Electrochemistry The study of the interchange between chemical
and electrical energy
Slide 3
Oxidation-Reduction Reactions Redox reactions involve the
transfer of electrons (e-) Reduction: gain e- Oxidation: lose e-
LEO the lion says, GER OIL RIG Use oxidation states to keep track
of the e-
Slide 4
Leo says Ger Lose electron oxidation Zn 2e - + Zn 2+ Gain
electron reduction 2e - + Cu 2+ Cu My name is Leo. Grr-rrrr
Slide 5
Assigning Oxidation States Specific rules for assigning Ox #s
Usually the same charge assigned by the PT H is almost always +1 O
is almost always -2 F is always -1 in compounds For elements (H 2,
O 2, F 2, Ca, K, etc ) the oxidation state always = 0 Some
exceptions do exist!
Slide 6
Assigning Oxidation Numbers Overall charge = sum of the
oxidation states of all atoms in it Neutral Compounds (e.g. H 2 O,
CO 2, CH 4 ) H 2 O : The overall charge is 2(1) + -2 = 0 CO 2 :
What is the oxidation state of C? Since C + 2 (O) = 0 C + 2(-2) =
0, thus CH 4 : Is C still +4? H is always +1 To remain neutral 4(1)
+ C = 0 C must = - 4 H = +1 and O = -2 C = +4
Slide 7
Assigning Oxidation Numbers Charged compounds (e.g. NO 3 -, CO
3 2- ) NO 3 - or (NO 3 ) - : What is the oxidation # of N? O is -2,
and the overall charge is -1 So N + 3(O) = -1 or N + 3(-2) = -1 N =
+ 5 (CO 3 ) 2- : What is the oxidation # of C? O is -2, and the
overall charge is -2 So C + 3(O) = -2 or C + 3(-2) = -2 C = +4 The
oxidation # of ions = charge of ions Mn 3+ has an oxidation # of +3
S 2- has an oxidation # of -2
Slide 8
Assigning Oxidation # Practice Assign oxidation numbers to each
atom Cl 2 Fe 2+ ClO 3 - ClO 4 - IO 2 - CrO 4 2- Fe 3 (PO 4 ) 2 CoSO
4 H 2 CO 3 Cl: 0 (element) Fe: 2+ (ion) O: 2-, 3(2-) + Cl = 1-Cl:
5+ O: 2-, 4(2-) + Cl = 1-Cl: 7+ O: 2-, 2(2-) + I = 1-I: 3+ O: 2-,
4(2-) + Cr = 2-Cr: 6+ Fe: 2+ (ion) PO 4 :3- (ion).O:2-, 4(2-) + P =
3-, P: 5+ Co: 2+ (ion) SO 4 :2- (ion).O:2-, 4(2-) + S = 2-, S: 6+
H: 1+ (ion) CO 3 :2- (ion).O:2-, 3(2-) + C = 2-, C: 4+
Slide 9
Assigning Oxidation Numbers Review Try theseMnO 4 -, Cr 2 O 7
2-, C 2 O 4 2- (MnO 4 ) - O = -2, so [4(-2) + Mn = -1] Mn = +7 (Cr
2 O 7 ) 2- O = -2, so [7(-2) + 2Cr = -2] 2Cr = 12, therefore (C 2 O
4 ) 2- O = -2, so [2C + 4(-2) = -2] 2C = 6, therefore Cr = +6 C =
+3
Slide 10
Oxidation-Reduction Reactions Two separate reactions occurring
simultaneously Oxidation: oxidation # of an atom increases e.g.
Fe(s) Fe 3+ (aq) Reduction: oxidation # of an atom is reduced e.g.
O 2 (g) O 2- (aq) When occurring together Fe(s) + O 2 (g) Fe 3+
(aq) + O 2- (aq) This is the redox reaction responsible for rust!
But, how do we balance this? (ox # goes from 0 +3) (oxidation #
goes from 0 -2)
Slide 11
Balancing by Half-Reactions *in acidic solution 1.Assign
oxidation states for each element. 2.Write separate half-reactions
for the reduction/oxidation reactions. 3.Balance all the atoms
EXCEPT O and H. 4.Balance the oxygen with water (H 2 O). 5.Balance
the hydrogen with hydrogen ions (H + ). 6.Balance the charge with
electrons. 7.Multiply each half-reaction by an appropriate number
to make the electrons equal. 8.Combine both reactions into one and
cancel the e -
Slide 12
Balancing by Half-Reactions * in acidic solution CH 3 OH (aq) +
Cr 2 O 7 2- (aq) CH 2 O(aq) + Cr 3+ (aq) 1.Assign oxidation states.
