ELECTROCHEMISTRY Electrochemistry involves the relationship between electrical energy and chemical energy. OXIDATION-REDUCTION REACTIONS SPONTANEOUS REACTIONS Can extract electrical energy from these. Examples: voltaic cells, batteries. NON-SPONTANEOUS REACTIONS Must put in electrical energy to make them go. Examples: electrolysis, electrolytic cells. QUANTITATE REACTIONS

OXIDATION-REDUCTION Oxidation = loss of electrons. Occurs at ANODE An oxidizing agent is a substance that causes oxidation (and is itself reduced). Reduction = gain of electrons. At CATHODE A reducing agent is a substance that causes reduction (and is itself oxidized). LEO goes GER LEO: Loss of Electrons = Oxidation GER: Gain of Electrons = Reduction

Rules for determining Oxidation States For pure elements, oxidation state = 0 (examples: S 8, P 4, O 3, F 2, K, Be) In compounds, some elements have “common” oxidation numbers: Assign others by difference (using charge on ion) Alkali metals (Na +, K +..)+1 Alkaline earths (Mg 2+, Ba 2+ …)+2 F = -1, O = -2 (but -1 in peroxides) Cl, Br, I = -1 (except if bonded to O or halogen) H = +1 or -1 (depends what it is bonded to)

1. Write incomplete half-reactions. 2. Balance each half-reaction separately. a. Balance atoms undergoing redox. b. Balance remaining atoms i. Add H 2 O to balance oxygens. ii. Add H + to balance hydrogens. 3. Balance charges by adding electrons. 4.Multiply half-reactions to cancel electrons. 5. Add the half-reactions. 6. In basic solutions, add OH  to neutralize H + BALANCING REDOX REACTIONS

The half-reactions for Sn 2+ (aq) + 2Fe 3+ (aq)  Sn 4+ (aq) + 2Fe 2+ (aq) Sn 2+ (aq)  Sn 4+ (aq) +2e - 2Fe 3+ (aq) + 2e -  2Fe 2+ (aq) Oxidation: electrons are products. Reduction: electrons are reagents. Half Reactions

In Acid solution.: C + HNO 3  NO 2 + CO 2 + H 2 O Basic solution.: PbO 2 + Cl  + OH   Pb(OH) 3  + ClO  Balancing Redox Reactions

The energy released in a spontaneous redox reaction is used to perform electrical work. Voltaic or galvanic cells are devices in which electron transfer occurs via an external circuit. Voltaic cells are spontaneous. If a strip of Zn is placed in a solution of CuSO 4, Cu is deposited on the Zn and the Zn dissolves by forming Zn 2+. Zn 0 (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu 0 (s) VOLTAIC CELL

E 0 cell = 1.10 V

Voltaic cells consist of Anode: Zn(s)  Zn 2+ (aq) + 2e - Cathode: Cu 2+ (aq) + 2e -  Cu(s) The two solid metals are the electrodes (cathode and anode). Salt bridge: (used to complete the electrical circuit): Anions and cations move to compensate excess charge.

Zn is oxidized to Zn 2+ and 2e . The electrons flow to the anode where they are used in the reduction reaction. We expect the Zn electrode to lose mass and the Cu electrode to gain mass. “Rules” of voltaic cells: 1. At the anode electrons are products. Oxidation occurs at the anode. 2. At the cathode electrons are reagents. Reduction occurs at the cathode 3. Electrons cannot swim!

VOLTAIC CELLS Cell voltage (EMF or E cell ) is the measure of rxn spontaneity E cell : Intensive property, energy per electron Cell voltage depends on: the individual species undergoing oxidation and reduction and their concentrations The more spontaneous a reaction, the higher the voltage (more positive) the higher the K eq the more negative the  G 0

STANDARD POTENTIAL FOR AN ELECTROCHEMICAL CELL The standard potential for an electrochemical cell is the potential (voltage) generated when reactants and products of a redox reaction are in their standard states. Standard States: T = 25°C. Gases, P = 1 atm. [Solutes] = 1M Solids, liquids = pure

