Electrochemistry. Chapter 17 Electrochemistry GCC CHM152. Ox # examples. Redox: LEO the lion goes GER. Oxidation Numbers (Chapter 4).

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1 Chapter 17 Electrochemistry GCC CHM152 Electrochemistry Electrochemistry is the study of batteries and the conversion between chemical and electrical energy. Based on redox (oxidation-reduction) reactions in which one substance gains electrons and another loses electrons. These two processes MUST happen together. Single replacement, combustion, combination and decomposition rxns are all examples of redox reactions. Oxidation Numbers (Chapter 4). Free elements: ox # = 0 (H 2 (g), Hg(l), etc.) Ions in binary ionic compounds: ox # = charge. Ex. For Al 2 S 3, ox # for Al = +3, ox # for S = -2. H: ox # = +1, except when H is with alkali metals it is -1 (LiH, NaH, etc.) O: ox # = -2, except in peroxides it is -1 (H 2 O 2, K 2 O 2 ) The sum of the oxidation numbers of all the atoms in a molecule = 0 For polyatomic ions, the sum of the oxidation numbers must equal the charge on the ion. Ox # examples Find ox numbers for all the atoms in HCO 3 - H: ox # = +1 O: ox # = -2 1H + 1C + 3O = C + 3x(-2) = -1 Ox # for C = +4 What is the ox # for Cr in Cr 2 O 7 2-? 2Cr + 7O = -2 2Cr + 7(-2) = -2 Cr = +6 Redox: LEO the lion goes GER Oxidation: loss of electrons; ox #, more + Reduction: gain of electrons; ox #, more oxidizing agent: substance that is reduced; it caused oxidization of other substace. reducing agent: substance that is oxidized; it caused reduction of other substance. Zn(s) + Cu(NO 3 ) 2 (aq) Zn(NO 3 ) 2 (aq) + Cu(s) Identify atom oxidized, atom reduced, oxidizing agent and reducing agent. Zn(s) + Cu(NO 3 ) 2 (aq) Zn(NO 3 ) 2 (aq) + Cu(s) Zn(s) Zn 2+ (aq) + 2e- Zn(s) dissolves Cu 2+ (aq) + 2e- Cu(s) Cu plates on electrode 1

2 Redox Reactions What is oxidized/reduced, give the specific atom. Agents: give the whole substance as an answer. MnO 4- (aq) + 8H + (aq) + 5Fe 2+ (aq) Mn 2+ (aq) + 5Fe 3+ (aq) + 4H 2 O(l) For this reaction: What is oxidized? What is reduced? What is the oxidizing agent? What is the reducing agent? Balancing Redox Reactions Cr 3+ (aq) + Be(s) Cr(s) + Be 2+ (aq) First, break it up into two half reactions, the oxidation and reduction half reactions. Balance charge by adding electrons (e-) to the more positive side for each reaction. Oxidation: Be(s) Be 2+ (aq) + 2e - Reduction: Cr 3+ (aq) + 3e - Cr(s) Balancing Redox Rxns contd Balance electrons by multiplying each half reaction by an integer so that the # electrons gained = # electrons lost. Oxidation: (Be(s) Be 2+ (aq) + 2e - )x3 Reduction: (Cr 3+ (aq) + 3e - Cr(s))x2 2Cr 3+ (aq) + 3Be(s) 2Cr(s) + 3Be 2+ (aq) Types of Cells Galvanic cell (also called voltaic cell) Electrochemical cell in which a spontaneous reaction generates electricity. Electrolytic cell Electrochemical cell in which an electrical current is used to drive a nonspontaneous reaction. Batteries Galvanic Cells: Figure17.2 Batteries are examples of galvanic cells we use in everyday life. For a Voltaic cell the voltage keeps dropping as the spontaneous reaction proceeds right. Battery is dead when E cell = 0 (at equilibrium) Bigger batteries only last longer, they don t have more volts. Volts are determined by the chemical reaction that occurs. 2

