1. (Applications of Redox)
Electrochemistry
(Applications of Redox)

2. Unit Essential Questions
What does electrochemistry study? How are cell potentials calculated?

3. Review- Redox Redox = oxidation/reduction reaction.
What occurs during a redox reaction?
Changes in oxidation states; oxidation = losing e-, reduction = gaining e-.
OIL- RIG
Oxidation Involves Loss
Reduction Involves Gain
LEO-GER
Lose Electrons Oxidation
Gain Electrons Reduction

4. Practice
Which of the reactions below exhibits oxidation and which exhibits reduction? Cu+2(aq) + 2e-  Cu(s) Zn(s)  Zn+2(aq) + 2e- What is the reducing agent? What is the oxidizing agent?
reduction
oxidation
Zn(s)
Cu+2

5. Electrochemical Cells
The two half reactions can be combined from the previous example:
Cu+2(aq) + Zn(s)  Cu(s) + Zn+2(aq)
Notice the e- are not shown because they cancelled out on both sides.
The e- are directly transferred from the copper to the zinc.
These e- can be used to do work if they are indirectly transferred between substances!

6. Electrochemical Cells
Two half reactions are separated in two different beakers with a wire connecting them.
Wire allows the current (e-) to travel between beakers.
Flow of e- through the wire can be used to do work.
Electrochemical cells can be used to produce electricity from a redox reaction or can use electricity to produce a redox reaction.

7. Galvanic (Voltaic) Cells
Section 17.1
Galvanic (Voltaic) Cells

8. What is a Galvanic Cell?
A type of electrochemical cell that allows chemical energy to be changed into electrical energy.
Chemical energy comes from change in oxidation states (redox reaction).
Wire used in the galvanic cell allows the electrical energy to be used for work.

9. Cu+2(aq) + Zn(s)  Cu(s) + Zn+2(aq)
Making a Galvanic Cell
Continue with previous example:
Cu+2(aq) + Zn(s)  Cu(s) + Zn+2(aq)
Zinc metal is placed in one beaker and copper metal is placed in the other.
These are the electrodes- the part that conducts e- in the redox reaction.
Zinc sulfate solution is added to the zinc metal and copper (II) sulfate is added to the copper metal.
Called electrode compartments.

10. Electrodes are connected with a wire so the reaction can start
Electrodes are connected with a wire so the reaction can start. But nothing happens! Why?
Charge would build up, and solutions (electrode compartments) must remain neutral!
Negativeb/c lose Cu+2 ions
Positive b/c form Zn+2 ions Zn Cu
Zn+2
SO4-2
Cu+2
SO4-2

11. Salt bridge is added to maintain neutrality.
Galvanic Cell e-
Salt bridge is added to maintain neutrality. e- e-
NO3- K+ Zn Cu
SO4-2
SO4-2
Cu+2
Zn+2

12. Salt Bridge
Any electrolyte can be used to keep charges neutral, as long as the ions don’t interfere with the redox reaction.
KNO3 used in previous example.
K+ ions flow into the copper beaker (to make up for the negative charge).
NO3- ions flow into the zinc beaker (to make up for the positive charge).
Salt bridge allows the circuit to be complete and the redox reaction to occur.

13. Galvanic Cell Components
A voltmeter can be attached to the wire between the electrodes to measure the current that can do work.
Oxidation half reaction is always shown in the left beaker and the reduction half reaction is always shown in the right beaker.
Anode = oxidation half reaction,
Cathode = reduction half reaction
Also have anode and cathode compartments.

14. Notice the e- travel from the anode to the cathode!
Galvanic Cell e-
Notice the e- travel from the anode to the cathode! e- e-
NO3- K+ Zn Cu
SO4-2
SO4-2
Cu+2
Zn+2
anode
cathode

15. Inert Electrodes
Solid electrodes that do not participate in the redox reaction can be used.
These electrodes are only there to complete the circuit/allow the electricity to flow.
They supply/accept e- as needed.
Graphite and platinum are often used.