HALF-CELL POTENTIAL The potential associated with the half-reaction. Rules for half-cell potentials: 1.The sum of two half-cells potentials in a cell equals the overall cell potential: E° cell = E° 1/2 (oxid) + E° 1/2 (reduc) 2. For any half-reaction: E° 1/2 (oxid) =  E° 1/2 (reduc) 3.Standard half-cell is a hydrogen electrode: H 2 (g,1atm) = 2H + (aq, 1M) + 2e  E° 1/2 (oxid) = E° 1/2 (reduc) = 0 V

E 0 cell = 0.76 V

E 0 cell = 1.10 V

E 0 1/2 (Zn  Zn 2+ ) = ? E 0 1/2 (Cu 2+  Cu) = ?

The more positive E  red the stronger the oxidizing agent on the left. The more negative E  red the stronger the reducing agent on the right. A species that is higher and to the left will spontaneously oxidize one that is lower and to the right in the table. Example: F 2 will oxidize H 2 or Li; Ni 2+ will oxidize Al(s). Oxidizing and Reducing Agents

Are the following reactions spontaneous? If so, evaluate E 0 cell Cu(s) + Cl 2 (g)  Cu 2+ (aq) + 2Cl  (aq) 2Cu 2+ (aq) + 2H 2 O  Cu(s) + O 2 (g) + 4H + (aq)

RELATIONSHIP BETWEEN  G AND E  G =  nFE At Standard State:  G° =  nFE° n = number of electrons transferred in a balanced redox reaction F = Faraday = 96,500 coulomb/mole e - 1 coulomb = 1 Amp-sec 1 J = 1 Amp-sec-V = 1 coulomb-V 1 F = 96,500 J/V-mole e 

Example: Is this reaction spontaneous? Cd(s) + 2H +  Cd 2+ + H 2 (g) Cd e   Cd(s) E o red =  0.403V 2H + + 2e   H 2 E o red = 0 Reaction goes as written (reduction) for more positive redox couple. H + cathode: 2H + + 2e   H 2 Cd anode: Cd  Cd e  E o cell = E o red (cathode) + E o oxid (anode) E o cell = =  G o =  nFE o =  2 x 96,500 J/V-mol x 0.403V =  8.3 x 10 4 =  83 kJ/mol Yes: the reaction is spontaneous!

 G° =  RT log K eq  G° =  nFE° E° = RT log K eq nF R = J/K-mole F = 96,500 J/V-mole e  At 25°C = 298K: E° = (0.0592) log K eq n For electrochemical cell at equilibrium:  G = 0

Effect of Concentration Standard states: 1M solution, 1 atm gas pressure What if concentrations are different?

Effect of Concentration So for a half reaction: aA + bB + n e   cC + dD

Examples 1.What is E cell for a fuel cell running in air (P O2 = 0.2 atm), at pH = 2, with P H2 = 1 atm? 2.What is the half cell potential of the Ag/Ag+ redox couple (E 0 = V) in a 1 M NaCl solution that contains solid AgCl (K sp = 1.1 x )? O 2 (g) + 4H + (aq) + 4e   2H 2 O(l) E o = V 2H + + 2e   H 2 E o = 0

Lead/Acid Battery DURING DISCHARGE Anode: Pb(s) + SO 4 2  (aq)  PbSO 4 (s) + 2e  Cathode: PbO 2 (s) + SO 4 2  (aq) + 4H + + 2e   PbSO 4 (s) + 2H 2 O Overall: Pb(s) + PbO 2 (s) + 2H 2 SO 4  2PbSO 4 (s) + 2H 2 O Batteries

Lead-Acid Battery

DRY CELL Anode: Zn(s) Zn(s)  Zn 2+ (aq) + 2e  Cathode: NH 4 Cl + MnO 2 + graphite paste 2NH 4 + (aq) + 2MnO 2 (s) + 2e   Mn 2 O 3 (s) + 2NH 3 (aq) + H 2 O In ALKALINE CELL, NH 4 Cl is replaced by KOH. Provides up to 50% more energy. Zn is used as a powder mixed with the electrolyte. Batteries

