e- e-
e are available to do work
Zn (s) --> Zn+2 + 2e
Cu e --> Cu(s)
Anode (Zn)
Is oxidized
Cathode (Cu)
Is reduced
Porous
barrier e- e-  +
Reducing Agent Zn
Oxidizing Agent Cu+2

27. The Complete Picture Zn (s) + Cu+2 --> Cu(s) + Zn+2
Red. at cathode: Cu e --> Cu(s)
Oxid. at anode: Zn (s) --> Zn+2 + 2e
2e
2e
2e
salt bridge
NaNO3 Zn Cu
Anode 
Cathode 
NO3 Cu+2
SO4-2 Na+
NO3 Zn+2
SO Na+
CuSO4
Solution
(electrolyte)
ZnSO4
Solution
All + ions flow this way
All  ions flow this way

28. While Operating . . .
The Zn electrode is oxidized and loses mass, while [Zn+2] increases in solution.
The Cu electrode is reduced, gains mass, while the [Cu+2] decreases in solution.
Salt bridge allows ions to flow in both directions. ions move toward the anode
+ions move toward the cathode
Salt bridge contains an electrolyte that doesn’t react with anything in the cell.
Salt bridge connects the 2 cells keeping + and  ions neutral.
The actual charge on an electrode is zero.

29. Cr2O7-2 + 14H+ + 6I- ---> 2Cr+3 +3I2 + 7H2O
For the above reaction, a simple voltaic cell is assembled.
Which is the anode? Which is the cathode?
I is the anode ; Cr2O7-2 is the cathode
What reaction occurs at the anode?
6 I ---> 3I2 + 6e
What reaction occurs at the cathode?
Cr2O H e ---> 2Cr H2O
What is the direction of electron flow?
Electrons flow from the anode, I , to the cathode, Cr2O7-2
What is the sign at each electrode?
I Anode is () and Cr2O7-2, cathode is ()

30. Cell Potential Ecell
Electrons flow from anode to cathode because there is a potential energy difference between them.
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

31. The magnitude of Ecell depends upon:
Specific reactions at each electrode
Concentrations of reactants / products
Temperature
The value of Ecell
Is what the voltmeter reads.
Is the difference between the 2 half cell potentials (which we look up in tables)
Eºcell = Eºcathode  Eºanode

32. 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.
Platinum wire
H2 in
H+ Cl-
1 M HCl
Platinum
electrode

33. H2 in Cathode Anode H+ Cl- Zn+2 SO4-2 1 M ZnSO4 1 M HCl 0.76
For example
Zn ---> Zn+2 + 2e-
0.76
H2 in
Cathode
Anode
H+ Cl-
Zn+2 SO4-2
1 M ZnSO4
1 M HCl

34. Cell Potential
The total cell potential is the sum of the potential at each electrode.
Eºcell = Eºcathode + Eºanode
We can look up reduction potentials in the table. Tables are for reduction potential. The oxidation reaction must be reversed, so we change its sign.
Since we always change the sign of the oxidation reaction, the formula is expressed as:
Eºcell = Eºreduction  Eºoxidation OR
Eºcell = Eºcathode  Eºanode

35. Calculating Ecell Co(s) --> Co+2 (aq) + 2e Ered = 0.277 V
AgCl(s) + e --> Ag(s) + Cl Ered = V
What reaction occurs at the anode?
What reaction occurs at the cathode?
What is the standard cell potential, Ecell?
Oxidation Co(s) --> Co+2 (aq) + 2e
Reduction AgCl(s) + e --> Ag(s) + Cl
Eºcell = Eºcathode  Eºanode
Eºcell =  (0.277) = V
Note: multiplying a reaction by a coefficient doesn’t effect the Value of Eo.

36. Cell Potential
Determine the cell potential for a voltaic cell based on the redox reaction:
Cu(s) + Fe+3(aq) ® Cu+2(aq) + Fe+2(aq)
Fe+3(aq) + e-® Fe+2(aq) Eºred = V
Cu+2(aq)+2e- ® Cu(s) Eºred = V
Eºcell = Eºcathode  Eºanode
Eºcell = 0.77  (+0.34) = V
Remember, to operate a cell must
have a (+) cell potential.

37. In Summary
A voltaic cell operates on a driving force called the cell potential, Ecell or EMF.
An operating voltaic cell must have a + Ecell value.
In any voltaic cell, the reaction at the cathode (reduction occurs) has a more + E°red value than does the reaction at the anode (oxidation occurs).
The strongest oxidizing agents (they gain electrons) have the most + E°red values.
Standard Reduction Potential tables can be used to mix and match the strongest voltaic cells possible.

38. Spontaneity of Redox Reactions
A complete circuit always flows in a one way direction.
A Direct Current (DC) results.
The one-way direction is the direction of the spontaneous reaction.
To be spontaneous, Ecell must be (+)
If Ecell is (-), it will not work!!
The stronger oxidizing agent has the greater “driving force” and forces the reducing agent to run in reverse as an oxidation reaction

39. Activity Series and Eºred
“The ions of the metal below will oxidize the solid metal above.” The activity series is experimentally determined with the strongest reducing agents listed at the top.
The activity series is really a list of oxidation potentials - the flipped up-side-down list of our standard reduction potentials. The Ered table has the strongest oxidizing agents listed at the top.
We now know why certain metal ions react with other metals: potential energy exists as a driving force, + Ecell
And, why certain metals don’t react,  Ecell

40. Working Voltaic Cells-Why ?
Recall that ∆G, Gibbs Free Energy determines spontaneity.
∆G = spontaneous reaction
+ Ecell = spontaneous redox reaction
Michael Faraday determined the relationship between Free Energy and cell potential:
∆G = nFE
where n = no. of electrons transferred (always (+) )
F = Faraday’s constant: 96,500 J/V*mol e-
E = Ecell or emf

41. DG makes the cell operational
DGº = -nFEº
if E º = +, then DGº 
spontaneous
if E º = , then DGº 
nonspontaneous, the reverse is spontaneous.