16. Alternative Galvanic Cell
* A porous disk is used instead of a salt bridge to maintain the circuit/neutrality.
* Porous disk allows ions to flow between solutions.

17. Cell Potential Oxidizing agent ‘pulls’ the e- from reducing agent.
The pull/driving force = cell potential Ecell, or electromotive force (emf).
Unit = volts (V)
1 volt = 1 joule of work/coulomb of charge
Measured with a voltmeter

18. Section 1 Homework
Pg. 830 #15, 25

19. Standard Reduction Potential
Section 17.2
Standard Reduction Potential

20. Standard Reduction Potentials
Show the number of volts produced from a half reaction.
All values are based on the reduction half reaction (thus the name standard reduction potential).
This value is for substances in their standard states: 25°C(298K), 1M, 1atm, and pure solid electrodes.
Standard hydrogen electrode- all other values based off of H!

21. AP Practice Question 2BrO3-(aq) + 12H+(aq) + 10e-  Br2(aq) + H2O (l)
Which of the following statements is correct for the above reaction?
The BrO3- is oxidized at the anode.
Br goes from a -1 to a 0 oxidation state.
Br2 is oxidized at the anode.
The BrO3- is reduced at the cathode.

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

23. H2 in H+ Cl- Zn+2 SO4-2 1 M ZnSO4 1 M HCl 0.76V
Since H = 0, this value is assigned to the oxidation of Zn.
0.76V
H2 in
Anode
Cathode
H+ Cl-
Zn+2 SO4-2
1 M ZnSO4
1 M HCl

24. 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.
E ºcell = E ºZn® Zn+2 + E ºCu+2 ® Cu
Look up reduction potentials in a table.
Since potentials are given in terms of reduction, the sign is switched for the oxidation reaction since the reaction is the reverse of reduction.

25. Cell Potential Zn(s) + Cu+2 (aq) ® Zn+2(aq) + Cu(s)
Reduction potentials are as follows:
Zn+2 + 2e-  Zn E = -0.76V
Cu+2 + 2e-  Cu E = 0.34V
But Zn is not being reduced- it’s being oxidized! So the sign must be changed:
E ºcell = 0.76V V = 1.10V
Alternate formula: E ºcell = E ºcathode - E ºanode
Cell potentials are always > 0 because they run spontaneously in the direction that produces a positive potential.

26. Cell Potential
Note that even if a half reaction must be multiplied by an integer to keep e- transfer equal, the standard reduction potential associated with that half reaction IS NOT multiplied by that same integer!
This is because the standard reduction potential depends upon the reduction that is occurring, not how many times it occurs.
Like how the density of a sample of a substance is always the same, regardless of the size. It only depends on identity!

27. Practice: 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
Eºcell = 0.43V

28. Cell/Line Notation
Shorthand way of representing galvanic cells. Format:
solid½Aqueous½½Aqueous½solid
Anode on the left, cathode on the right.
Single line separates different phases at each electrode.
Double line indicates porous disk or salt bridge.
Concentrations in ( ) may be added after the aqueous species, if known.

29. Cell/Line Notation Examples: (1) Mg(s)½Mg+2(aq)½½Al+3(aq)½Al(s)
(2) Zn(s)½Zn+2(1M)½½Cu+2(1M)½Cu(s)
If an inert electrode is used:
Ag+(aq) + Fe+2(aq)  Fe+3(aq) + Ag(s)
*Pt electrode used for Fe+2 or Fe+3
Pt(s)½Fe+2(aq),Fe+3(aq)½½ Ag+(aq)½Ag(s)

30. Practice: Cell/Line Notation
Given the cell reaction below, write the cell/line notation.
Ni(s) + 2Ag+(aq)  Ni+2(aq) + 2Ag(s)
Ni(s)½Ni+2(aq)½½Ag+ (aq)½Ag

31. Summing up Galvanic Cells
Reaction always runs spontaneously in the direction that produces a positive cell potential (E ºcell > 0).
Four things for a complete description:
Cell Potential
Direction of flow
Designation of anode and cathode
Species present in all components- electrodes and ions.

