Electrochemistry. Chapter 18 Electrochemistry and Its Applications. Redox Reactions. Redox Reactions. Redox Reactions

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1 John W. Moore Conrad L. Stanitski Peter C. Jurs Chapter 18 Electrochemistry and Its Applications Stephen C. Foster Mississippi State University Electrochemistry Electrochemistry is the study and use of e flow in chemical reactions. Redox reactions generate (and use) e Those e can be harnessed (batteries). Corrosion is an electrochemical reaction. Applied e flow: Can drive reactantfavored reactions toward products. Rechargeable batteries, electrolysis, and electroplating Redox Reactions Oxidation Number Refresher Pure element = 0. Monatomic ion = charge of ion. (ox. numbers in a species) = overall charge. Element ox. no. Exceptions? F 1 None Cl, Br, I 1 Interhalogens H +1 Metal hydrides = 1 O Metal peroxides = 1 Halogen oxides Redox Reactions Oxidation & reduction (Redox) always occur together. Reduction = gain of e = decrease in ox. no. Oxidation = loss of e = increase in ox. no. + e HCl(aq) + Mg(s) H (g) + MgCl (aq) e H + is reduced, Mg is oxidized. Redox Reactions Give oxidation numbers for each atom. Identify the oxidizing and reducing agents: 6 Fe + + Cr O H 3 O + 6 Fe 3+ + Cr H O Species Ox. number Explanation Fe + + charge on ion Cr O 7 O = ; Cr = +6 O is usually ; (Cr) + 7() = H 3 O + O = ; H = +1 O is usually ; H is usually +1 Fe charge on ion Cr charge on ion H O O = ; H = +1 O is usually, H is usually +1 Fe + Fe 3+ oxidation Cr(+6) Cr 3+ reduction Fe + = reducing agent Cr O 7 = oxidizing agent Using HalfReactions to Understand Redox Redox reactions split into half reactions: 1

2 Using HalfReactions to Understand Redox Halfreactions may include different numbers of e : Al(s) Al 3+ (aq) + 3 e Zn + (aq) + e e must balance in the full reaction. Zn(s) [ Al(s) Al 3+ (aq) + 3 e ] 3[ Zn + (aq) + e Zn(s) ] Al(s) + 3 Zn + (aq) Al 3+ (aq) + 3 Zn(s) Balancing Redox Equations Redox in acidic or basic solutions are harder (H O, H 3 O + or OH are often omitted ). Balance: H 3 AsO 4 + I HAsO + IO 3 which occurs in aqueous acidic solution. Balancing Redox Equations in Acidic Solution H 3 AsO 4 + I HAsO + IO 3 (acidic solution) (i) What is oxidized? Reduced? I (I = 0) IO 3 (I = +5) oxidation H 3 AsO 4 (As = +5) HAsO (As = +3) reduction (ii) Write unbalanced halfreactions: H 3 AsO 4 HAsO I IO 3 (iii) Balance atoms (except H and O). H 3 AsO 4 HAsO I IO 3 Balancing Redox Equations in Acidic Solution (iv) Balance O (add H O as needed). H 3 AsO 4 HAsO + H O I + 6 H O IO 3 (v) Balance H (add H + as needed). H 3 AsO 4 + H + HAsO + H O I + 6 H O IO H + (vi) Balance charges (add e ). H 3 AsO 4 + H + + e HAsO + H O I + 6 H O IO H e zero charge (1) + 1(+1)} 10(1) 0 = Balancing Redox Equations in Acidic Solution (vii) Equalize e and add. 5 [ H 3 AsO 4 + H + + e HAsO + H O ] 1 [ I + 6 H O IO H e ] 5 H 3 AsO H e + I + 6 H O H + 5 HAsO + 10 H O + IO H e 4H O 5 H 3 AsO 4 + I 5 HAsO + 4 H O + IO 3 + H + (viii) Make H 3 O + (H O + H + ). Add H O if needed. 5 H 3 AsO 4 + I 5 HAsO + H O + IO 3 + H 3 O + Balancing Redox Equations in Basic Solution Balance the following (basic conditions): N + S S + N H 4 (i) Oxidized? Reduced? N (N = 0) N H 4 (N = ) reduction S (S = ) S (S = 0) oxidation (ii) Unbalanced halfreactions: N N H 4 (iii) Balance (except H and O). N N H 4 S S S S