Dry Cell Alkaline Battery

Rechargeable Nickel-Cadmium Battery Anode: Cd metal Cd(s) + 2OH  (aq)  Cd(OH) 2 (s) + 2e  Cathode: NiO 2 (s) NiO 2 (s) + 2H 2 O + 2e   Ni(OH) 2 (s) + 2OH  (aq) Overall: Cd(s) + NiO 2 (s) + 2H 2 O  Cd(OH) 2 (s) + Ni(OH) 2 (s) Batteries

FUEL CELLS H 2 -O 2 Fuel Cell - expensive but light. Used in spacecraft. Anode: 2H 2 (g) + 4OH  (aq)  4H 2 O(l) + 4e  Cathode: O 2 (g) + 2H 2 O(l) + 4e   4OH  (aq) Overall: 2H 2 (g) + O 2 (g)  2H 2 O(l) E 0 cell = 1.23 V Batteries

Corrosion Differential aeration mechanism

CATHODIC PROTECTION OF IRON Corrosion

CATHODIC PROTECTION OF IRON Corrosion

ELECTROLYSIS Electrolysis: Driving non-spontaneous reactions by applying electrical energy. An electrolysis cell consists of two electrodes in either aqueous solution (of ions) or in a molten salt e.g. molten NaCl. The anode is where oxidation occurs. Anions migrate to the anode and lose electrons. The cathode is where reduction occurs. Cations migrate to the cathode and gain electrons.

ELECTROLYSIS OF MOLTEN NaCl Cathode: 2Na + + 2e   2Na(l) Anode: 2 Cl   Cl 2 (g) + 2e  ___________________________ 2 Na Cl   2Na(l) + Cl 2 (g)

ELECTROLYSIS OF NaCl, cont. Electrolysis of aqueous NaCl: Cathode:H 2 O + 2e   H 2 (g) + 2OH  Anode: 2 Cl   Cl 2 (g) + 2e  ___________________________________ 2 H 2 O + 2 Cl   H 2 (g) + Cl 2 (g) + 2OH  so 2 Na OH  = 2 NaOH is left behind It is easier to reduce H 2 O than Na + The easiest (least non-spontaneous) reaction happens

ELECTROLYSIS - examples What products will form when an aqueous solution of ZnBr 2 is electrolyzed? What products will form when aqueous AgNO 3 is electrolyzed?

ELECTROLYSIS OF AQUEOUS Na 2 SO 4 Cathode: 4H 2 O + 4e   2H 2 (g) + 4OH  Anode: 2H 2 O  O 2 (g) + 4H + + 4e  _____________________________ 6H 2 O  2H 2 (g) + O 2 (g) + 4H + + 4OH  or 2H 2 O  2H 2 (g) + O 2 (g)

COMMERCIAL APPLICATIONS OF ELECTROLYSIS Production of metals – Na, Al. Purification of Metals – Cu. Electroplating.

Purification of Copper Cathode: thin sheet of pure copper Cu e   Cu(s) Anode:impure copper Cu(s)  Cu e  As the reaction proceeds, Cu moves from anode to cathode.

ELECTROLYSIS CALCULATIONS 1 mole of e - = 1 Faraday = 96,500 Coulombs = charge on 1 mole of e - 1 Ampere = 1 coulomb/second 1 coulomb = 1 Amp-sec Electromotive Force (EMF) force that cases electrons to flow (voltage) 1 Watt = 1 Amp-Volt 1 Joule = 1 coul-Volt = 1 Amp-sec-Volt = 1 Watt-sec 1 kW-hour = (1000 Watt)(3600 sec) = 3.6 x 10 6 Watt-sec = 3.6 x 10 6 Joules

ELECTROLYSIS CALCULATION Electrolysis gives 1.00g of Cu from CuSO 4 Reaction is: Cu e   Cu How many Faradays (F) of charge are required? How many Coulombs is this?

ELECTROLYSIS CALCULATION If 1.00g of Cu is obtained in 1 hour, how many amps are required? If 2 amps were used, how long would it take to produce 1g?