Electrochemistry Redox Oxidation States Voltaic Cells Balancing Equations Concentration Cells Nernst Equation CorrosionBatteriesElectrolysis Equilibrium Cell EMF 9/12/2015

Oxidation and Reduction Metal undergoes corrosion, it loses electrons to form cations:Ca(s) +2H + (aq)  Ca 2+ (aq) + H 2 (g) Voltaic Cell:Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Oxidized: atom, molecule, or ion becomes more positively charged. –Oxidation is the loss of electrons. Reduced: atom, molecule, or ion becomes less positively charged. –Reduction is the gain of electrons. Oxidation-Reduction Reactions

Metal undergoes corrosion, it loses electrons to form cations: Ca(s) +2H + (aq)  Ca 2+ (aq) + H 2 (g) Voltaic Cell: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

Oxidation and Reduction Oxidation-Reduction Reactions

Oxidation Numbers Oxidation numbers are assigned by a series of rules: 1.If the atom is in its elemental form, the oxidation number is zero. e.g., Cl 2, H 2, P 4. 2.For a monatomic ion, the charge on the ion is the oxidation state. 3.Nonmetal usually have negative oxidation numbers: (a)Oxidation number of O is usually –2. The peroxide ion, O 2 2-, has oxygen with an oxidation number of –1. (b)Oxidation number of H is +1 when bonded to nonmetals and –1 when bonded to metals. (c)Halogens generally -1, oxid. # of F is always –1. 4.The sum of the oxidation numbers for the atom is the charge on the molecule (zero for a neutral molecule). Oxidation-Reduction Reactions

The Periodic Table Alkali metals Alkaline Earth metals Halogens Noble Gases Chalcogens Transition Metals Inner Transition Metals

Ions and Ionic Compounds Predicting Ionic Charge

Common (Type II) Cations IonSystemic Name Older Name Fe 3+ Iron(III)Ferric Fe 2+ Iron(II)Ferrous Cu 2+ Copper(II)Cupric Cu + Copper(I)Cuprous Co 3+ Cobalt(III)Cobaltic Co 2+ Cobalt(II)Cobaltous Sn 4+ Tin(IV)Stannic Sn 2+ Tin(II)Stannous Pb 4+ Lead(IV)Plumbic Pb 2+ Lead(II)Plumbous

Oxidation Numbers Oxidation-Reduction Reactions 1. Determine the oxidation numbers (oxidation states) on the Sulfur atom for:(a) H 2 S(b) S 8 (c) SCl 2 (d) Na 2 SO 3 (e) SO Determine o.s. for the Redox reaction in Ni-Cd Batteries: Cd(s) + NiO 2 (s) + 2H 2 O(l)  Cd(OH) 2 (s) + Ni(OH) 2 (s) 3.Determine o.s. for the following Redox reaction: Al(s) + MnO 4 - (aq) + 2H 2 O(l)  Al(OH) 4 - (aq) + MnO 2 (s) HW-Answers

Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end. Conservation of charge: electrons are not lost in a chemical reaction. Half Reactions Half-reactions are a convenient way of separating oxidation and reduction reactions. Balancing Redox Reactions

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

Balancing Equations by the Method of Half Reactions 1. Write down the two half reactions. 2. Balance each half reaction: a. First with elements other than H and O. b. Then balance O by adding water. c. Then balance H by adding H +. d. Finish by balancing charge by adding electrons. 3. Multiply each half reaction to make the number of electrons equal. 4. Add the reactions and simplify. 5. Check for balanced atoms and charges! Balance: MnO 4 - (aq) + C 2 O 4 2- (aq)  Mn 2+ (aq) + CO 2 (g) (Acidic) Balancing Redox Reactions

Balance: MnO 4 - (aq) + C 2 O 4 2- (aq)  Mn 2+ (aq) + CO 2 (g) (Acidic)

Balancing Equations for Reactions Occurring in Basic Solution Use OH - and H 2 O rather than H + and H 2 O. Follow same method as in Acidic Solution, but OH - is added to “neutralize” the H + used. Consider: CN - (aq) + MnO 4 - (aq)  CNO - (aq) + MnO 2 (s) [ Basic Solution ] Balancing Redox Reactions

CN - (aq) + MnO 4 - (aq)  CNO - (aq) + MnO 2 (s) [ Basic Solution ] Solution Key

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+. Voltaic Cells Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

Flow of electrons from anode to cathode is spontaneous. Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode. Potential difference: difference in electrical potential. Measured in volts. One volt is the potential difference required to impart one joule of energy to a charge of one coulomb: Cell EMF

Electromotive force (emf) is the force required to push electrons through the external circuit. Cell potential: E cell is the emf of a cell. For 1M solutions at 25  C (standard conditions), the standard emf (standard cell potential) is called E  cell. Cell EMF

Standard Reduction (Half-Cell) Potentials Convenient tabulation of electrochemical data. Standard reduction potentials, E  red are measured relative to the standard hydrogen electrode (SHE). Cell EMF

Standard Reduction (Half-Cell) Potentials Reactions with E  red < 0 are spontaneous oxidations relative to the SHE. The larger the difference between E  red values, the larger E  cell. In a voltaic (galvanic) cell (spontaneous) E  red (cathode) is more positive than E  red (anode). Cell EMF Calculate E o cell for: 2 Al(s) + 3 I 2 (s)  2 Al 3+ (aq) + 6 I - (aq)