3 Balancing Redox Equations in Basic Solution (iv) Balance O (add H O as needed). N N H 4 S S (v) Balance H (add H + as needed). N + 4 H + N H 4 S S (vi) Balance charges (add e ). N + 4 H e N H 4 S S + e Balancing Redox Equations in Basic Solution (vi) Equalize e and add. 1 [ N + 4 H e N H 4 ] [ S S + e ] N + 4 H e + S N H 4 + S + 4 e N + 4 H + + S N H 4 + S (vii) Make H O (H + + OH ). Add OH. N + 4 H OH + S N H 4 + S + 4 OH N + 4 H O + S N H 4 + S + 4 OH Electrochemical Cells Cu + (aq) + Zn(s) Cu(s) + Zn + (aq) Electrochemical Cells Cu + (aq) + Zn(s) Cu(s) + Zn + (aq) e e Linked oxidation and reduction reactions. e move across an external conductor. e e Electrodes (anode & cathode) Allow e to pass in and out of solution. A salt bridge (or porous barrier) is required... Also called a voltaic cell or a battery. A battery is strictly a series of linked voltaic cells. Anode (oxidation) Salt bridge Cathode (reduction) Electrochemical Cells Salt bridge Contains a salt solution (e.g. K SO 4 ). Ions pass into the cells (restricts bulk flow). Stops charge buildup. Electrochemical Cells Zn SO 4 K SO 4 porous plug Zn + Cu + K + Cu SO 4 released as Zn Zn + K + released as Cu + Cu 3

4 Electrochemical Cells Zinc is removed: Zn Zn + + e Oxidation at the anode (both vowels). Zn supplies e. Anode has charge. Copper is deposited: Cu + + e Reduction at the cathode (both consonants). Cu + accepts e. Cathode has + charge. Cu Electrochemical Cells A compact notation: Zn(s) Zn + (aq) Cu + (aq) Cu (s) anode cell cathode cell Current flows from anode to cathode. = phase boundary. = salt bridge. Details (e.g. concentration) listed after each species. Electrical work = charge x ΔE p = (number of e ) ΔE p SI Units Charge: 1 coulomb (C) = 1 ampere x second = 1 As Potential: 1 volt (V) = 1 J C 1 Voltage depends on cell chemistry size. Cell voltage varies if conditions vary. A standard voltage ( E ) occurs if: All [solute] = 1 M. or saturated if the solubility < 1 M. All gases have P = 1 bar. All solids are pure. Charge depends on n reactants size. E cell is positive = product favored reaction (E cell < 0 is reactant favored) Standard hydrogen electrode (SHE) Pt H (1bar), 1M H 3 O + Absolute voltages cannot be measured. They are measured relative to a standard electrode. E = 0 V (oxidation & reduction). H 3 O + (aq, 1M) + e H (g, 1 bar) + H O (l) 4