3 Galvanic Cells Oxidation occurs at the anode - both vowels mass of anode decreases - it is dissolving as metal atoms lose electrons to form ions in solution Reduction occurs at the cathode - both consonants mass of cathode increases as metal ions are reduced to form atoms that plate onto the cathode. External Circuit - electrons flow from the anode to the cathode via an external wire. Salt bridge - soluble salt solution in a bridge that connects the two half cells; ions flow through the bridge to complete the electrical circuit. Dr. Lisa s cell Mnemonic Fat red cat eats electons! Anorexic ox spits them out! Cathode is reduced (gains e - ) & mass (plating) Anode is oxidized (loses e - ) & mass (dissolves) e- e- What happens in Salt Bridge? Migration of ions maintains charge neutrality in both compartments: Anions move into the anode where excess + charge builds up as metal cations are formed by oxidation. Cations move into the cathode where excess (-) charge builds up as the metal cations are reduced to form neutral metal atoms. Short-hand Notation Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) Anode anode ion cathode ion cathode indicates a phase boundary between two phases in the same cell. denotes the salt bridge between the cells. anode is always on the left side, and the cathode on the right side. Electrodes are always on the two ends Shorthand Notation for Galvanic Cells Anode half-reaction: Zn(s) Zn 2+ (aq) + 2e- Cathode half-reaction: Cu 2+ (aq) + 2e- Cu(s) Overall cell reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Salt bridge Anode half-cell cathode half-cell Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) Inert Electrodes Pt and graphite are typically electrodes for gas phase and aqueous reactions. Ex. Standard Hydrogen Electrode (SHE) uses Pt electrode. (figure 17.4) SHE consists of Pt electrode in contact with 1 M H + solution and H 2 gas at 1 atm pressure H 2 (g) 2H + (aq) + 2e - Phase boundary Chapter 17/ Pearson Education, Inc. 3

4 Galvanic Cells - Pt electrode Figure 17.4 Electromotive Force emf = E cell = cell voltage: Electromotive force is the cell potential measured in volts. This is the driving force that pushes electrons away from the anode and towards the cathode. Joules = Coulombs x Volts E cell is measured in volts: V = J/C coulomb - the quantity of charge that passes a point in 1 sec when a current of 1 ampere flows. Free Energy and Emf DG = -nfe n = number of moles of e- transferred E = Emf of cell F = Faraday's constant Coulombs J 1F 1mol e V mol e spontaneous reaction: DG < 0 and E > 0 nonspontaneous reaction: DG > 0 and E < 0 At equilibrium: DG = 0 and E = 0 Standard Cell Potential, E o Gases at 1 atm, Solutions at 1 M, Temperature at 298 K (25 o C) Standard potential for any galvanic cell is the sum of the half-cell potentials. E o cell = E o ox + E o red All cell potentials are compared to hydrogen (SHE: standard hydrogen electrode) 2H + + 2e - H 2 (g) E o red = 0 V Reduction Potentials We can use the table of Standard Reduction Potentials for reduction half reactions to determine the cell potential of a galvanic cell. Table 17.1: All potentials are listed as reduction potentials. Oxidation potentials: reverse the reaction and E ( anode) E ( cathode) ox red Calculating Cell potentials E cell E red ox ( cathode) E ( anode) For the oxidation reaction at the anode, make sure you reverse the reaction and change the sign for E ox, then add the reduction and oxidation potentials. 4

5 Strength of Oxidizing Agents Oxidizing Agents: F 2 (g) + 2e - 2F - (aq) E o red = 2.87 V F 2 has the most positive E o red value. F 2 is the easiest to reduce. (It wants e- the most!) Thus F 2 is strongest oxidizing agent. As E o red, strength ox agent Strength of Reducing Agents Li + (aq) + e - Li(s) E o red = V Li has the most negative E o red value, so it is the easiest to oxidize. Li(s) is strongest reducing agent. Li(s) wants to lose electrons, so reverse reaction occurs: Li(s) Li + (aq) + e - E o ox = V Table 17.1 info continued Active metals tend to be good reducing agents. (red agent = ox = lose e-) Active nonmetals tend to be good oxidizing agents. (ox agent = red = gain e-) E is intensive; it does not depend on # of moles involved. Don't need to multiply E by factor if coefficients of reaction are changed. Cell Potentials Based on their E o red values, determine the best oxidizing agent, best reducing agent, worst oxidizing agent, and worst reducing agent. Au e - Au(s) E o red = 1.50 V Br 2 (l) + 2e - 2Br - (aq) E o red = 1.07 V Pb e - Pb(s) E o red = V Ni e - Ni(s) E o red = V Cell Potentials Ni(s) is the easiest to oxidize, best reducing agent. Au 3+ is the easiest to reduce, best oxidizing agent. Au(s) is the hardest to oxidize, worst reducing agent. Ni 2+ is the hardest to reduce, worst oxidizing agent. 5

