1. OAKTON
Community College
Electrochemistry

2. Redox: Transfer of Electrons
General College Chemistry I
Lecture 4, June 23 2

3. Redox: Transfer of Electrons
General College Chemistry I
Lecture 4, June 23 3

4. Oxidation Numbers What is Chemistry?
Oxidation Numbers =Charge
Oxidation Numbers
What is Chemistry?
General College Chemistry I
Lecture 4, June 23 4 4

5. Oxidation Numbers =Charge Redox View Points
General College Chemistry I
Lecture 4, June 23

6. Oxidation Numbers =Charge Oxidation Numbers
General College Chemistry I
Lecture 4, June 23

7. Recognizing Redox Reactions
General College Chemistry I
Lecture 4, June 23

8. Electromagnetic Radiation
Types of Electrochemical Cells: Electrolytic Cell
Electromagnetic Radiation
An electrolytic cell uses electrical energy to drive a nonspontaneous reaction (D G > 0).
In the cell reaction, an external source supplies free energy to convert lower energy reactants into higher energy products. Thus, the surroundings do work on the system. Electroplating and the recovery of metals from ores utilize electro- lytic cells.

9. Electromagnetic Radiation
Types of Electrochemical Cells: Voltaic Cell
Electromagnetic Radiation
A voltaic cell (also called a galvanic cell) uses a spontaneous reaction (D G < 0) to generate electrical energy.
In the cell reaction, some of the difference in free energy between higher energy reactants and lower energy products is converted into electrical energy, which operates the load (surroundings)—flashlight, MP3 player, car starter motor, and so forth. Thus, the system does work on the surroundings. All batteries contain voltaic cells.

10. Electromagnetic Radiation
Types of Electrochemical Cells
Electromagnetic Radiation
Both cells have several similarities:
Two electrodes, which conduct the electricity between cell and surroundings, dip into an electrolyte
An electrode is identified as either anode or cathode depending on the half-reaction that takes place there:
The oxidation half-reaction occurs at the anode; electrons leave the oxidation half-cell at the anode.
The reduction half-reaction occurs at the cathode; electrons enter the reduction half-cell at the cathode.

11. Electromagnetic Radiation
Types of Electrochemical Cells
Electromagnetic Radiation
The spontaneous rxn occurs in the cell
e-’s flow from – to + (“get to go where they want to go”)
Anode = where ox occurs
Cathode = where red occurs
Salt bridge prevents charge buildup (which would stop flow)

12. Electromagnetic Radiation
Types of Electrochemical Cells
Electromagnetic Radiation
Anode and Oxidation start with vowels while the words Cathode and Reduction start with consonants
Alphabetically, the A in anode comes before the C in cathode, and the O in oxidation comes before the R in reduction
ANode = OXidation
AN OX
REDuction = CAThode
RED CAT

13. Electromagnetic Radiation
A Voltaic Cell Based on the Zinc-Copper Reaction.
Electromagnetic Radiation
The spontaneous reaction between zinc and copper(II) ion. When zinc metal is placed in a solution of Cu2+ ion (left), zinc is oxidized to Zn2+, and Cu2+ is reduced to copper metal (right). (The very finely divided Cu appears black.)

14. Electromagnetic Radiation
Operation of Voltaic Cell

15. Electromagnetic Radiation
A Voltaic Cell Using Inactive Electrodes
Electromagnetic Radiation
In the anode half-cell, I- ions are oxidized to solid I2, and the released electrons flow into the graphite electrode (C) and through the wire.
From the wire, the electrons enter the graphite electrode in the cathode half-cell and reduce MnO4- ions to Mn2+ ions.
A KNO3 salt bridge is used.

16. Inert Platinum Electrode
Used when neither redox species in a half reaction (or electrode) is a neutral metal. (i.e.both are solution species)
You used graphite in place of Pt in lab for Fe3+/Fe2+ and I2/I-cells. A cheaper “inert” electrode.
Chapter 18, Figure 18.4
Inert Platinum Electrode
Neither is a neutral metal

17. Electromagnetic Radiation the cathode compartment
Notation for a Voltaic Cell
Electromagnetic Radiation
Components of
the anode compartment
Components of
the cathode compartment
Phase boundary
Half-cells are
physically separated
half--cell components
Half-cell components usually appear in the same order as in the half-reaction.
Ions in the salt bridge are not part of the reaction so they are not in the notation.

