Chapter 20 Electrochemistry

Transcription

1 Chapter 20 Electrochemistry 20.1 Oxidation States and Oxidation-Reduction Reactions An oxidation occurs when an atom or ion loses electrons. A reduction occurs when an atom or ion gains electrons. One cannot occur without the other (redox reaction). Examples: Combustion, corrosion of metals, electroplating, etc. Zn: 0 +2 lost e s, increase ox # (oxidized) H: +1 0 gained e s, decreased ox # (reduced) Zn (s) + 2 HCl (aq) ZnCl 2 (aq) + H 2 (g) Net ionic: Zn (s) + 2 H+ Zn 2+ + H 2 (g) Ox. #: Oxidizing agent causes oxidation of something else (it gets reduced in the process). HCl here. Reducing agent causes reduction of something else (it gets oxidized in the process). Zn here. Redox Reactions To determine if an oxidation-reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity. Again: Oxidation loss of e s, increase in oxidation number (becomes more positive). Reduction gain of e s, decrease in oxidation number (becomes more negative). Oxidizing agent the substance containing the element that is reduced. Reducing agent the substance containing the element that is oxidized. Oxidation Numbers Elements in natural elemental state have oxidation number = 0. The oxidation number of a monatomic ion is the same as its charge. Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. Oxygen has an oxidation number of 2, except in the peroxide ion in which it has an oxidation number of 1. Hydrogen is 1 when bonded to a metal, +1 when bonded to a nonmetal. K2O vs K2O2 ; BaO vs BaO2 KH ; NaH ; CaH2 1

2 Fluorine always has an oxidation number of 1. The other halogens have an oxidation number of 1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions. The sum of the oxidation numbers in a neutral compound is 0. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. Problem 1 Determine the oxidation numbers: Li2O H3PO4 MnO4 Cr2O7 2 C7H8 Problem 2 Identify what s oxidized and what s reduced. Cu(s) + 2 NO3 + 4H + Cu NO2 + 2 H2O 20.2 Balancing Redox Reactions The task of balancing redox reactions can be complicated: 1) many are written as net ionic equations. 2) many have elements in multiple compounds. The main principle is that electrons are transferred (i.e. if we can find a method to keep track of the electrons it will allow us to balance the equation). We ll use the Method of Half-Reactions. Half-Reactions We generally split the redox reaction into two separate half-reactions a reaction just involving oxidation or reduction the oxidation half-reaction has electrons as products the reduction half-reaction has electrons as reactants 3 Cl2 + I + 3H2O 6 Cl + IO3 + 6 H oxidation: I IO3 + 6 e reduction: Cl2 + 2 e 2 Cl 2

3 Balancing Redox Reactions by the Half-Reaction Method The reaction is broken down into two half-reactions, one for oxidation and another for reduction Each halfreaction is balanced for its atoms. Then the two half-reactions are adjusted so that the electrons lost and gained will be equal when combined. Balancing redox reactions in acidic solution 1. Assign oxidation states and determine element oxidized and element reduced. Fe 2+ + MnO4 Fe 3+ + Mn Separate into oxidation & reduction half-reactions. ox: Fe 2+ Fe 3+ red: MnO4 Mn Balance half-reactions by mass a) first balance atoms other than O and H b) balance O by adding H2O to side that lacks O c) balance H by adding H + to side that lacks H Fe 2+ Fe 3+ MnO4 Mn 2+ MnO4 Mn H2O MnO4 + 8H + Mn H2O 4. Balance each half-reaction with respect to charge by adjusting the numbers of electrons a) electrons on product side for oxidation b) electrons on reactant side for reduction Fe 2+ Fe e MnO4 + 8H + Mn H2O Balance electrons between half-reactions Fe 2+ Fe e 5 MnO4 + 8H e Mn H2O MnO4 + 8H e Mn H2O 6. Add half-reactions, canceling electrons and common species 5 Fe 2+ 5 Fe e MnO4 + 8H e Mn H2O 7. Check that numbers of atoms and total charge are equal 5 Fe 2+ + MnO4 + 8H + Mn H2O + 5 Fe 3+ 3