5 Using Standard Cell Potentials Tabulated as reductions. Reduction Half Reaction E (V) F (g) + e F (aq) +.87 H O (aq) + H 3 O + e 4 H O(l) MnO 4 (aq)+8 H 3 O e Mn + (aq) + 1 H O(l) Cl (g) + e Cl (aq) Br (g) + e Br (aq) Ag + (aq) + e Ag(s) Cu + (aq) + e Cu(s) H 3 O + (aq) + e H (g) + H O(l) 0.00 Ni + (aq) + e Ni(s) 0.5 Fe + (aq) + e Fe(s) 0.44 Zn + (aq) + e Zn(s) Al 3+ (aq) + 3 e Al(s) 1.66 Li + (aq) + e Li(s) Cu + (aq) + e Cu(s) reduction Zn(s) Zn + (aq) + e oxidation The overall voltage: E cell = E Zn+, reduction + E Cu, oxidation If an equation is reversed, E 1 x E E oxidation = E reduction. E tables are reduction values, so: E cell = E reduction + E oxidation E cell = E cathode E anode Reaction Process E red (Table) E Zn Zn + + e oxidation 0.76 V V Cu + + e Cu reduction V V Or E cell = E cathode E anode E cell = 0.34 (0.76) V = 1.10 V E cell = E oxid + E red = V = 1.10 V What is E for a Ni(s) Ni + Ag + Ag(s) cell? Reduction Half Reaction E (V) Ag + (aq) + e Ag(s) Ni + (aq) + e Ni(s) 0.5 Anode written on the left; cathode the right. E = E cathode E anode = (0.5) V = 1.05 V If Ni(s) Ni + (aq, 1M) is connected to SHE, the Ni electrode loses mass over time. E cell = 0.5 V. Is Ni oxidized or reduced? What is E for the Ni half cell? Ni loses mass: Ni(s) Ni + (aq) + e Ni is oxidized (anode). Since: E cell = E cathode E anode 0.5 V = E SHE E anode = 0 E anode E anode = 0.5 V (Note: this is the tabulated reduction value) Using Standard Cell Potentials 1. Tabulated halfcell E are reductions.. Reactions can be reversed. 3. More positive E = easier reduction. 4. Less positive E = easier oxidation (for the reverse reaction). 5. A left species, will oxidize any right below it. 6. E depends on [reactant] and [product], but not on n reactant or n product (i.e not stoichiometric coefficient). 5

6 Using Standard Cell Potentials Will Zn(s) react with a 1 M iron(iii) solution? If so what is E for the reaction? Reduction Half Reaction E (V) F (g) + e F (aq) +.87 Fe 3+ (aq) + e Fe + (aq) H 3 O + (aq) + e H (g) + H O(l) +0.0 Zn + (aq) + e Zn(s) Using Standard Cell Potentials Will Zn react with a 1 M iron(iii) solution? If so what is E for the reaction. Fe 3+ + e Fe + E = V Zn Zn + + e E = 1(0.763 V) Fe 3+ + Zn Fe + + Zn + E cell = V Note x (Fe 3+ reaction) to balance e. Yes! Fe 3+ ( left ) is higher than Zn ( right ). E (Fe 3+ ) is not doubled. Using Standard Cell Potentials a) Will Al(s) react with a 1 M tin(iv) solution? b) Will 1 M Fe + react with Sn(s)? Sn 4+ + e Sn + (s) V Sn + + e Sn (s) 0.14 V Fe + + e Fe (s) 0.44 V Al e Al (s) 1.66 V (a) Yes left (Sn 4+ ) above right (Al). (b) No left (Fe + ) below right (Sn). Using Standard Cell Potentials What s the voltage of: Al(s) Al 3+,1M Sn +,1M Sn(s)? Sn + + e Sn E = 0.14 V Al Al e E = 1(1.66 V) Balance e : 3(Sn + + e Sn) E = 0.14 V (Al Al e ) E = V 3 Sn + + Al 3 Sn + Al 3+ E cell = +1.5 V E and Gibbs Free Energy E and Gibbs Free Energy Productfavored reactions: ΔG < 0. Spontaneous cell reactions: E cell > 0. ΔG = n F E cell Cu + + Zn(s) Cu(s) + Zn + Spontaneous. E cell = 1.10 V with n = moles of e transferred, F = Faraday constant = charge/(mol of e ). = (e charge) x (Avogadro s number). = ( x C)(6.014 x 10 3 mol 1 ). F = 96,485 C/mol = 96,500 C/mol (3 sig. fig.) ΔG = nfe cell = mol (96500 C/mol)(1.10 V) =.1 x10 5 J (1 J = 1 C V) = 1 kj 6