18. Electromagnetic Radiation
Skillbuilder
Draw a diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode

19. Electromagnetic Radiation
Skillbuilder: Quick Quiz
NOTE: Every electrode compartment has one oxidizing agent and one reducing agent (this pair is called the redox “couple”)
If an electrode has Ni(s) and Ni2+ ions in it, which species is the oxidizing agent and which the reducing agent (of the pair)?
Oxidizing agent is ___
Ni2+ (b/c it’s “more positive”; it has “room for an electron”
Reducing agent is ___
Ni (b/c it’s “more negative”; it has an electron to give)

20. Cell Potential Why Do Electrons Flow?
Electrode in each half-cell have its own individual potential = standard electrode potential
The overall standard cell potential Eºcell is the difference between the two standard electrode potentials:
Eºcell = Eºfinal – Eºinitial therefore,
Eºcell = Eºcathode - Eºanode

21. Electromagnetic Radiation
“Competition for the electrons”.
Which oxidizing agent “wants them more”?
Electromagnetic Radiation
Who is the (possible) oxidizing agent on the left? _____
Who is the (possible) oxidizing agent on the right? _____
Zn2+
Cu2+
Hint: The two “players” are Zn and Zn2+
Hint: The two “players” are Cu and Cu2+.
Who wins? (Which one “got” the electrons?) ____
Cu2+
 Cu2+ “pulled harder”
Cu e-  Cu(s)
So…which of the half reactions shown at the right is more favorable (greater tendency to happen)?
Zn e-  Zn(s)
By how much?.....

22. Electromagnetic Radiation Where reduction takes place
Standard Reduction Potential
Electromagnetic Radiation
Reducing Cu2+ is more favorable than reducing Zn2+ …by 1.10 V! (Measure it w/voltmeter!)
We define a “standard reduction potential”, E°red, for every reduction half-reaction such that:
E°cell = E°red(cathode) - E°red(anode)
Where reduction takes place
The more positive the “E” (Ecell, Ered, or Eox), the more favorable the process

23. Electromagnetic Radiation
Standard Reduction Potential
Electromagnetic Radiation
Reducing Cu2+ is more favorable than reducing Zn2+ …by 1.10 V! (Measure it w/voltmeter!)
E°cell = E°red(cathode) - E°red(anode)
1.10 V = E°red(Cu2+/Cu) - E°red(Zn2+/Zn)
NOTE:
If E°red(Zn2+/Zn) were 0 V, E°red(Cu2+/Cu) would be V
If E°red(Zn2+/Zn) were -1.0 V, E°red(Cu2+/Cu) would be V
If E°red(Zn2+/Zn) were +1.0 V, E°red(Cu2+/Cu) would be V
 The “zero” is arbitrary, but must be chosen / agreed upon!

24. Electromagnetic Radiation
Standard Hydrogen Electrode
Electromagnetic Radiation
This electrode (SHE) was ultimately the one chosen by the scientific community to be the “zero” of potential.
2 H+ + 2 e-  H2 (g); E°red = 0.0 V
Upshot:
One can determine any E°red experimentally by just setting up a cell where one of the half cells is SHE!

25. Electromagnetic Radiation
Determning a Standard Reduction Potential Using the SHE
Electromagnetic Radiation
Determining a Standard Reduction Potential using the SHE
0.76 V = E°red(SHE) - E°red(Zn2+/Zn)
 0.76 V = 0 - E°red(Zn2+/Zn)
 E°red(Zn2+/Zn) = V
Both as reductions
Ecell = Ecathode - Eanode
The reduction of H+ is more favorable than the reduction of Zn2+….
by 0.76 V!
H+ (not Zn2+) gets reduced

26. Compare Potential Energy
Cell Potential
Compare Potential Energy
The electrode in any half-cell with a lesser tendency to undergo reduction (or greater tendency to undergo oxidation) is negatively charged relative to the SHE and, there- fore, has a negative E°.
The electrode in any half-cell with a greater tendency to undergo reduction is positively charged relative to the SHE and therefore has a positive E°.