4 Q. Balance the following redox reaction in acidic solution. MnO4 + C2O4 2 Mn 2+ + CO2 Q. Balance the following redox reaction in acidic solution. I + Cr2O7 2 Cr 3+ + I2 4

5 Balancing in Basic Solution If a reaction occurs in a basic solution, one can balance it as if it occurred in acid. There is a change though: to balance half-reactions by mass in base, we need to add OH to neutralize the H + in the equation and create water in its place (add OH to both sides). If this produces water on both sides, you might have to subtract water from each side. Example: Redox in basic solution I (aq) + MnO4 (aq) I2(aq) + MnO2(s) Mass Balance of each half-reaction (as usual): I (aq) I2(aq) MnO4 (aq) MnO2(s) 2 I (aq) I2(aq) 4 H + (aq) + MnO4 (aq) MnO2(s) + 2 H2O(l) NEW: Neutralize H + with OH. 4 H + (aq) + 4 OH (aq) + MnO4 (aq) MnO2(s) + 2 H2O(l) + 4 OH (aq) 4 H2O(aq) + MnO4 (aq) MnO2(s) + 2 H2O(l) + 4 OH (aq) 2 H2O(l) + MnO4 (aq) MnO2(s) + 4 OH (aq) Balance the electrons on each half-reaction: 2 I (aq) I2(aq) + 2 e MnO4 (aq) + 2 H2O(l) + 3 e MnO2(s) + 4 OH (aq) Balance the electrons between half-reactions: 2 I (aq) I2(aq) + 2 e 3 MnO4 (aq) + 2 H2O(l) + 3 e MnO2(s) + 4 OH (aq) 2 Add the reactions: 6 I (aq) 3 I2(aq) + 6 e 2 MnO4 (aq) + 4 H2O(l) + 6 e 2 MnO2(s) + 8 OH (aq) 6 I (aq)+ 2 MnO4 (aq) + 4 H2O(l) 3 I2(aq)+ 2 MnO2(s) + 8 OH (aq) 5

6 Q. Balance the following equation in basic solution: I + CrO4 2 I2 + Cr Voltaic Cells When we talk about the current of a liquid in a stream, we are discussing the amount of water that passes by in a given period of time. When we discuss electric current, we are discussing the amount of electric charge that passes a point in a given period of time... Current and Redox Reactions Since redox reactions involve the transfer of electrons, they have the potential to generate an electric current. Here, the electrons are flowing directly between atoms. However, in order to use that current (to do work), we need to separate the place where the half-reactions are occurring, so as to force the electrons to move through a wire. We call such a setup a voltaic cell. If the reaction occurs directly, it produces heat instead. Electrochemistry Electrochemistry is the study of redox reactions that produce or require an electric current. The conversion between chemical energy and electrical energy is carried out in an electrochemical cell. 6

7 Spontaneous redox reactions take place in a voltaic cell. aka galvanic cells (batteries fall in this category) Nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy. Electrochemical Cells Oxidation and reduction half-reactions are kept separate in half-cells. The electrons flow through a wire along with ions flow through a solution constitutes an electric circuit. Requires a conductive solid electrode to allow the transfer of electrons. Requires ion exchange between the two half-cells of the system. electrolyte Electrodes Anode Cathode electrode where oxidation occurs (vowels rule) anions are attracted to it connected to positive end of battery in an electrolytic cell loses weight in electrolytic cell electrode where reduction occurs (consonants rule) cations are attracted to it connected to negative end of battery in an electrolytic cell gains weight in electrolytic cell electrode where plating takes place in electroplating The cell Anions migrate toward anode Cations migrate toward cathode The salt bridge is required to complete the circuit and maintain charge balance (the build up of charge would stop the flow of electrons). Remember: The electrons flow from the anode (where they are produced due by oxidation) to the cathode (where they are used up by the reduction rxn). For a spontaneous cell (voltaic or galvanic): Anode is and Cathode is (the opposite is true for nonspontaneous cells electrolytic cells). 7