7 ΔG, E cell, and K Since: ΔG = RT ln K = nfe cell R T R T E cell = ln K = (.303) log K n F n F At 98 K: E cell = V ln K or, n V E cell = log K n ΔG, E cell, and K Determine K for: Cu + + Zn(s) Cu(s) + Zn + E cell = V log K n 1.10 V = V log K log K = K = = 1.5 x mol e E cell = 1.10 V Effect of Concentration on Cell Potential E values apply if [solute] = 1 M (or saturated). Other conditions: E cell = E cell At 98 K: E cell = E cell E cell = E cell RT ln Q nf V n V n Nernst equation log Q ln Q Effect of Concentration on Cell Potential What is the voltage for: Cu + + Zn(s) Cu(s) + Zn + if [Cu + ] = 0.1 M and [Zn + ] = 5.0 M. E cell = 1.10 V. E = E E = 1.10 = 1.05 V log log [Zn + ] [Cu + ] Concentration Cells Concentration dependence leads to: Zn Zn + (dilute) Zn + (conc.) Zn E cell 0 V Concentration Cells E = E log [Zn + ] dilute [Zn + ] conc Example Zn Zn + (0.01M) Zn + (1M) Zn anode = oxidation cathode = reduction E = log Zn(s) Zn + (0.01M) + e Zn + (1M) + e Zn(s) Zn + (1M) Zn + (0.01M) (net reaction) E = 0.0 log E = V 7

8 Common Batteries Primary battery One time use. Not easily rechargeable Secondary battery Rechargeable battery. Primary Batteries Alkaline Battery: Zn(s) + OH (aq) ZnO(aq) + H O(l) + e MnO (s) + H O(l) + e MnO(OH)(s) + OH (aq) Overall Zn(s) + H O(l) + MnO (s) ZnO(aq) + MnO(OH)(s) E cell = 1.54 V when new. Primary Batteries Mercury battery Secondary Batteries LeadAcid Battery (high capacity, high current). Pb(s) + HSO 4 (aq) + H O(l) PbSO 4 (s) + H 3 O + (aq) + e PbO (s) + 3 H 3 O + (aq) + HSO 4 (aq) + e PbSO 4 (s) + 5 H O Pb + PbO (s) + H 3 O + + HSO 4 (aq) PbSO 4 (s) + 4 H O Net E = V Zn(s) + OH ZnO(aq) + H O(l) + e HgO(s) + H O(l) + e Hg(l) + OH (aq) Zn(s) + HgO(s) Hg(l) + ZnO(aq) E cell = 1.35 V Insoluble PbSO 4 stays on the electrodes The reaction is reversed by recharging. LeadAcid Storage Battery 6 cells in series (1 V). Secondary Batteries NickelCadmium (Nicad). Cd(s) + OH (aq) Cd(OH) (s) + e [NiO(OH)(s) + H O(l) + e Ni(OH) (s)+ OH (aq)] Cd(s) + NiO(OH)(s) + H O Cd(OH) (s) + Ni(OH) (s) Insoluble (Rechargable ) net: E = V 8

9 Secondary Batteries Nickelmetal hydride (NiMH). Doesn t use toxic Cd. MH(s) + OH (aq) M(s) + H O(l) + e NiO(OH)(s) + H O(l) + e Ni(OH) (s) + OH (aq) MH(s) + NiO(OH)(s) M(s) + Ni(OH) (s) E cell =+1.4 V M is a metal alloy in KOH Secondary Batteries Lithium Ion. Low mass, high energy density. Li(s) (in polymer) Li + (in polymer) + e Li + ( in CoO ) + e + CoO LiCoO Li(s) + CoO (s) LiCoO (s) E cell = 3.4 V Fuel Cells Convert bond energy into electricity. ProtonExchange Membrane (PEM) fuel cell. H H + + e ½ O + H + + e H O Graphite electrodes. Pt catalyst coated on both sides of the membrane. H in gases flow through channels H out Anode V e e H + H + H + exchange membrane Pt catalyst O in Cathode H O out Electrolysis Electrolytic cell: Applied voltage forces a reaction to occur. e.g. electrolysis of molten NaCl: Na + + e Na (l) Cl Cl (g) + e Na + + Cl Cl (g) + Na(l) E =.714 V E = V E = 4.07 V Na(l) and Cl (g) produced if > 4.1 V is applied. However, melting NaCl takes lots of energy Electrolysis Aqueous solutions? Other reactions can occur. Consider KI(aq): Electrolysis reductions oxidations K + (aq) + e K(s) E =.95 V H O(l) + e H (g) + OH (aq) E = 0.88 V I (aq) I (aq) + e E = V 6 H O(l) O (g) + 4 H 3 O + (aq) + 4 e E = 1.9 V E ox 1 x E red The most positive E half reactions occur H O + I H + I + OH E cell = V Brown = I ; Purple = OH (phenolphthalein) 9