27. Electromagnetic Radiation
Standard Reduction Potentials
Electromagnetic Radiation
Could use this info to predict that this direct reaction would occur:
Use these values to:
predict which reactions are spontaneous at standard state and to find any E°cell!
E°cell = E°red(cat) - E°red(an)
2 H+(aq) + 2 e H2(g)
E°cell = 0 – (-0.76) = V
H+ gets reduced; Zn2+ does not (Zn gets oxidized):
2 H+ + Zn → H2(g) + Zn2+
is spontaneous: E°cell > 0
Zn2+(aq) + 2 e Zn(s)

28. Electromagnetic Radiation
Types of Electrochemical Cells
Electromagnetic Radiation
Standard Electrode Potential
Refers only to species on the left side of the arrow. E.g., F2, is a better ox agent than H2O2 which is better ox agent than Au3+ (but all of these species are very good oxidizing agents relative to most!)
Refers only to species on the right side of the arrow. E.g., F-, is a poorer red agent than H2O which is poorer red agent than Au(s) (but all of these are very poor reducing agents relative to most!) 28

29. Standard Electrode Potential
Table 18.1 (continued)
Standard Electrode Potential
Click to add notes

30. Skillbuilder
Calculate Ecell for the reaction at 25 C 2 IO3– + 12 H I−→ 6 I2 + 6 H2O
separate the reaction into the oxidation and reduction half-reactions
ox (anode): 2 I−(s)  I2(aq) + 2 e−
red (cathode):
IO3−(aq) + 6 H+(aq) + 5 e−  ½ I2(s) + 3 H2O(l)
find the E for each half-reaction and Ecell
Eanode= −0.54 V
Ecathode= V
Ecell = 1.20 V-(-0.54) = 0.66 V
Is the reaction spontaneous?

31. Calculate Ecell for the reaction at 25 C
Skillbuilder
Calculate Ecell for the reaction at 25 C
separate the reaction into the oxidation and reduction half-reactions
find the E for each half-reaction and Ecell

32. Predicting the Spontaneous Direction of a Redox Rxn
Consider the two reduction half-reactions:
Since the Mn half-reaction has the more negative electrode potential, it repels electrons and proceeds in the reverse direction (oxidation).
Since the Ni half-reaction has the more positive (or least negative) electrode potential, it attracts electrons and proceeds in the forward direction (reduction).

33. Predicting the Spontaneous Direction of a Redox Rxn
Another way to predict the spontaneity of a redox reaction is to note the relative positions of the two half-reactions:
Any reduction half-reaction listed will be spontaneous when paired with the reverse of any half-reaction that appears below it in table.

34. State the better oxidizing agent, determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow
Ni2+(aq) + 2 e Ni(s) V
Mn2+(aq) + 2 e Mn(s) V
The better oxidizing agent is:___
Ni2+
(Because its Ered is more positive)
So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.
Ni2+
right
Chapter 18, Figure 18.8
Mn/Ni2+ Electrochemical Cell
cath
So the Ni electrode must be ______ive
posit
E°cell = _____ - _____
-0.23 V
-1.18 V
OR E°cell = _____ + _____
-0.23 V
-(-1.18 V)
= _______
+0.95 V

35. 1 M Fe2+ 1 M Pb2+ e Salt bridge Pb(s) Fe(s)
State the better oxidizing agent, determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow
Fe2+(aq) + 2 e Fe(s) V
Pb2+(aq) + 2 e Pb(s) V e
The better oxidizing agent is:___
Pb(s)
Fe(s)
Pb2+
So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.
Pb2+
Salt
bridge
Chapter 18, Figure 18.10
Mn/Ni2+ Electrochemical Cell
left
cath
So the Pb electrode must be ______ive
1 M Fe2+
1 M Pb2+
posit
E°cell = _____ - _____
-0.13 V
-0.45 V
= _______
+0.32 V

36. Determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow
Fe2+(aq) + 2 e Fe(s) V
Mg2+(aq) + 2 e Mg(s) V
The better oxidizing agent is:___
Fe2+
So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.
Fe2+
Chapter 18, Figure 18.8
Mn/Ni2+ Electrochemical Cell
right
cath
So the Fe electrode must be ______ive
posit
E°cell = _____ - _____
-0.45 V
-2.37 V
= _______
+1.92 V

37. Skillbuilder
A spontaneous reaction will take place when a reduction half-reaction is paired with an oxidation half-reaction lower on the table
If paired the other way, the reverse reaction is spontaneous
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) spontaneous
Cu2+(aq) + 2 e−  Cu(s) Ered = V higher on a table
Zn2+(aq) + 2 e−  Zn(s) Ered = −0.76 V lower on the table
should go reverse!
Cu(s) + Zn2+(aq)  Cu2+(aq) + Zn(s) nonspontaneous