6 Cell Potentials Positive E o cell means the reaction is productfavored and spontaneous. A negative E o cell means the reaction won t happen in the forward direction. Therefore, we want two half reactions that yield the most positive E o cell value. Assign rxn with more + E o red as cathode! E o cell = E o red + E o ox Cell Potentials in Reactions What will E o cell be if we react Ni(s) with Au 3+? What will E o cell be if Pb 2+ reacts with Br -? Calculate E o cell for the following cell Ni(s) Ni 2+ (aq) Br 2 (l), Br - (aq) Pt(s) Which are spontaneous? Example: Cl 2 (g) + Zn(s) For a spontaneous reaction (+ E o cell), write the appropriate anode and cathode reactions: Cl 2 (g) + 2 e - 2 Cl - (aq) E red = 1.36 V Zn 2+ (aq) + 2 e - Zn (s) E red = V Cl 2 (g) has more + E red, so it s the cathode reaction Cathode: Cl 2 (g) + 2 e - 2 Cl - (aq) E red = 1.36 V Zn reaction is reversed for the anode: Anode: Zn (s) Zn 2+ (aq) + 2 e - E ox = V The E ox reaction is flipped so the sign is the opposite! E o cell = = 2.12 V Cell Potentials of Reactions If we were to make a Galvanic cell from the following metals, which would act as the anode and which as the cathode? Refer to table Write the half-cell reactions and the overall balanced reaction for each cell. Calculate E o cell for each one. Also calculate E o cell for each one. Fe(s) and Ni(s) Cu(s) and Ag(s) The Nernst Equation: Calculate emf for nonstandard conditions Recall: DG = DG o + RT ln Q Plugging in DG = - nfe and DG o = - nfe o Nernst Equation: At 298 K: -nfe = -nfe o + RT ln Q 2.303RT E E logq nf RT E E lnq nf J R K mol V E E logq n, F J V mol e Both [ ] s & P s can be plugged into Q; V = volts (unit) Standard emf & K At equilibrium, E = 0 and Q = K (plug these conditions into Nernst Equation) RT 0 E RT lnk E lnk nf nf J R K mol At 298 K, E, F J V mol e V log K n 6

7 Calculations Calculate DG o for the Ni(s) + Au 3+ reaction. (#4a) Calculate K at 25 o C for Ni(s) + Br 2 cell (#4c): Ni(s) Ni 2+ (aq) Br 2 (l), Br - (aq) Pt(s) K e nfe RT or K 10. ne V Cell Potentials Summary Positive E o cell value DG o is negative K is large Reaction is product-favored Negative E o cell value DG o is positive K is small Reaction is reactant-favored The Nernst Equation A galvanic cell utilizes the following reaction: 2Ag + + Pb(s) Pb Ag(s) a) Write the half reactions and calculate E o cell. b) Write the short-hand notation for this cell. c) Calculate the cell potential if [Pb(NO 3 ) 2 ] = 0.88 M and [AgNO 3 ] = 0.14 M. Corrosion The oxidative deterioration of a metal (i.e., solid metal converted to ions). Rust formation is the corrosion of iron. Metals can be plated with non-reactive metals to protect them (chromium, tin, or zinc are common). Electrolysis Electrolytic Cell: Electrical energy from an external source (outlet or a battery) is used to force a nonspontaneous redox reaction to proceed. Molten salt: 2NaCl(l) 2Na(s) + Cl 2 (g) Cathode: 2Na + (l) + 2e- 2Na(s) Anode: 2Cl - (l) Cl 2 (g) + 2e- E cell = - 4 V This reaction naturally wants to run in reverse direction. We need more than 4 volts to drive this reaction forward. Salts don t normally decompose into elements. Electrolysis of Water Water doesn t naturally decompose into hydrogen and oxygen. 2H 2 O(l) 2H 2 (g) + O 2 (g) E cell = V Anode: 2H 2 O(l) O 2 (g) + 4H + (aq) + 4e - Cathode: 4H 2 O(l) + 4e - 2H 2 (g) + 4OH - (aq) The electrodes are the still the same for electrolysis: (oxidation at anode, reduction at cathode). 7

8 Electrolysis Calcs Used to find mass or volume of product produced by passing current through cell. Current: measured in Amps A = C/s ampere: unit of electric current; rate of flow of e-. charge = current x time coulombs = amps x seconds Electrolysis Calculations How many grams of Cu can be collected in 1.00 hour by a current of 1.62 A from a CuSO 4 solution? Reduction reaction: Cu e- Cu(s) Calculate coulombs: Current (C/s) x time (s) = C C mol e - mol solid g solid 1.62C 1 mole 1molCu 63.5g 3600s = 1.92 g Cu s 96,500C 2mole mol Worked Examples 17.10, 17.11; Problems 17.22, Batteries Household batteries Lead storage (i.e., car batteries) Dry-cell (or Laclanche) batteries 9V battery sum of 1.5 Volts Ni-Cad Batteries - rechargeable Positive electrode: NiOOH Negative electrode: Cd 8

9 Lithium Ion Batteries Positive electrode: Lithium cobalt oxide Negative electrode: Carbon Fuel Cell Used for space travel, and in hydrogen fuel cars Reactants are stored external to the cell and introduced to the electrodes as they are needed. The reactants are usually gaseous, such as O 2 and H 2, or O 2 and CH 4, or O 2 and NH 3. H 2 + 2OH - 2H 2 O + 2e - E o = 0.83 V O 2 + 2H 2 O + 4e - 4OH - E o = 0.40 V 2H 2 + O 2 2H 2 O E o = 1.23 V Fuel cells in action Fuel Cell 9

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