10 Electrolysis Aqueous oxidation will not occur. Aqueous reduction will not occur. Reduction Half Reaction E (V) F (g) + e F (aq) +.87 H O (aq) + H 3 O+ e 4 H O(l) Cl (g) + e Cl (aq) O (g) + 4 H 3 O + (aq) + 4 e 6 H O(l) +1.9 Br (g) + e Br (aq) Ag + (aq) + e Ag(s) H 3 O + (aq) + e H (g) + H O(l) 0.00 Ni + (aq) + e Ni(s) 0.5 Fe + (aq) + e Fe(s) 0.44 Zn + (aq) + e Zn(s) H O(l) + e H (g) + OH (aq) Al 3+ (aq) + 3 e Al(s) 1.66 Na + (aq) + e Na(s).714 K + (aq) + e K(s).95 Electrolysis Summary Metal ions are reduced if E red > 0.8 V Aqueous Na +, K +, Mg +, Al 3+ cannot be reduced. Anions can be oxidized if E red < +1. V Aqueous F cannot be oxidized. In practice E required > E calculated. Overvoltage is needed Cl (aq) can be oxidized to Cl (g). Electrolysis of Brine Chlorine is produced from NaCl(aq) by the chloralkali process. Cl (aq) Cl (g) + e H O(l) + e OH (aq) + H (g) Counting Electrons Cu + (aq) + e Cu(s) mol e 1 mol Cu. Also charge = current x time 1 coulomb = 1 ampere x 1 second 1 C = 1 A s NaOH(aq) is 1 30% NaOH by weight. Counting Electrons Determine the mass of Cu plated onto an electrode from a Cu + solution by the application of a 10. A current for 10. minutes. Cu + (aq) + e Cu(s) Charge = 10. A x 10. min x (60 s/min) = 6.0 x 10 3 C Counting Electrons What is the cost to produce 14. g of Al (mass of a soda can) by the reduction of Al 3+. Assume V = 4.0 V and 1 kwh of electricity costs 10 cents. Al e 14. g 1 mol Al 3 mol e C 6.98 g 1 mol Al 1 mol e = 1.5 x 105 C Al 6.0 x10 1 mol e g 3 C 1 mol Cu C mol e 1 mol Cu =.0 g Energy used = (1.5 x10 5 C)(4.0V) 1 J 1 C x 1V = 0.17 kwh (or 1.7 cents) 1 kwh 3.60 x 10 6 J 10

11 Corrosion: ProductFavored Reactions Oxidation of metals by the environment. Anode: M(s) M n+ + n e Cathode: often involve water and/or O O (g) + H O(l) + 4 e 4 OH (aq) H O(l) + e OH (aq) + H (g) Anode and cathode must be electrically connected. (The metal itself acts as the conductor). Corrosion: ProductFavored Reactions Iron rusts Anode: [Fe(s) Fe + + e ] Cathode: O (g) + H O(l) + 4 e 4 OH (aq) Fe(s) + O (g) + H O(l) Fe(OH) (s) Iron(II) hydroxide is converted to rust by O and H O: Rust (Fe O 3 xh O(s); x = 4) Corrosion: ProductFavored Reactions Iron nails corroding: Blue = Fe + indicator Purple = OH indicator Stress points corrode quickly. Corrosion Protection Anodic inhibition Paint or coat the surface. Form thin metal oxide coat: Fe(s) + Na CrO 4 (aq) + H O(l) Fe O 3 (s) + Cr O 3 (s) + 4 NaOH(aq) Insoluble coating; impervious to O and H O Cathodic protection Attach a more reactive metal which will corrode first. Corrosion Prevention Iron can also be galvanized (coated with Zn): Zn(OH) (insoluble film) forms on the surface. Zn Zn + + e E = V Fe Fe + + e E = 0.44 V 11

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