38. Skillbuilder
Decide whether each of the following will be spontaneous as written or in the reverse direction
F2(g) + 2 I−(aq)  I2(s) + 2 F−(aq)
Reduction Half-Reaction
F2(g) + 2e− 2 F−(aq)
IO3−(aq) + 6 H++ 5e− I2(s) + 3H2O(l)
Ag+(aq) + 1e− Ag(s)
I2(s) + 2e− 2 I−(aq)
Cu2+(aq) + 2e− Cu(s)
Cr3+(aq) + 1e− Cr2+(aq)
Mg2+(aq) + 2e− Mg(s)
spontaneous as written
Mg(s) + 2 Ag+(aq)  Mg2+(aq) + 2 Ag(s)
spontaneous as written
Cu2+(aq) + 2 I−(aq)  I2(s) + Cu(s)
spontaneous in reverse
Cu2+(aq) + 2 Cr2+(aq)  Cu(s) + 2 Cr3+(aq)
spontaneous as written

39. Skillbuilder F− I− I2 Cr3+ F− I− I2 Cr3+
Which of the following materials can be used to oxidize Cu without oxidizing Ag? F− I− I2
Cr3+ F− I− I2
Cr3+
Reduction Half-Reaction
F2(g) + 2e− 2 F−(aq)
IO3−(aq) + 6 H++ 5e− I2(s) + 3H2O(l)
Ag+(aq) + 1e− Ag(s)
I2(s) + 2e− 2 I−(aq)
Cu2+(aq) + 2e− Cu(s)
Cr3+(aq) + 1e− Cr2+(aq)
Mg2+(aq) + 2e− Mg(s)

40. Predicting whether a Metal Will Dissolve in an Acid
Metals dissolve in acids if the reduction of the metal ion is easier than the reduction of H+(aq)
Metals whose ion reduction reaction lies below H+ reduction on the table will dissolve in acid
as a single displacement reaction
Almost all metals will dissolve in HNO3
having N reduced rather than H
Au and Pt dissolve in HNO3 + HCl
NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(l)

41. Skillbuilder
Which of the following metals will dissolve in HC2H3O2(aq)? Write the reaction. Ag Cu
2 Fe(s) + 6 HC2H3O2(aq) → 2 Fe(C2H3O2)3(aq) + 3 H2(g)
2 Cr(s) + 6 HC2H3O2(aq) → 2 Cr(C2H3O2)3(aq) + 3 H2(g) Ag Cu Fe Cr
Reduction Half-Reaction
Au3+(aq) + 3e− Au(s)
Ag+(aq) + 1e− Ag(s)
Cu2+(aq) + 2e− Cu(s)
2H+(aq) + 2e− H2(g)
Fe3+(aq) + 3e− Fe(s)
Cr3+(aq) + 3e− Cr(s)
Mg2+(aq) + 2e− Mg(s)

42. E°cell, DG° and K For a spontaneous reaction
one that proceeds in the forward direction with the chemicals in their standard states
DG° < 1 (negative)
E° > 1 (positive)
K > 1
DG° = −RTlnK = −nFE°cell
n is the number of mole of electrons
F = Faraday’s Constant = 96,485 C/mol e−

43. Calculate DG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Skillbuilder
Calculate DG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Given:
Find:
I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
DG°, (J)
Conceptual Plan:
Relationships:
E°ox, E°red
E°cell
DG°
Solve:
ox: 2 Br−(aq) → Br2(l) + 2 e− E° = 1.09 V
red: I2(l) + 2 e− → 2 I−(aq) E° = V
tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 V
Answer:
because DG° is +, the reaction is not spontaneous

44. Skillbuilder Calculate DG for the reaction at 25C
2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)
Reduction Half-Reaction
Ered, V
F2(g) + 2e− 2 F−(aq)
+2.87
IO3−(aq) + 6 H++ 5e− ½I2(s) + 3H2O(l)
+1.20
Ag+(aq) + 1e− Ag(s)
+0.80
I2(s) + 2e− 2 I−(aq)
+0.54
Cu2+(aq) + 2e− Cu(s)
+0.34
Cr3+(aq) + 1e− Cr2+(aq)
−0.50
Mg2+(aq) + 2e− Mg(s)
−2.37