8 Cell Notation It is a shorthand description of a voltaic cell: Electrode electrolyte electrolyte electrode The oxidation half-cell goes on the left, and the reduction half-cell on the right. Single = phase barrier if multiple electrolytes in same phase, a comma is used rather than Double line = salt bridge Voltaic Cell Anode = Zn(s) the anode is oxidized to Zn 2+ Cathode = Cu(s) Cu 2+ ions are reduced at the cathode Zn(s) Zn 2+ (aq) Cu 2+ (aq) Cu(s) Electrodes Many times the anode is made of the metal that is oxidized and the cathode is made of the same metal as is produced by the reduction. If no solids are involved in the reaction, we may use an inert electrode. Examples are: - A redox reaction that involves the oxidation or reduction of an ion to a different oxidation state. - The oxidation or reduction of a gas. Inert Electrode An inert electrode is one that does NOT participate in the reaction, but just provides a surface for the transfer of electrons to take place on Think of a half-reaction that involves reducing the Mn oxidation state from +7 to +2. Since both ions are in solution, we need to use an electrode that will provide a surface for the electron transfer without reacting with the MnO4. Platinum works well because it is extremely nonreactive and conducts electricity. Fe(s) Fe 2+ (aq) MnO4 - (aq), Mn 2+ (aq), H + (aq) Pt(s) 8

9 Inert Electrode e e Fe 3+ (aq) + e 2+ Fe (aq) Reduction + H 2 (g) 2H (aq) Oxidation + 2e Pt(s) H2(aq),2H + (aq) Fe 3+ (aq), Fe 2+ (aq) Pt(s) Q. For a voltaic cell with the overall rxn: Ni (s) + 2Ag + Ni Ag (s) 1. Give the half-reactions. 2. Label oxidation and reduction. 3. Draw the cell. 4. Label anode and cathode. 5. Identify + and electrodes. 6. Direction of e s flow? 7. Direction of ions (K + (aq), NO 3(aq) ) flow from the salt bridge. 8. Cell notation? 9

10 20.4 Cell Potentials Under Standard Conditions A waterfall (analogy): The higher it is, the more potential energy the water has, and the more energy the water will posses as it hits the bottom (Voltage). The flow rate is the amount (volume) of water going over the edge of the fall each second (Current). Flow Rate Current is the number of electrons that flow through the system per second unit = Ampere 1 A of current = 1 Coulomb of charge flowing by each second 1 A = x electrons per second Electrode surface area dictates the number of electrons that can flow (i.e. batteries produce larger currents) Potential Difference The difference in potential energy between the reactants and products is the potential difference unit = Volt (1V = 1J 1C) the voltage needed to drive electrons through the external circuit Amount of force pushing the electrons through the wire is called the electromotive force, emf. Cell Potential The difference in potential energy between the anode and the cathode in a voltaic cell is called the cell potential (emf or voltage). The cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode The cell potential under standard conditions is called the standard emf, E cell. 25 C, 1 atm for gases, 1 M concentration of solution. sum of the cell potentials for the half-reactions. Spontaneous reactions have voltages. Ecell depends on the reaction, temperature, and concentrations. Assume 25 unless stated otherwise. For Zn/Cu galvanic cell, E cell = 1.10 V (std. cond.) 10

11 A half-reaction with a strong tendency to occur has a large positive half-cell potential. When two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occur. Which Way Will Electrons flow? Under standard conditions, zinc has a stronger tendency to oxidize than copper Zn Zn e E = Cu Cu e E = 0.34 Therefore the electrons flow from zinc; making zinc the anode Standard Reduction Potential We cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another halfreaction We select as a standard half-reaction the reduction of H + to H2 under standard conditions, which we assign a potential difference = 0 V standard hydrogen electrode, SHE 2 H + (aq, 1M) + 2e H2(g, 1 atm) Zn(s) Zn 2+ (1 M) H + (1 M) H2(g) Pt(s) Half-Cell Potentials Standard Reduction Potentials compare the tendency for a particular reduction half-reaction to occur relative to the reduction of H + to H2 under standard conditions Half-reactions with a stronger tendency toward reduction than the SHE have a + value for E red Half-reactions with a stronger tendency toward oxidation than the SHE have a value for E red For an oxidation half-reaction, E oxidation = E reduction (i.e. reverse reaction and change the sign of E ) Ag + + e Ag Ag Ag + + e E = V E = 0.80 V 11