C -2 H 4 + O 2- + (Cr 2 6+ O 7 2- ) 2- C 0 H 2 + O 2- + Cr 3+ 2.
Write separate half-reactions for the reduction and oxidation
reactions. (only keep charges that are changing) Ox: C -2 H 4 O C 0
H 2 O (C is going from -2 to 0) Red: (Cr 2 6+ O 7 ) 2- Cr 3+ (Cr is
being reduced from +6 to +3)
Slide 13
Ox: C 2- H 4 O C 0 H 2 O 3. For each half reaction, balance all
the atoms EXCEPT O and H. 4. Balance the oxygen by adding water (H
2 O). 5.Balance the hydrogen by adding hydrogen ions (H + )
6.Balance the charge by adding electrons. use the oxidation state
as a guide Balancing the Oxidation Carbon is already balanced! + 2H
+ + 2e- Oxygen is already balanced!
Slide 14
On to the reduction (Cr 2 6+ O 7 ) 2- Cr 3+ 3.Balance all
elements except H and O 4.Balance O by adding H 2 O, if necessary
5.Balance H by adding H +,if necessary 6.Balance charge by adding
e- Remember, you only care about the charges that are changing 2 +
7H 2 O 14H + + 6e- +
Slide 15
Adding Half-Reactions *in acidic solution Now add the two
reactions together Ox: CH 4 O CH 2 O + 2H + + 2e- Red: 6e- + 14H +
+ Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 7. Multiply each half-reaction by an
appropriate number to make the electrons equal. CH 4 O CH 2 O + 2H
+ + 2e- 6e- + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 3CH 4 O 3CH 2 O +
6H + + 6e- 3 ( )
Slide 16
6e- + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 3CH 4 O 3CH 2 O + 6H
+ + 6e- 3CH 4 O + + Cr 2 O 7 2- 3CH 2 O + 2Cr 3+ + 7H 2 O and the
reaction is now balanced! 8H + Adding Half-Reactions *in acidic
solution 8. Combine both reactions into one and cancel.
Slide 17
Practice Balancing Redox Reactions Unbalanced reaction (in
acid): MnO 4 + Fe 2+ Mn 2+ + Fe 3+ Balanced Reduction
half-reaction: 8H + + MnO 4 + 5e Mn 2+ + 4H 2 O Balanced Oxidation
half-reaction: Fe 2+ Fe 3+ + e Balanced overall reaction: 8H + +
MnO 4 + 5Fe 2+ Mn 2+ + 5Fe 3+ + 4H 2 O 5( )
Slide 18
Balancing by Half-Reactions *in basic solution 1.Assign
oxidation states. 2.Write separate half-reactions for the
reduction/oxidation reactions. 3.Balance all the atoms EXCEPT O and
H. 4.Balance the oxygen by adding water (H 2 O). 5.Balance the
hydrogen by adding H +. 6.Balance the charge by adding electrons.
7.Multiply each half-reaction by an appropriate number to make the
electrons equal. 8.Combine both reactions into one and cancel.
9.Add OH - to both sides to cancel out H + and create H 2 O.
Simplify further, if necessary.
