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Electrochemistry Chapter 19 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Electrochemical processes are oxidation-reduction reactions in which: • the energy released by a spontaneous reaction is converted to electricity or • electrical energy is used to cause a nonspontaneous reaction to occur Redox = Reduction - Oxidation Redox Reaction - A reaction where one species gains electrons and one species loses electrons. Must have both processes occurring. Remember Bronsted acid-base reactions where protions are transferred. One has to give, the other has to take. 0 0 2+ 2- 2Mg (s) + O2 (g) 2Mg O2 + 4e- 2MgO (s) 2Mg2+ + 4e- Oxidation half-reaction (lose e-) 2O2- Reduction half-reaction (gain e-) Reduction Involves Gain of electrons. (oxidation number ) OIL RIG Oxidation Involves Loss of electrons. (oxidation number ) Oxidizing agent - reactant of reduction O 0 -2 O2- Reducing agent - reactant of oxidation Mg 0 +2 Mg2+ Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. Number assigned to each atom to estimate its excess charge or deficiency. There is ALWAYS a change in oxidation numbers in a redox reaction. Rules for Determination of Oxidation Number 1. Free elements (uncombined state) and purely covalent compounds all have atoms with an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 2. In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 3. The oxidation number of oxygen is usually –2. In H2O2 and O22- it is –1. 4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1. • The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. What are the oxidation numbers of all the atoms in HCO3- ? HCO3From rules we know: O = -2 H = +1 Can calculate C: 3x(-2) + 1 + ? = -1 C = +4 4.4 What are the oxidation numbers of Mn in the following compounds and ions ? Mn - element oxidation # = 0 Mn2+ - ion oxidation # = +2 metal colorless MnO2 - compound oxidation # = +4 brown +4 + (2x(-2)) = 0 MnO4- - ion oxidation # = +7 purple +7 + (4x(-2)) = -1 MnO4-2 - ion oxidation # = +6 green +6 + (4x(-2)) = -2 Mn can have at least 5 oxidation states. Balancing Redox Reactions Reaction can be balanced in acidic, neutral or basic solutions. MnO4- + HSO3- Mn2+ + SO4-2 acid MnO4- + HSO3- MnO2 + SO4-2 neutral MnO4- + HSO3- • Balance mass. MnO42- + SO4-2 base • Balance charge. • Balance number of electrons exchanged (treat electrons as a reactant). Balancing Redox Equations The oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution? 1. Write the unbalanced equation for the reaction ion ionic form. Fe2+ + Cr2O72- Fe3+ + Cr3+ 2. Separate the equation into two half-reactions. +2 +3 Fe2+ Oxidation: +6 Reduction: Cr2O7 Fe3+ +3 2- Cr3+ Oxidation and reduction must take place simultaneously, but we can consider them separately for simplicity. Can look in tables of standard reduction potentials for half reaction or build the half reaction ourselves. 6e- + 14H+ + Cr2O72Fe2+ 2Cr3+ + 7H2O Fe3+ + 1e- NOTE: We reversed the direction of the oxidation reaction (table is of reduction potentials.). 