45. Skillbuilder Calculate DG for the reaction at 25C
2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)
Given:
Find:
2IO3−(aq)+12H+(aq)+10I−(aq) → 6 I2(s) + 6 H2O(l)
G°, (J)
Conceptual Plan:
Relationships:
E°ox, E°red
E°cell
G°
Solve:
ox : 2 I−(s) → I2(aq) + 2 e− Eº = 0.54 V
red: IO3−(aq) + 6 H+(aq) + 5 e− → ½ I2(s) + 3 H2O(l) Eº = 1.20 V
tot: 2IO3−(aq) + 12H+(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V
Answer:
because DG° is −, the reaction is spontaneous in the forward direction under standard conditions

46. Skillbuilder
Calculate K at 25 °C for the reaction Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq)
Given:
Find:
Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq) K
Conceptual Plan:
Relationships:
E°ox, E°red
E°cell K
Solve:
ox: Cu(s) → Cu2+(aq) + 2 e− E° = 0.34 V
red: 2 H+(aq) + 2 e− → H2(aq) E° = V
tot: Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) E° = −0.34 V
Answer:
because K < 1, the position of equilibrium lies far to the left under standard conditions

47. Skillbuilder Calculate K for the reaction at 25C
2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)
Reduction Half-Reaction
Ered, V
F2(g) + 2e− 2 F−(aq)
+2.87
IO3−(aq) + 6 H++ 5e− ½I2(s) + 3H2O(l)
+1.20
Ag+(aq) + 1e− Ag(s)
+0.80
I2(s) + 2e− 2 I−(aq)
+0.54
Cu2+(aq) + 2e− Cu(s)
+0.34
Cr3+(aq) + 1e− Cr2+(aq)
−0.50
Mg2+(aq) + 2e− Mg(s)
−2.37

48. Skillbuilder Calculate K for the reaction at 25C
2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l)
Given:
Find:
2 IO3−(aq)+12 H+(aq)+10 I−(aq) → 6 I2(s) + 6 H2O(l) K
Conceptual Plan:
Relationships:
E°ox, E°red
E°cell K
Solve:
ox : 2 I−(s) → I2(aq) + 2e− Eº = 0.54 V
red: IO3−(aq) + 6 H+(aq) + 5e− → ½ I2(s) + 3H2O(l) Eº = 1.20 V
tot: 2IO3−(aq) + 12H+(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V
Answer:
because K >> 1, the position of equilibrium lies far to the right (“product favored [at equilibrium]”)

49. Standard vs. Nonstandard Cell
Cell Potential under Nonstandard Conditions
Zn + Cu2+  Zn2+ + Cu; Q = ??
Chapter 18, Figure 18.11
Cell Potential and Concentration
(T = 25C)
**Always write out the balanced redox equation before using the Nernst Equation or predicting whether Ecell should increase or decrease!!**

50. Calculate Ecell at 25 °C for the reaction
Skillbuilder
Calculate Ecell at 25 °C for the reaction
Given:
Find:
3Cu(s) + 2MnO4−(aq) + 8H+(aq) → 2MnO2(s) + Cu2+(aq) + 4H2O(l)
[Cu2+] = M, [MnO4−] = 2.0 M, [H+] = 1.0 M
Ecell
Conceptual Plan:
Relationships:
E°ox, E°red
E°cell
Ecell
Solve:
ox: Cu(s) → Cu2+(aq) + 2 e− }x3 E° = 0.34 V
red: MnO4−(aq) + 4 H+(aq) + 3 e− → MnO2(s) + 2 H2O(l) }x2 E° = V
tot: 3Cu(s) + 2MnO4−(aq) + 8H+(aq) → 2MnO2(s) + Cu2+(aq) + 4H2O(l)) E° = V
Check:
units are correct, Ecell > E°cell as expected because [MnO4−] > 1 M and [Cu2+] < 1 M

51. Calculate Ecell at 25 °C for the reaction
Skillbuilder
Calculate Ecell at 25 °C for the reaction
Given:
Find:
2 IO3−(aq)+12 H+(aq)+10 I−(aq) → 6 I2(s) + 6 H2O(l)
[H+] = 0.10 M, [IO3−] = 0.10 M, [I−] = 0.10 M
Ecell
Conceptual Plan:
Relationships:
E°ox, E°red
E°cell
Ecell
Solve:
ox : 2 I−(s) → I2(aq) + 2e− Eº = −0.54 V
red: IO3−(aq) + 6 H+(aq) + 5e− → ½ I2(s) + 3H2O(l) Eº = 1.20 V
tot: 2IO3−(aq) + 12H+(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V
Check:
units are correct, Ecell < E°cell as expected because all the ions are reactants and < 1M