In a voltaic (galvanic) cell (spontaneous) E  red (cathode) is more positive than E  red (anode) since More generally, for any electrochemical process A positive E  indicates a spontaneous process (galvanic cell). A negative E  indicates a nonspontaneous process. Spontaneity of Redox Reactions

EMF and Free-Energy Change Can show that under Standard Conditions:  G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E o is the emf of the cell. Since n and F are positive, if  G > 0 then E < 0. Spontaneity of Redox Reactions Calculate E o and ΔG o for: (a)4Ag(s) + O 2 (g) + 4H + (aq)  4Ag + (aq) + 2H 2 O(l) (b)2Ag(s) + ½O 2 (g) + 2H + (aq)  2Ag + (aq) + H 2 O(l)

Calculate E o and ΔG o for: (a)4Ag(s) + O 2 (g) + 4H + (aq)  4Ag + (aq) + 2H 2 O(l) (b)2Ag(s) + ½O 2 (g) + 2H + (aq)  2Ag + (aq) + H 2 O(l)

Calculate E o and ΔG o for: (a)4Ag(s) + O 2 (g) + 4H + (aq)  4Ag + (aq) + 2H 2 O(l) (b)2Ag(s) + ½O 2 (g) + 2H + (aq)  2Ag + (aq) + H 2 O(l)

Calculate E o and ΔG o for: (a)4Ag(s) + O 2 (g) + 4H + (aq)  4Ag + (aq) + 2H 2 O(l) (b)2Ag(s) + ½O 2 (g) + 2H + (aq)  2Ag + (aq) + H 2 O(l)

The Nernst Equation The Nernst equation can be simplified by collecting all constants together using a temperature of 298 K: n is number of moles of electrons. Effect of Concentration on Cell EMF Calculate the emf for the Zn-Cu Voltaic cell with [Cu 2+ ] = 1.50 M and [Zn 2+ ] = M.

Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

Concentration Cells Effect of Concentration on Cell EMF

Cell EMF and Chemical Equilibrium A system is at equilibrium when  G = 0. From the Nernst equation, at equilibrium and 298 K (E = 0 V and Q = K eq ): Effect of Concentration on Cell EMF

A battery is a self-contained electrochemical power source with one or more voltaic cell. When the cells are connected in series, greater emfs can be achieved. BatteriesBatteries

Lead-Acid Battery A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V. Cathode: PbO 2 on a metal grid in sulfuric acid: PbO 2 (s) + SO 4 2- (aq) + 4H + (aq) + 2e -  PbSO 4 (s) + 2H 2 O(l) Anode: Pb: Pb(s) + SO 4 2- (aq)  PbSO 4 (s) + 2e - BatteriesBatteries

Lead-Acid Battery The overall electrochemical reaction is PbO 2 (s) + Pb(s) + 2SO 4 2- (aq) + 4H + (aq)  2PbSO 4 (s) + 2H 2 O(l) for which E  cell = E  red (cathode) - E  red (anode) = ( V) - ( V) = V. Wood or glass-fiber spacers are used to prevent the electrodes from touching. BatteriesBatteries

Alkaline Battery Anode: Zn cap: Zn(s)  Zn 2+ (aq) + 2e - Cathode: MnO 2, NH 4 Cl and C paste: 2NH 4 + (aq) + 2MnO 2 (s) + 2e -  Mn 2 O 3 (s) + 2NH 3 (aq) + 2H 2 O(l) The graphite rod in the center is an inert cathode. For an alkaline battery, NH 4 Cl is replaced with KOH. BatteriesBatteries

Fuel Cells Direct production of electricity from fuels occurs in a fuel cell. On Apollo moon flights, the H 2 -O 2 fuel cell was the primary source of electricity. Cathode: reduction of oxygen: 2H 2 O(l) + O 2 (g) + 4e -  4OH - (aq) Anode: 2H 2 (g) + 4OH - (aq)  4H 2 O(l) + 4e - BatteriesBatteries

Corrosion of Iron Since E  red (Fe 2+ ) < E  red (O 2 ) iron can be oxidized by oxygen. Cathode: O 2 (g) + 4H + (aq) + 4e -  2H 2 O(l). Anode: Fe(s)  Fe 2+ (aq) + 2e -. Dissolved oxygen in water usually causes the oxidation of iron. Fe 2+ initially formed can be further oxidized to Fe 3+ which forms rust, Fe 2 O 3.xH 2 O(s). CorrosionCorrosion

Corrosion of Iron Oxidation occurs at the site with the greatest concentration of O 2. Preventing Corrosion of Iron Corrosion can be prevented by coating the iron with paint or another metal. Galvanized iron is coated with a thin layer of zinc. CorrosionCorrosion

Preventing Corrosion of Iron To protect underground pipelines, a sacrificial anode is added. The water pipe is turned into the cathode and an active metal is used as the anode. Often, Mg is used as the sacrificial anode: Mg 2+ (aq) +2e -  Mg(s), E  red = V Fe 2+ (aq) + 2e -  Fe(s), E  red = V CorrosionCorrosion

Electrolysis of Aqueous Solutions –In electrolytic cells the anode is positive and the cathode is negative. (In galvanic cells the anode is negative and the cathode is positive.) ElectrolysisElectrolysis

Electroplating Active electrodes: electrodes that take part in electrolysis. Example: electrolytic plating. ElectrolysisElectrolysis

Electrochemistry Redox Oxidation States Voltaic Cells Balancing Equations Concentration Cells Nernst Equation CorrosionBatteriesElectrolysis Equilibrium Cell EMF Oxidation is the loss of electrons