3. Balance the atoms other than O and H in each half-reaction. Cr2O72- 2Cr3+ 4. For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. Cr2O7214H+ + Cr2O72- 2Cr3+ + 7H2O 2Cr3+ + 7H2O 5. Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe2+ 6e- + 14H+ + Cr2O72- Fe3+ + 1e2Cr3+ + 7H2O 19.1 Balancing Redox Equations 6. If necessary, equalize the number of electrons in the two halfreactions by multiplying the half-reactions by appropriate coefficients. 6Fe2+ 6Fe3+ + 6e6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 7. Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. Oxidation: 6Fe2+ 6Fe3+ + 6eReduction: 6e- + 14H+ + Cr2O72- 14H+ + Cr2O72- + 6Fe2+ 2Cr3+ + 7H2O 6Fe3+ + 2Cr3+ + 7H2O Overall reaction should be balanced if halfreactions are properly balanced. Balancing Redox Equations 8. Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O 9. For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation. 19.1 Balancing Redox Reactions Balance reaction in acidic solution. MnO4- + HSO3Mn2+ + SO4-2 Reduction: MnO4Mn2+ add H2O to balance O atoms. MnO4MnO4- + 8 H+ add H+ to balance H atoms. add electrons to balance charge. Mn2+ + 4 H2O Mn2+ + 4 H2O -1 + 8 +2 + 0 need 5 e-’s MnO4- + 8 H+ + 5 eMn2+ + 4 H2O Oxidation: HSO3SO4-2 SO4-2 add H2O to balance O atoms. HSO3- + H2O add H+ to balance H atoms. HSO3- + H2O add electrons to balance charge. HSO3- + H2O SO4-2 + 3 H+ -1 + 0 -2 + 3 need 2 e-’s SO4-2 + 3 H+ + 2 e- equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 2x(MnO4- + 8 H+ + 5 eMn2+ + 4 H2O) 5x(HSO3- + H2O SO4-2 + 3 H+ + 2 e- ) 2 MnO4- + 16 H+ + 10 e- + 5 HSO3- + 5 H2O 2 Mn2+ + 1 8 H2O + 5 SO4-2 + 15 H+ + 10 e3 2 MnO4- + H+ + 5 HSO3- Check balance: H Mn 1+5=3x2 2=2 O 8 + 15 = 3 + 20 2 Mn2+ + 3 H2O + 5 SO4-2 S 5=5 Charge - 2 + 1 - 5 = 4 - 10 -6 = -6 Balance reaction in neutral solution. MnO4- + HSO3MnO2 + SO4-2 Reduction: MnO4- MnO2 Choose half-reaction with neutral product from table. MnO4- + 4 H+ + 3 eOxidation: HSO3- MnO2 + 2 H2O SO4-2 Use same half-reaction as before. HSO3- + H2O SO4-2 + 3 H+ + 2 e- equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 2x(MnO4- + 4 H+ + 3 e3x(HSO3- + H2O MnO2 + 2 H2O) SO4-2 + 3 H+ + 2 e- ) 2 MnO4- + 8 H+ + 6 e- + 3 HSO3- + 3 H2O 2 MnO2 + 4 H2O + 3 SO4-2 + 9 H+ + 6 e1 1 2 MnO4- + 3 HSO32 MnO2 Check balance: S H 3=2+1 Charge -2 Mn 2=2 O 8 + 9 = 4 + 1 + 12 + H2O + 3 SO4-2 + H+ 3=3 -3=0+0-6+1 -5 = -5 Balance reaction in basic solution. MnO4- + HSO3MnO42- + SO4-2 Reduction: MnO4MnO42add electrons to balance charge. MnO4- + eOxidation: HSO3- MnO42SO4-2 Use same half-reaction as before. HSO3- + H2O SO4-2 + 3 H+ + 2 e- equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 2x(MnO4- + e1x(HSO3- + H2O 2 MnO4- + 2 e- + HSO3- + H2O MnO42-) SO4-2 + 3 H+ + 2 e- ) 2 MnO42- + SO4-2 + 3 H+ + + 2 e- 2 MnO4- + HSO3- + H2O Check balance: H 1+2=3 Mn 2=2 O 8+3+1=8+4 2 MnO42- + SO4-2 + 3 H+ S 1=1 Charge -2 - 1 = - 4 - 2 + 3 -3 = -3 But, are we done ??? Solution needs to be basic and we are generating 3 H+ 2 MnO4- + HSO3- + H2O 3x(OH- + H+ 2 MnO4- + HSO3- + 3 OH- Check balance: H 1+3=4 Mn 2=2 O 8+3+3=8+4+2 2 MnO42- + SO4-2 + 3 H+ H2O) Neutralize H+ 2 MnO42- + SO4-2 + 2 H2O S 1=1 Charge -2 - 1- 3 = -4 - 2 + 0 -6 = -6 Redox Chemistry and Electrical Energy Electrical Conduction moving electrical charges In Metals: electric charges that move are electrons in the metal electric potential difference is what causes electrons to move (similar to gravitational forces causing mass to move or chemical potential driving reactions) resistance interferes with the flow of electrons - usually atomic motion circuit - electrons move in circles, must have a place to go and other electrons to replace them electric current is the rate charges move - # charges / time Symbol Units Charge Q Coulombs (c) Faradays (F) Potential Difference Resistance E R Volts (V) Ohms () Current I Amperes (A) I= 1 A = 1 c / 1 sec Q time c = A . sec 1 Faraday of charge = charge of 1 mole of electrons = 96,500 c Energy = 1 J = (1 V) . (1 c) E=IR Ohm’s Law Conduction is good if potential is high and resistance is low. In Solutions (or liquids ionic compounds at temperatures high enough to melt): electric charges that move are ions cations (positively charged) move towards anything negatively charged anions (negatively charged) move towards anything positively charged Electrochemical Cell - A circuit that includes electrolytic and metallic conductors Electrode - connects metal conductor to electrolytic conductor Electrochemical Cells anode oxidation cathode reduction spontaneous redox reaction Attracts anions Attracts cations 19.2 Electrochemical Cells The difference in electrical potential between the anode and cathode is called: • cell voltage • electromotive force (emf) • cell potential Cell Diagram Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq) [Cu2+] = 1 M & [Zn2+] = 1 M anode cathode Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s) Use || to separate half cells Use | to separate reactants/phases in each half cell 19.2 Salt Bridge Electrolytic connection between two half cells to complete the circuit Tube containing a solution of inert salt (usually KNO3) Two Types of Cells Electrochemical Cell - need “direct current source” = an electron pump (ex: battery) • Electrons are forced in one direction, independent of sponteneity • Electrical energy is used to cause a nonspontaneous reaction to occur - drive the desired chemistry • Electrons are added to the cathode by electron pump causing reduction. Two Types of Cells Voltaic Cell or Galvanic Cell - passively electric (no need of “direct current source” = an electron pump) • Electrons move because of spontaneous reaction • Use chemistry to get desired energy • Electrons are taken from the cathode by reduction, causing electrons to flow to it (from outside it appears ) • Can use as a direct current source for electrolytic cell. How do we know if reaction is spontaneous? What is the electric potential? How do concentrations effect the process? Electromotive Force (emf) is the electric potential of cell E (emf) units = volts (V) emf is the potential difference between the anode and the cathode Standard Electrode Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Zn (s) Cathode (reduction): 