Slide 19
Balancing by Half-Reactions (in basic solution) Lets balance a
previous example in basic solution Remember, it is all the same
steps up to this point 3CH 4 O + 8H + + Cr 2 O 7 2- 3CH 2 O + 2Cr
3+ + 7H 2 O 3CH 4 O + + Cr 2 O 7 2- 3CH 2 O + 2Cr 3+ + 7H 2 O + 8OH
- 3CH 4 O + H 2 O + Cr 2 O 7 2- 3CH 2 O + 2Cr 3+ + 8OH - + 8OH - 8H
2 O
Slide 20
Practice Balancing Basic Redox Rxns Unbalanced reaction: ClO +
Zn Cl - + Zn 2+ Balanced Reduction half-reaction: 2e - + 2H + + ClO
- Cl - + H 2 O Balanced Oxidation half-reaction: Zn Zn 2+ + 2e -
Balanced overall reaction (acidic): 2H + + ClO + Zn Zn 2+ + Cl - +
H 2 O Balanced overall reaction (basic): H 2 O + ClO + Zn Zn 2+ +
Cl - + 2OH -
Slide 21
Redox Vocabulary ClO + Zn Cl - + Zn 2+ Oxidized species (atom,
ion, molecule, or compound) whichever species is being oxidized (
oxidation #) Reduced species whichever species is being reduced (
oxidation #) Oxidizing agent whichever species CAUSES oxidation to
occurin other words, the oxidizing agent IS the reduced species
Reducing agent whichever species CAUSES reduction to occurin other
words, the reducing agent IS the oxidized species
Slide 22
ClO + Zn Cl - + Zn 2+ Oxidized species Reduced species Reducing
Agent Oxidizing Agent ClO - Zn Notice that these terms only ever
apply to reactants
Slide 23
Sample electrochemical processes: 1. Corrosion e.g. 4 Fe (s) +
3 O 2(g) 2 Fe 2 O 3(s) 2. Biological processes e.g. C 6 H 12 O 6 +
6 O 2 6 CO 2 + 6 H 2 O 3. Batteries (Galvanic or Voltaic cells)
Electrochemical cells that produce a current (flow of electrons) as
a result of a redox reaction 4. Electrolytic cells Electrical
energy is used to produce chemical change Used to prepare or purify
metals (such as sodium, aluminum, copper) Electrochemistry The
study of the interchange of chemical and electrical energy
Slide 24
Chemical Change Electron Flow Copper: Cu (s), Cu 2+ (aq) Cu (s)
Cu 2+ (aq) + 2e - G rxn = G f (Cu 2+ ) = 65.6 kJ Silver: Ag (s), Ag
+ (aq) Ag (s) Ag + (aq) + e - G rxn = G f (Ag + ) = 77.2 kJ Cu (s)
Cu 2+ (aq) + 2e - G = +65.6 kJ Ag + (aq) + e- Ag (s) G = -77.2 kJ
Cu (s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag (s) G = -88.8 kJ Spontaneous
w max = -88.8 kJ Cu 2+ in solution 2( ) 2( ) Ag(s)
Slide 25
Harnessing the Energy Separate the half-reactions Creates a
galvanic or voltaic cell CuAg 1 M CuSO 4 Cu (s) Cu 2+ (aq) + 2e - 1
M AgNO 3 Ag + (aq) + e - Ag (s) Luigi Galvani Alessandro Volta Cu
2+ SO 4 2- Ag + NO 3 - KNO 3(aq) K+K+ NO 3 - e-e- salt bridge
Oxidation Reduction AnodeCathode cathode and reduction begin with
consonants anode and oxidation begin with vowels (produces
electrons) + (attracts electrons) Red Cat
Slide 26
Standard Reduction Potentials The cell potential, cell,can be
determined from the standard reduction potentials ( red) for the
half- reactions: Reduction potential = tendency for reduction to
happen Positive red spontaneous reduction reaction Negative red
non-spontaneous reduction (use reverse reaction) Standard ( o ) =
standard conditions (1 M solutions)
Slide 27
Standard Reduction Potentials Half-Reaction (V) F 2 + 2e - 2F -
2.87 Au 3+ + 3e - Au 1.50 Ag + + e - Ag 0.80 Cu 2+ + 2e - Cu 0.337
2H + + 2e - H 2 0.00 Ni 2+ + 2e - Ni-0.28 Zn 2+ + 2e - Zn-0.763 Al
3+ + 3e - Al-1.66 Li + + e - Li-3.05 = 0 Standard Hydrogen
Electrode > 0 Spontaneous reduction (Oxidizing Agents!) < 0
Non-Spontaneous reduction (Reducing Agents!) Spontaneous oxidation
(reverse rxn) Ni Ni 2+ + 2e - +0.28 Zn Zn 2+ + 2e - +0.763 Al Al 3+
+ 3e - +1.66 Li Li + + e - +3.05 Remember: an oxidation CANNOT
happen without a reduction!