52. Cell Potential under Nonstandard Conditions
Concentration Cells
Cell Potential under Nonstandard Conditions
It is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different.
Electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution:
oxidation of the electrode in the less concentrated solution will increase the ion concentration in the solution – the less concentrated solution has the anode
reduction of the solution ions at the electrode in the more concentrated solution reduces the ion concentration – the more concentrated solution has the cathode

53. Concentration Cell Concentration Cell
When the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)
When the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow
Cu(s) + Cu2+(aq) (2.0 M) Cu2+(aq) (0.010 M) + Cu(s)

54. Electrochemical Cell In voltaic cells
anode is the source of electrons and has a (−) charge
cathode draws electrons and has a (+) charge
In electrolytic cells
electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery
electrons are forced toward the anode, so it must have a source of electrons, the − terminal of the battery

55. Electrolysis
Electrolysis is the process of using electrical energy to break a compound apart.
Electrolysis is done in an electrolytic cell (opposite of the spontaneous process); electrolysis use electrical NRG to overcome the NRG barrier of a non-spontaneous reaction.
Electrolytic cells can be used to separate elements from their compounds:
2 H2(g) + O2(g)  2 H2O(l) spontaneous
2 H2O(l)  2 H2(g) + O2(g) electrolysis
Some applications are (1) metal extraction from minerals and purification, (2) production of H2 for fuel cells, (3) metal plating

56. Electroplating In electroplating, the work piece is the cathode
Cations are reduced at cathode and plate to the surface of the work piece
The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution.

57. Stoichiometry of Electrolysis
In an electrolytic cell, the amount of product made is related to the number of electrons transferred
essentially, the electrons are a reactant
The number of moles of electrons that flow through the electrolytic cell depends on the current and length of time
1 Amp = 1 Coulomb of charge/second
1 mole of e− = 96,485 Coulombs of charge
Faraday’s constant

58. Skillbuilder
Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-rxn Au3+(aq) + 3 e− → Au(s)
Given:
Find:
3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min
mass Au, g
Conceptual Plan:
Relationship s:
t(s), amp
charge (C)
mol e−
mol Au
g Au
Solve:
Check:
units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e−

59. Skillbuilder
Calculate the amperage required to plate 2.5 g of gold in 1 hour (3600 sec) Au3+(aq) + 3 e− → Au(s)

60. Skillbuilder
Calculate the amperage required to plate 2.5 g of gold in 1 hour (3600 sec) Au3+(aq) + 3 e− → Au(s)
Given:
Find:
3 mol e− : 1 mol Au, mass = 2.5 g, time = 3600 s
current, A
Conceptual Plan:
Relationships:
g Au
mol Au
mol e−
charge
amps
Solve:
Check:
units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e−

61. Cu2+/Cu & Fe3+/Fe2+ 0.32 V Cu2+/Cu +0.45 V -0.13 V Zn2+/Zn & Ag+/Ag
Determined by the color of the leads when voltmeter gave positive reading.
Electrons flow away from (-) electrode  Oxidation occurs there. RA loses e-s
Cu2+/Cu & Fe3+/Fe2+
0.32 V
Cu2+/Cu
Cu  Cu2+ + 2e-
+0.45 V
Fe3+ + e- Fe2+
-0.13 V
Zn2+/Zn & Ag+/Ag
1.50 V
Zn2+/Zn
Zn  Zn2+ + 2e-
1.50 V
Ag+ + e- Ag
0.0 V
Cu2+/Cu & Zn2+/Zn
1.05 V
Zn2+/Zn
Zn  Zn2+ + 2e-
1.50 V
Cu2+ + 2e- Cu
-0.45 V
**Circle the species (not the “electrode”!) that is the better oxidizing agent**

62. From Text Table 0.0 V 1.50 V -0.45 V -0.13 V 0.0 V 0.80 V 0.80 V
(See next slide)
Cu2+ + 2e- Cu
Zn  Zn2+ + 2e-
Fe3+ + e- Fe2+
Ag+ + e- Ag
0.0 V
1.50 V
-0.45 V
-0.13 V
From Table A.1:
0.0 V
Ag+ + e- Ag
0.80 V
0.80 V
-0.13 V
Fe3+ + e- Fe2+
0.67 V
0.77 V
-0.45 V
Cu2+ + 2e- Cu
0.35 V
0.34 V
-1.50 V
Zn2+ + 2e-  Zn
-0.70 V
-0.76 V
Flip
Zn2+ + 2e-  Zn
-1.50 V