42. Practice Problem DGº = -nFEº DGº = -(2)(96,500J/V•mol)(0.78V)
Calculate DGº for the following reaction:
Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq)
Fe+2(aq) + 2e-® Fe(s) Eº = 0.44 V
Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V
Eºcell = Eºcathode  Eºanode
Eºcell =  (0.44)
Eºcell = V
DGº = -nFEº
DGº = -(2)(96,500J/V•mol)(0.78V)
DGº = 150,540J/mol = 150.4 kJ/mol

43. Practice Problem DGº = -nFEº DGº = -(6)(96,500J/V•mol)(-0.12V)
3Ni Cr(OH)3 (s) + 10 OH ---->
3 Ni (s) + 2CrO42 + 8 H2O
Calculate ∆G for this reaction.
3Ni e ----> 3 Ni (s) E = 0.25V
CrO42 + 4H2O + 3e ---> Cr(OH)3 + 5 OH
E = 0.13V
Eºcell = 0.25  (0.13) = -0.12V)
DGº = -nFEº
DGº = -(6)(96,500J/V•mol)(-0.12V)
DGº = 69,480 J/mol  69 kJ/mol

44. Why don’t batteries last forever?
As a voltaic operates, reactants are consumed and products are generated.
As the conc. of reactants decreases, so does the Ecell.
At some point the battery goes “dead” and the Ecell = 0
Walther Nernst determined the relationship between the conc. and EMF

45. The Nernst Equation E = Eº  RT lnQ nF
DG = DGº + R T lnQ (standard free energy change)
-nFE = -nFEº + RTlnQ (substituting DGº = -nFEº)
E = Eº  RT lnQ nF
At standard conditions, RT/F = V
So, E = Eº  V log Q n
Q is the reaction quotient, [prod]/[react]
Remember, we leave solids out of the equilibrium expression

46. The Nernst Equation
Remember, always determine n by balancing the redox reaction.
What it means:
Equation relates concentrations to the emf (voltage).
If concentration can give us voltage,
then from voltage we can tell
concentration.

47. Equilibrium and Cell emf
When E =0, the cell reaction has reached equilibrium and no net reaction is occurring in the voltaic cell.
Recall that at equilibrium, ∆G = 0 and
Q = Keq
Substituting, simplifying and rearranging the Nernst Equation gives us:
nE
log Keq = (at 25C)
0.0592
Thus, the equilibrium constant for a redox reaction can be determined from the E value.

48. Equilibrium and Cell emf
As a voltaic operates, reactants are consumed and products are generated. When “dead”, the cell at E =0 has reached equilibrium, ∆G =0.
Where’s the equilibrium?
Qualitatively - we can predict direction of change in E from LeChâtelier.
E = Eº  V log Q n

49. E = Eº  0.0592 V log Q E = Eº  0.0592 V log [Al+3]2 Practice Problemn
2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)
Predict if Ecell will be greater or less than Eºcell
if [Al+3] = 1.5 M and [Mn+2] = 1.0 M
E = Eº  V log [Al+3]2
n [Mn+2]3
if [Al+3] = 1.0 M and [Mn+2] = 1.5M
if [Al+3] = 1.5 M and [Mn+2] = 1.5 M
Less than Ecell
Greater
greater

50. Practice Problem
Calculate the emf at 298K generated by a cell for the equation below when [Cr2O7-2]=2.OM, [H+]=1.0M, and [Cr+3]=1.0x10-5M.
Cr2O7-2(aq) + 14H+(aq)+ 6I-(aq)  Cr+3 (aq) + 3I2(s) + 7H2O(l)
Solve for Q
Q = 5.0 x 10-11
Figure out n
n = 6
Plug into Nernst Equation and get 0.89 V

51. Batteries are Voltaic Cells
Car batteries are lead storage batteries.
Pb +PbO2 +H2SO4 ®PbSO4(s) +H2O
Dry Cell
Zn + NH4+ +MnO2 ® Zn+2 + NH3 + H2O+ MnO(OH) (s)
Alkaline
Zn +MnO2 ® ZnO+ Mn2O3 (in base)
NiCad
NiO(OH) s + Cd + 2H2O ® Cd(OH)2 +Ni(OH)2

52. Corrosion Rusting - spontaneous oxidation.
Most structural metals have reduction potentials that are less positive than O2 .
Fe+2 +2e- ® Fe Eº= 0.44 V
O2 + 2H2O + 4e- ® 4OH- Eº= 0.40 V
Fe+2 + O2 + H2O ® Fe2 O3 + H+
Reaction happens in two places.

53. Salt speeds up process by increasing conductivity
Water O2
Fe+2
Rust
Fe2O3
Fe+2
Fe+2 e-
Iron Dissolves in Water: Fe ® Fe+2

54. 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.

55. Electrolysis Running a voltaic cell backwards.
Put a voltage bigger than the potential and reverse the direction of the redox reaction.
Used for electroplating.

56. 1.10 e- e- Zn Cu
1.0 M Zn+2
1.0 M Cu+2
Anode
Cathode

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

58. Calculations for 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
How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag+?

59. q = I x t and q t = I = 5.00 amp I q = moles x Fn
How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag+?
q = I x t and q
t = I = 5.00 amp I
q = moles x Fn
q = mol(96,500) (1)
q = 13,862 C
13,862 C
t = = 2772 sec = 46.2 min.
5.00 amp