Slide 28
Cell Potential cell = reduction + oxidation Ag + (aq) + e - Ag
(s) = 0.80 V Cu 2+ (aq) + 2 e - Cu (s) = 0.34 V Reduction
reaction:2(Ag + (aq) + e - Ag (s) ) = +0.80 V Oxidation reaction:Cu
(s) Cu 2+ (aq) + 2 e - = - 0.34 V Cu (s) + 2 Ag + (aq) Cu 2+ (aq) +
2 Ag (s) cell = +0.46 V The cell MUST be + and thus spontaneous for
Galvanic cells
Slide 29
Brain Warmup Half-Reaction (V) Ag + + e - Ag 0.80 Cu 2+ + 2e -
Cu 0.34 Zn 2+ + 2e - Zn-0.76 Al 3+ + 3e - Al-1.66 What is for each
of the following reactions? Which reaction(s) are spontaneous? 1.3
Ag + (aq) + Al (s) 3 Ag (s) + Al 3+ (aq) 2.Cu 2+ (aq) + Zn (s) Cu
(s) + Zn 2+ (aq) 3.2 Al 3+ (aq) + 3 Zn (s) 2 Al (s) + 3 Zn 2+ (aq)
2.46 V 1.10 V -0.90 V Spontaneous? Y Y N Zn can reduce Cu 2+, but
not Al 3+
Slide 30
Line Notation for Galvanic Cells CuAg 1 M CuSO 4 Cu (s) Cu 2+
(aq) + 2e - 1 M AgNO 3 Ag + (aq) + e - Ag (s) Cu 2+ SO 4 2- Ag + NO
3 - K+K+ e-e- Oxidation Reduction Anode () Cathode (+) Anode always
on the left Cu (s) Cu 2+ (1 M) Ag + (1 M) Ag (s) Cathode always on
the right
Slide 31
Practice Time Given the following information, draw a galvanic
cell. Fe(s) Fe 2+ (1 M) Au 3+ (1 M) Au(s) Be sure to include the
following: Anode/Cathode reactions Balanced overall reaction with
potential Complete circuit (external wire with e- flow direction,
salt bridge) Label all parts of the cell (solution, electrode,
etc.)
Slide 32
Fe(s) Fe 2+ (1 M) Au 3+ (1 M) Au(s) FeAu 1 M Fe 2+ Fe (s) Fe 2+
(aq) + 2e - 1 M Au 3+ Au 3+ (aq) + 3e - Au (s) Fe 2+ Au 3+ anions
e-e- Oxidation Reduction AnodeCathode 3Fe(s) + 2Au 3+ (aq) 3Fe 2+
(aq) + 2Au(s) cations cell = +0.440V (Fe rxn) + 1.50 V (Au rxn) =
1.94 V
Slide 33
Practice Time Given the following information, draw a galvanic
cell. C(gr) Cr 2+ (1 M), Cr 3+ (1 M) Cl - (1M) Cl 2 (g) Pt(s) Be
sure to include the following: Anode/Cathode reactions Balanced
overall reaction with potential Complete circuit (external wire
with e- flow direction, salt bridge) Label all parts of the cell
(solution, electrode, etc.)
The First Battery Alessandro Volta (1745 1827) Generated
electricity by putting a layer of cardboard soaked in brine between
discs of copper and zinc a voltaic cell. When he made a pile of
these cells, he increased the amount of electricity generated. This
was the first battery a collection of cells. Newmark, CHEMISTRY,
1993, page 46 Zinc disc Copper disc
Slide 36
Car Battery (Lead storage battery) Anode: Pb + HSO 4 - PbSO 4 +
H + + 2e - Cathode: PbO 2 + HSO 4 - + 3H + + 2e - PbSO 4 + 2H 2 O
Cell: Pb + PbO 2 + 2H + + 2 HSO 4 - 2 PbSO 4 + 2H 2 O 2 volts per
cell, 6 cells to a battery 12 volt battery Alkaline Battery Anode:
Zn Zn 2+ + 2e - Cathode: 2 MnO 2 + H 2 O + 2e - Mn 2 O 3 + 2OH -
cell = 1.5 volts Lemon Battery Anode: Zn Zn 2+ + 2e - Cathode: Cu
2+ + 2e - Cu Cell reaction: Zn + Cu 2+ Zn 2+ + Cu cell = 1.1 volts
(if all goes well) Want more volts? Link cells in series...