32. Section 2 Homework
I will do #31, part 25, with you.
Pg. 830 #27, 31

33. Review: Balancing Redox Rxns. Using Half Rxn. Method
Write separate half reactions.
For each half reaction balance all species except H and O.
Balance O by adding H2O to one side.
Balance H by adding H+ to one side.
Balance charge by adding e- to the more positive side.

34. Review: Balancing Redox Rxns. Using Half Rxn. Method
Multiply equations by a number to make electrons equal.
Add equations together and cancel identical species. Reduce coefficients to smallest whole numbers.
Check that charges and elements are balanced.

35. In Basic Solution Add enough OH- to both sides to neutralize the H+.
Any H+ and OH- on the same side form water. Cancel out any H2O’s on both sides.
Simplify coefficients, if necessary.

36. MnO4-(aq) + H+(aq) + C2O4-2(aq)  Mn+2(aq) + H2O(l) + CO2(g)
AP Practice Question
What is the coefficient of H+ when the following reaction is balanced?
MnO4-(aq) + H+(aq) + C2O4-2(aq)  Mn+2(aq) + H2O(l) + CO2(g) 16 2 8 5

37. Homework Balance the following equation in acidic and basic solution:
NO3- + Mn  NO + Mn+2
Balanced equations:
3Mn + 8H+ + 2NO3-  2NO + 4H2O + 3Mn+2
3Mn + 4H2O + 2NO3-  2NO + 8OH- + 3Mn+2

38. Cell Potential, Electrical Work, and Free Energy
Section 17.3
Cell Potential, Electrical Work, and Free Energy

39. Linking Cell Potential to ΔG
Cell potential is directly related to the difference in free energy between reactants and products.
This is shown in the following equation:
ΔG° = -nFE
n = moles of e- in redox/half reaction
F = Faraday’s constant = 96,485C/mole-
E = cell potential (V = J/C)
Units of G = J
Coulomb = unit of electric charge

40. Linking Cell Potential to ΔG
ΔG° = -nFE
Why negative?
Galvanic cells are spontaneous!
If ΔG° was positive, the redox reaction would be nonspontaneous.
ΔG° can be calculated for half reactions and for entire redox reactions.

41. Cu+2(aq) + Fe(s)  Cu(s) + Fe+2(aq)
Sample Calculation
Calculate ΔG° for the reaction:
Cu+2(aq) + Fe(s)  Cu(s) + Fe+2(aq)
Look up half reactions and determine individual E values. Then calculate E for the cell.
E = 0.78V = 0.78J/C
Make sure e- are correct!
ΔG° = -(2mole-)(96,485C)(0.78J) = -1.5x105J
Process is spontaneous.
mole C

42. Mn+2(aq) + IO4-(aq)  IO3-(aq) + MnO4-(aq)
Practice Problem
Consider the following reaction:
Mn+2(aq) + IO4-(aq)  IO3-(aq) + MnO4-(aq)
Calculate the value of E cell and ΔG°. Be sure to balance the e- (you can use the half reactions and balance them from there).
E cell = 0.09V
ΔG° = -90kJ

43. Section 3 Homework
Pg. 831 #: 37

44. Dependence of Cell Potential on Concentration
Section 17.4
Dependence of Cell Potential on Concentration

45. Qualitative Understanding
The following reaction is under standard conditions:
Cu(s) + 2Ce+4(aq)  Cu+2(aq) + 2Ce+3(aq)
What if [Ce+4] was greater than 1.0M?
Use LeChatelier’s principle!
This shifts the rxn. forward, which means more products are formed and therefore more e- are flowing/transferring to allow them to form.
Cell potential increases.

46. Another Way of Looking at It
Cu(s) + 2Ce+4(aq)  Cu+2(aq) + 2Ce+3(aq)
What if [Ce+4] was greater than 1.0M?
Use LeChatelier’s principle!
The rxn. shifting more towards the products means the rxn. is more spontaneous, so –ΔG increases.
Recall that –ΔG = -nFE°cell, so if -ΔG is increasing, so must E°cell because n and F are constant for this rxn.