2e- + 2H+ (1 M) Zn (s) + 2H+ (1 M) Zn2+ (1 M) + 2eH2 (1 atm) Zn2+ + H2 (1 atm) 19.3 Standard Electrode Potentials Standard reduction potential (E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Reduction Reaction 2e- + 2H+ (1 M) H2 (1 atm) Defines E0 = 0 V Use as a reference to measure all other potentials. Standard hydrogen electrode (SHE) 19.3 Standard Electrode Potentials 0 = 0.76 V Ecell 0 ) Standard emf (Ecell 0 0 = E0 Ecell cathode - Eanode reduction oxidation Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) 0 = E 0 + - E 0 2+ Ecell H /H2 Zn /Zn Convention: 0 0.76 V = 0 - EZn2+/Zn E° > 0 spontaneous reaction 0 EZn 2+/Zn = -0.76 V Zn2+ (1 M) + 2e- Zn E0 = -0.76 V 19.3 Standard Electrode Potentials 0 = 0.34 V Ecell 0 0 = E0 Ecell cathode - Eanode 0 = E 0 2+ 0 Ecell Cu /Cu – EH +/H 2 0 2+ 0.34 = ECu /Cu - 0 0 2+ ECu /Cu = 0.34 V Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation): H2 (1 atm) Cathode (reduction): 2e- + Cu2+ (1 M) H2 (1 atm) + Cu2+ (1 M) 2H+ (1 M) + 2eCu (s) Cu (s) + 2H+ (1 M) 19.3 • E0 is for the reaction as written • The half-cell reactions are reversible • The sign of E0 changes when the reaction is reversed (E° red = -E°oxid) • Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0 19.3 Strongest oxidizing agent • The more positive E0 the greater the tendency for the substance to be reduced Zero reference point Strongest reducing agent 0 = 0.76 V Ecell 0 = 0.34 V Ecell Combine Zn (s) Zn2+ (1 M) + 2e- 2e- + Cu2+ (1 M) Zn (s) + Cu2+ (1 M) Cu (s) Cu (s) + Zn2+ (1 M) 0 = 0.76 V + 0.34 V = 1.10 V Ecell What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V Cd is the stronger oxidizer Cr3+ (aq) + 3e- Cr (s) Anode (oxidation): E0 = -0.74 V Cr3+ (1 M) + 3e- x 2 Cr (s) Cathode (reduction): 2e- + Cd2+ (1 M) 2Cr (s) + 3Cd2+ (1 M) Cd will oxidize Cr Cd (s) x3 3Cd (s) + 2Cr3+ (1 M) 0 0 = E0 Ecell cathode - Eanode 0 = -0.40 – (-0.74) Ecell 0 = 0.34 V spontaneous Ecell 19.3 Spontaneity of Redox Reactions Sponteneity DG < 0 Energy = Q E = -nFEcell Total charge DG = -nFEcell n = number of moles of electrons in reaction J = F = 96,500 = 96,500 C/mol V • mol 0 DG0 = -RT ln K = -nFEcell DG0 0 -nFEcell 0 Ecell (8.314 J/K•mol)(298 K) RT ln K = ln K = nF n (96,500 J/V•mol) 0 Ecell 0.0257 V ln K = n 0 Ecell = 0.0592 V log K n Spontaneity of Redox Reactions DG0 = -RT ln K DG = -nFEcell 19.4 What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq) 0 Ecell = 0.0257 V ln K n Oxidation: Reduction: 2e- + 2Ag 2Ag+ + 2e- Fe2+ Fe n=2 0 0 E0 = EFe 2+/Fe – EAg + /Ag E0 = -0.44 – (0.80) E0 = -1.24 V 0 Ecell xn -1.24 V x 2 = exp K = exp 0.0257 V 0.0257 V K = 1.23 x 10-42 19.4 The Effect of Concentration on Cell Emf DG = DG0 + RT ln Q DG = -nFE DG0 = -nFE 0 -nFE = -nFE0 + RT ln Q Nernst equation E = E0 - RT ln Q nF Can look at concentration effects / nonstandard conditions At 298 K E= E0 0.0257 V ln Q n E= E0 0.0592 V log Q n 19.5 Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M? Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq) Oxidation: Reduction: Cd 2e- + Cd2+ + 2e- Fe2+ 2Fe n=2 0 0 E0 = EFe 2+/Fe – ECd2+/Cd E0 = -0.44 – (-0.40) E0 = -0.04 V 0.0257 V ln Q n 0.010 0.0257 V ln E = -0.04 V 2 0.60 E = 0.013 V E = E0 - E>0 Spontaneous 19.5 Concentration Cells have the same half-reaction in each cell, but at different concentrations. Will the following reaction occur spontaneously at 250C if [Ag+] = 0.10 M and [Ag+] = 0.010 M? Ag+ (aq) + Ag (s) Ag (s) + Ag+ (aq) Ag (s) | Ag+ (0.10 M) || Ag+ (0.010 M) | Ag (s) Oxidation: Reduction: Ag e- + Ag+ Ag+ + eAg Agoxid + Ag+red 0 0 E0 = EAg + /Ag – EAg+ /Ag n=1 Agred + Ag+oxid E0 = -0.7991 V – (-0.7991 V) E0 = -0.000 V 0.0257 V ln Q E = E0 n E = -0.000 V - 0.0257 V 1 E < 0 Non-spontaneous + [Ag oxid] ln 0.0257 V [Ag+red] = - ln 0.10 = - 0.0592 V 0.010 Batteries Dry cell Leclanché cell Anode: Cathode: Zn (s) 2NH+4 (aq) + 2MnO2 (s) + 2e- Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2eMn2O3 (s) + 2NH3 (aq) + H2O (l) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s) 19.6 Batteries Mercury Battery Anode: Cathode: Zn(Hg) + 2OH- (aq) HgO (s) + H2O (l) + 2eZn(Hg) + HgO (s) ZnO (s) + H2O (l) + 2eHg (l) + 2OH- (aq) ZnO (s) + Hg (l) 19.6 Batteries Lead storage battery Anode: Cathode: Pb (s) + SO2-4 (aq) PbSO4 (s) + 2e- PbO2 (s) + 4H+ (aq) + SO24 (aq) + 2e Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2(aq) 4 PbSO4 (s) + 2H2O (l) 2PbSO4 (s) + 2H2O (l) 19.6 Batteries Solid State Lithium Battery 19.6 Batteries A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: Cathode: 2H2 (g) + 4OH- (aq) O2 (g) + 2H2O (l) + 4e2H2 (g) + O2 (g) 4H2O (l) + 4e4OH- (aq) 2H2O (l) 19.6 Corrosion Dissolved oxygen in water causes oxidation E°red = -0.44 V E°red = 1.23 V Since E°red (Fe3+) < E°red (O2) Rust Fe2O3 Fe can be oxidized by oxygen 19.7 Cathodic Protection of an Iron Storage Tank E°red = -2.37 V E°red = 1.23 V Mg reduction potential is too negative to be overcome by oxygen. 19.7 Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. 19.8 Electrolysis of Water 19.8 Electrolysis and Mass Changes Quantitative Analysis How much current? time ? product? charge (C) = current (A) x time (s) 1 mole e- = 96,500 C 19.8 How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours? Anode: Cathode: 2Cl- (l) Ca2+ (l) + 2eCa2+ (l) + 2Cl- (l) Cl2 (g) + 2eCa (s) Ca (s) + Cl2 (g) 2 mole e- = 1 mole Ca C s 1 mol e- 1 mol Ca mol Ca = 0.452 x 1.5 hr x 3600 x x s hr 96,500 C 2 mol e= 0.0126 mol Ca = 0.50 g Ca 19.8 Consider the stoichiometryof the electrolytic cell: Cathode: e- + Ag+ Ag Anode: 2 H2O O2 (g) + 4 H+ + 4 e- What current (in amps) is required to convert 0.100 mole Ag+ to Ag in 10.0 min? 1 mole electrons = 1 F A = c/sec Q = nF I=Q/t Find Q Q = 0.10 mol Ag 1 mole electrons 1F 96,500 c = mole Ag mole electrons F Q = 9,650 c t = 10 min 60 seconds = 600 seconds 1 minute I = 9,6500 c / 600 s = 16 c/s = 16 A Stoichiometry of prodcuts at different electrodes What is the pH of the anode half cell (assume volume = 0.100 L) after 6.00 g of Ag plates out at the cathode? Cathode: e- + Ag+ Ag Anode: 2 H2O O2 (g) + 4 H+ + 4 eFind [H+] 6.00 g Ag 1 mol Ag 1 mole electrons 4 mol H+ 107.9 g Ag 1 mole Ag 1 mol electrons = 0.05567 mol H+ [H+] = 0.0556 mol = 0.56 M 0.10 L pH = 0.25 =