Batteries
Slide 37
Onion Battery? According to YouTube, you can charge your iPod
with just an onion and Gatorade As Michael Scott (from the Office)
points out, anyone can put anything on the Internet, so you know
the information is good
Slide 38
One Cell of a Lead Battery Zumdahl, Zumdahl, DeCoste, World of
Chemistry 2002, page 595
Slide 39
Mercury Battery Zumdahl, Zumdahl, DeCoste, World of Chemistry
2002, page 596
Slide 40
Common Alkaline Battery Zumdahl, Zumdahl, DeCoste, World of
Chemistry 2002, page 596
Slide 41
Cathodic Protection of an Underground Pipe Zumdahl, Zumdahl,
DeCoste, World of Chemistry 2002, page 598
Slide 42
Free Energy and Cell Potential G = w max = nF n= moles of e-
transferred F= Faradays constant = 96,485 C/mol e - = standard cell
potential (V or J/C) Michael Faraday Cu (s) Cu 2+ (1 M) Ag + (1 M)
Ag (s) cell = +0.46 V G = -nF cell G = -(2 mol e - )(96485 C/mol e
- )(0.46 V) G = -88,800 J or -88.8 kJ
Slide 43
3 Ag + (aq) + Al (s) 3 Ag (s) + Al 3+ (aq) Polishing Silver
with Aluminum Foil = 2.46 V 3 Ag 2 S + 2 Al + 3 H 2 O 6 Ag + Al 2 O
3 + 3 H 2 S Stinky!! (Add NaHCO 3 to neutralize H 2 S)
Slide 44
Reaction Quotient The reaction quotient (Q) sets up a ratio of
products and reactants For a reaction, A + 2B 3C + 4D [C] 3 [D] 4
[A] 1 [B] 2 Only include concentrations (aq) OR pressures (g)
Solids (s) and liquids (l) are not included Q =
Slide 45
Reaction Quotient practice Write the Q expression for the
following reaction CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Reaction
must be balanced first CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g )
(CO 2 )(H 2 O) 2 (CH 4 )(O 2 ) 2 Q =
Slide 46
Reaction Quotient practice Write the Q expression for the
following reaction Cu(s) + 2Ag + (aq) Cu 2+ (aq) + 2Ag(s) (Cu 2+
)(Ag) 2 (Cu)(Ag + ) 2 Is this correct? NO: Solids arent included in
the equation! (Cu 2+ ) (Ag + ) 2 Q =
Slide 47
Under Non-standard conditions: The Nernst Equation If we dont
have 1 M concentration or 1 atm pressure, we must take a different
approach G = G + RT lnQ G = -nF G = -nF -nF = -nF + RT lnQ Walther
Nernst cell = cell - R= 8.3145 J/mol K T= temperature (in K) n=
moles of e- transferred F= Faradays constant 96,485 C/mol e -
Slide 48
Practice with the Nernst Equation What will be the cell
potential of a Cu/Ag cell using 0.10 M Cu 2+ and 1.0 M Ag +
solutions at 25C? Cu (s) Cu 2+ (aq) + 2e - Ag + (aq) + e - Ag (s)
CuAg Cu 2+ SO 4 2- Ag + NO 3 - Cu (s) + 2 Ag + (aq) Cu 2+ (aq) + 2
Ag (s) cell = 0.46 V (-0.03 V) Cu (s) Cu 2+ (0.10 M) Ag + (1.0 M)
Ag (s) cell cell = 0.49 V 2( ) cell = cell -
Slide 49
3( ) 5( ) More Practice What will be the cell potential of a
Al/MnO 4 -,Mn 2+ cell using 0.5 M Al 3+, 1.5 M Mn 2+, 1.0 M MnO 4 -
and 2.0 M H + solutions at 15C? Al (s) Al 3+ (aq) + 3e - MnO 4 - +
8H + + 5e - Mn 2+ + 4 H 2 O AlPt Al 3+ H + MnO 4 -, Mn 2+ 5Al(s) +
3MnO 4 - + 24H + 3Mn 2+ + 5 Al 3+ + 12 H 2 O cell = 3.17 V (-0.03
V) Al (s) Al 3+ (0.5 M) Mn 2+ (1.5 M), MnO 4 - (1.0 M), H + (2.0 M)
Pt (s) cell cell = 3.20 V cell = cell - = 6.285 x 10 -9
Slide 50
Remove NH 3 .. NH 3 H 2 .. Add more N 2 .. Le Chateliers
principle N 2 (g) + 3 H 2 (g) 2 NH 3 (g) DisturbanceEquilibrium
Shift no shift When a system at equilibrium is disturbed, it shifts
to a new equilibrium that balances the disturbance Add a
solid/liquid Remove either reactant. Fritz Haber
Slide 51
In a chicken CaO + CO 2 CaCO 3 (eggshells) [ CaO ], shift [ CO
2 ], shift -- shift ; eggshells are thinner In summer, [ CO 2 ] in
a chickens blood due to panting. How could we increase eggshell
thickness in summer? -- give chickens carbonated water -- put CaO
additives in chicken feed -- air condition the chicken house TOO
much $$$ -- pump CO 2 gas into the chicken house would kill all the
chickens! I wish I had sweat glands.
