electrochemistry the study of the interchange between chemical and electrical energy

Electrochemistry The study of the interchange between chemical and electrical energy

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-

Leo says Ger Lose electron oxidation Zn 2e - + Zn 2+ Gain electron reduction 2e - + Cu 2+ Cu My name is Leo. Grr-rrrr

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!

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

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

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+

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

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)

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 -

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)

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!

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- +

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 ( )

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.

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( )

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.

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

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 -

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

ClO + Zn Cl - + Zn 2+ Oxidized species Reduced species Reducing Agent Oxidizing Agent ClO - Zn Notice that these terms only ever apply to reactants

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

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)

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

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)

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!

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

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+

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

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.)

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

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.)

C(gr) Cr 2+ (1 M), Cr 3+ (1 M) Cl - (1M) Cl 2 (g) Pt(s) CPt 1 M Cr 2+ /Cr 3+ Cr 2+ Cr 3+ + e - 1 M Cl -, 1 atm Cl 2 (g) Cl 2 (g) + 2e - 2Cl - Cr 2+ Cr 3+ Cl - Cl 2 anions e-e- Oxidation Reduction AnodeCathode cations

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

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

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

One Cell of a Lead Battery Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 595

Mercury Battery Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 596

Common Alkaline Battery Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 596

Cathodic Protection of an Underground Pipe Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 598

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

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)

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 =

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 =

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 =

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 -

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 -

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

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

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.

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

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

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 ??

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

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

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

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

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)

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 -

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

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

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

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

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

electrochemistry the study of the interchange between chemical and electrical energy

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