47. 2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
Practice
For the cell reaction:
2Al(s) + 3Mn+2(aq)  2Al+3(aq) + 3Mn(s)
E°cell = 0.48V
Predict whether E°cell is larger or smaller than E°cell for the following cases.
[Al+3] = 2.0M, [Mn+2] = 1.0M
[Al+3] = 1.0M, [Mn+2] = 3.0M
smaller
larger

48. Quantitative Understanding
We will not cover the quantitative side of how cell potential changes with concentration, but I want to mention the Nernst Equation:
Ecell = E°cell –(RT)lnQ = E°cell –(0.0592)logQ
Lets you calculate cell potential under non-standard conditions.
Not tested on the AP exam. nF n

49. Homework
Pg. 832 # 51

50. Section 17.7
Electrolysis

51. Electrolytic Cells
Electrolytic cells use electrolysis and are the opposite of galvanic cells.
Reactions are not spontaneous.
Electrolysis: a voltage bigger than the cell potential is applied to the cell to force the non spontaneous redox reaction to occur.
Can be used to decompose compounds.
Water can be broken into hydrogen & oxygen.
Can also be used for (electro)plating.
Forces metal ions in solution to plate out on electrodes in their solid form.

52. 1.10 e- e- Zn Cu 1.0 M Cu+2 1.0 M Zn+2 Anode Cathode Galvanic Cell
Cell potential = 1.10V
1.10 e- e- Zn Cu
1.0 M Zn+2
1.0 M Cu+2
Anode
Cathode

53. A battery >1.10V e- e- Zn Cu 1.0 M Zn+2 1.0 M Cu+2 Cathode Anode
Voltage
applied > 1.10V
Electrolytic Cell
A battery >1.10V e- e-   Zn Cu
1.0 M Zn+2
1.0 M Cu+2
Cathode
Anode

54. Electrolytic Cells
Relationship exists between current, charge, and time: I = q/t
I = current (A, amp)
q = charge (C)
t = time (s)
This formula can be used in conjunction with other conversion factors to solve for various items with respect to electrolytic cells.

55. Electrolytic Cells & Stoichiometry
The previous formula and stoichiometry can be used to determine the amount of chemical change that occurs when current is applied for a certain amount of time.
Answers the following questions:
How much of a substance will be produced?
How long will it take?
How much current is needed?

56. Electrolytic Cells & Stoichiometry
No set way to solve these problems each time! In addition to the formula, the following conversion factors may be used:
96,485C/mol e- (this is 1 Faraday)
Molar mass
Moles of electrons to moles of other species involved in the redox rxn.
Make sure moles in the half rxn. are correct!

57. Example
If liquid titanium (IV) chloride (acidified with HCl) is electrolyzed by a current of 1.000amp for 2.000h, how many grams of titanium will be produced?
Have: amps & time  can use both to find q from previously discussed formula:
I = q/t  1.000A = q/2.000h  convert h to s
1.000A = q/7,200.s
q = 7,200.As  q = 7,200.C  s  q = 7,200.C
*Now use stoichiometry to solve for grams! s

58. Example Cont.
If liquid titanium (IV) chloride (acidified with HCl) is electrolyzed by a current of 1.000amp for 2.000h, how many grams of titanium will be produced?
Use Faraday’s constant to get moles of e-, write balanced half rxn. to get moles of e- and moles of Ti: Ti+4  Ti + 4e-
7,200.C x 1mole- x 1molTi x 47.88gTi
96,485C
4mole-
1molTi
= g

59. Practice Problem
What mass of copper is plated out when a current of 10.0amps is passed for 30.0min through a solution containing Cu+2?
Solve for q: 10.0A = q/1,800s  q = 18,000C
18,000C (1mole-)(1molCu)(63.55gCu)
96,485C 2mole- 1molCu
=5.93gCu

60. Homework
In class- 77(a)
Given mass and I; from mass we can find charge, q. Then use formula to solve for time.
Pg. 834 #77(b&c),79(a&b)