Slide 52
Other applications with LeChateliers Principle N 2 + 3 H 2 2 NH
3 + heat Raising the temperature favors the endothermic reaction
(the reverse reaction) in which the rise in temperature is
counteracted by the absorption of heat. Increasing the pressure
favors the forward reaction in which 4 mol of gas molecules is
converted to 2 mol. Dorin, Demmin, Gabel, Chemistry The Study of
Matter 3rd Edition, page 532
Slide 53
AgCl + energy Ag o + Cl o shift to a new equilibrium: Then go
inside shift to a new equilibrium: Light-Darkening Eyeglasses
energy Go outside Sunlight more intense than inside light; GLASSES
DARKEN (clear) (dark) energy GLASSES LIGHTEN
Slide 54
Draw the following Galvanic Cell: Cu (s) Cu 2+ (aq) Cu 2+ (aq)
Cu (s) Cu 2+ (aq) + 2e - Cu (s) Cu (s) Cu 2+ (aq) + 2e - Oxidation
Reduction ox = -0.337 V Write the two half reactions and determine
the cell red = +0.337 V cell = 0 ??
Slide 55
Concentration Cells A voltage is generated just by a difference
in concentration between the two half-cells Cu (s) Cu 2+ (aq) (0.5
M) Cu 2+ (aq) (2.5 M) Cu (s) = - Cu 2+ (aq) + 2e - Cu (s) Cu (s) Cu
2+ (aq) + 2e - = 0 = - = 0.02 V = 0 - Cu(s) an + Cu 2+ (aq) cat Cu
2+ (aq) an + Cu(s) cat
Slide 56
Concentration Cells in Nature Living cells contain low-pH
vesicles surrounded by neutral-pH cytoplasm The [H + ]
concentration gradient creates a voltage pH = 3 pH = 7 cell = cell
- 4 H + + O 2 + 4e - 2 H 2 O = 1.23 V (10 -7 ) 4 (10 -3 ) 4 Q = ([H
+ ] outside ) 4 ([H + ] inside ) 4 = cell = -(0.059/4) log(10 -16 )
= 0.24 V
Slide 57
Using LeChatliers Cu (s) Cu 2+ (aq) (0.5 M) Cu 2+ (aq) (2.5 M)
Cu (s) Cu 2+ (aq) + 2e - Cu (s) Cu (s) Cu 2+ (aq) + 2e - Anode
Cathode Write the overall balanced equation Cu(s) an + Cu 2+ (aq)
cat Cu 2+ (aq) an + Cu(s) cat cell = + What effect would these
changes have on the cell voltage (or cell)? Increase [Cu 2+ ] cat
Increase [Cu 2+ ] an Decrease [Cu 2+ ] an Decrease [Cu 2+ ]
cat
Slide 58
Electrolytic Cells Galvanic cell (battery): spontaneous
reaction, o = + Electrolytic cell: non-spontaneous reaction, = - An
external power source is used to force the reaction to occur Used
for : Charging (rechargeable) batteries Producing or purifying
metals (aluminum, copper) Electroplating Charles Hall Discovered
how to produce aluminum by electrolysis
Slide 59
Electrolysis of Water Spontaneous reaction: formation of water
2 H 2(g) + O 2(g) 2 H 2 O (l) 0 0 +1 -2 As a galvanic cell:
Cathode: O 2 H 2 O(O 2(g) + 4 H + (aq) + 4 e - 2 H 2 O (l) ) Anode:
H 2 H 2 O(2 H 2(g) + 4 OH - (aq) 4 H 2 O (l) + 4 e - ) = 2.06 V
(voltage that can be used) Non-spontaneous reaction: electrolysis
of water 2 H 2 O (l) 2 H 2(g) + O 2(g) As an electrolytic cell:
Anode (oxidation):2 H 2 O (l) O 2(g) + 4 H + (aq) + 4 e - Cathode
(reduction):4 H 2 O (l) + 4 e - 2 H 2(g) + 4 OH - (aq) = -2.06 V
(the voltage added for rxn to occur)
Slide 60
Electrolysis of Water 2 H 2 O (l) 2 H 2(g) + O 2(g) = -2.06 V
Cathode (reduction): 4 H 2 O + 4 e - 2 H 2 + 4 OH - Produces: 2 mol
H 2 gas Base (OH - ) Anode (oxidation): 2 H 2 O O 2 + 4 H + + 4 e -
Produces: 1 mol O 2 gas Acid (H + ) Battery > 2.06 V + Pt
electrodes e- e- e -
Slide 61
60 s 1 min 31.998 g mol O 2 1 mol O 2 4 mol e - mol e - 96,485
C 3.0 C s Electrolysis Calculations How many grams of O 2(g) will a
3.0 amp power source produce in 5 minutes? Amperes (A) = electric
current = coulombs/second (C/s) Andr-Marie Ampre Oxygen
half-reaction: 2 H 2 O O 2 + 4 H + + 4 e - Time and Current Charge
(F) Moles of e - Moles of product Grams of product = 5.0 min 0.075
g O 2 3.0 C s
Slide 62
Electroplating Cathode Cu 2+ (aq) + 2e - Cu (s) Anode Cu (s) Cu
2+ (aq) + 2e - Battery + Cu Cu 2+ Cu 2+ e- e- e - How long will it
take a 15 amp power source to deposit 5.9 g of copper? mol Cu 63.55
g Cu s 15 C 96,485 C mol e - 2 mol e - mol Cu = 5.9 g Cu 1200 s or
20. min or 0.33 hr or 0.014 d or 3.8 x 10 -5 yr Time and Current
Charge (F) Moles of e - Moles of product Grams of product
Slide 63
Electric Vehicle Charging Stations Bloomington-Normal, IL (as
of 5/8/12) Level II charger = full charge 6-8 hrs Level III = 80%
charge 25 min Free of charge (so far) First hour parking free City
Hall Annex (6 units) Fire HQ (1 unit) Public Works (1 unit)
Marriott Deck (2 units) College Ave. Parking Deck (3 units)
Heartland C.C. (2 units) ISU (2 units) City of Bloomington
(2unitsLincoln Deck) IWU (2 units) CIRA (2 units) Commerce Bank (1
unit) Constitution Trail Center (2 units) Holiday Inn Express (1
unit) Advocate BroMenn (3 units) Mitsubishi commerical The New
Normal
Slide 64
Electric Vehicle Comparison: 2012 Nissan Leaf $36,000 MSRP base
99 mpg eq. 70 mi hwy range *Home charging = 20 hrs from dead
battery 220 V = 6-7 hrs Battery cells: 8 yr/100,000 mi warranty
Mitsubishi i-MiEV $30,000 MSRP base 112 mpg eq. 62 mi hwy range
*Home charging = 20 hrs from dead battery 220 V = 6-7 hrs Battery
cells: 8 yr/100,000 mi warranty Both eligible for $7500 Federal and
$3000 State tax credit www.fueleconomy.gov
Slide 65
Hybrid Vehicle Comparison: 2012 Chevy Volt $39,145 MSRP base 94
mpg eq. First 35 mi EV Then gas engine starts Gas engine = 37 mpg
(prem.) *Home charging = 8 hrs from dead battery 220 V = 4 hrs
Battery cells: 8 yr/100,000 mi warranty Toyota Prius Plug-in Hybrid
$35,760 MSRP base 95 mpg eq. First 11 mi EV Then gas engine starts
Gas engine = 50 mpg *Home charging = 6 hrs from dead battery 220 V
= 1.5 hrs Battery cells: 8 yr/100,000 mi warranty