12 Tendencies from the Table of Standard Reduction Potentials A redox reaction will be spontaneous when there is a strong tendency for the oxidizing agent to be reduced and the reducing agent to be oxidized higher on the table of Standard Reduction Potentials = stronger tendency for the reactant to be reduced. lower on the table of Standard Reduction Potentials = stronger tendency for the product to be oxidized. Predicting Spontaneity of Redox Reactions A spontaneous reaction will take place when a reduction half-reaction is paired with an oxidation half-reaction lower on the table If paired the other way, the reverse reaction is spontaneous Cu 2+ (aq) + 2 e Cu(s) Zn 2+ (aq) + 2 e Zn(s) E red = V E red = 0.76 V Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Cu(s) + Zn 2+ (aq) Cu 2+ (aq) + Zn(s) spontaneous nonspontaneous Calculating Cell Potentials under Standard Conditions E cell = E red (cathode) E red (anode) Because cell potential is based on the potential energy per unit of charge, it is an intensive property. Ag + + e Ag E = V 2Ag + + 2e 2Ag E = V Calculating the overall cell potential: Ni + Cu 2+ Ni 2+ + Cu E =? - Find the potential for each half reaction - Reverse anode reaction (change sign) - Add Ni e Ni E = 0.28 V Reverse!! Cu e Cu E = V Ni Ni e E = V Cu e Cu E = V Ni + Cu 2+ Ni 2+ + Cu E = V Spontaneous!! 12

13 Example: Calculate E cell for the reaction at 25 C Al(s) + NO3 (aq) + 4 H + (aq) Al 3+ (aq) + NO(g) + 2 H2O(l) Anode (ox): Al(s) Al 3+ (aq) + 3 e Cathode (red):no3 (aq) + 4 H + (aq) + 3 e NO(g) + 2 H2O(l) E red of Al 3+ = 1.66 V E red of NO3 = V E cell = E red (cathode) E red (anode) E cell = (0.96 V) ( 1.66 V) = V E cell = (+1.66 V) + (0.96 V) = V E cell = E oxidation + E reduction Q. If a voltaic cell is constructed from a Zn/Zn 2+ half-cell and a Al/Al 3+ half cell: a) Determine the overall reaction and E. b) Specify anode and cathode. Comparing Strengths of Oxidizing and Reducing Agents Example: Zn, Cu, H2, Ni. Cu e Cu 2H + + 2e H 2 Ni e Ni Zn e Zn E = V E = 0 V E = 0.28 V E = 0.76 V reductions Of these, reduction of Cu 2+ has the most + potential. Cu 2+ is the easiest to reduce. It is the best oxidizing agent on the list. Zn 2+ is the worst oxidizing agent. Order: Cu 2+, H +, Ni 2+, Zn 2+ 13

14 Cu e Cu 2H + + 2e H 2 Ni e Ni Zn e Zn E = V E = 0 V E = 0.28 V E = 0.76 V Reducing agents get oxidized. Reverse the above reactions and E values. Zn is the easiest to oxidize and Cu the hardest. Zn Zn e E = V Zn (s) is the best reducing agent on the list and Cu (s) the worst. Important note: Cu 2+, H +, Ni 2+, Zn 2+ can t be reducing agents because they cannot be oxidized. Their oxidation numbers cannot go higher. Zn, Ni, H 2, Cu can t be oxidizing agents because they cannot be reduced any further. In general: The best oxidizing agent has the most + reduction potential (F 2 (g) ), and the best reducing agent has the most + oxidation potential (Li (s) ). Comparing Strengths of Oxidizing and Reducing Agents The strongest oxidizers have the most positive reduction potentials. The strongest reducers have the most negative reduction potentials. The greater the difference between the two, the greater the voltage of the cell. Q. Sketch and label a voltaic cell in which one half-cell has Ag(s) immersed in 1 M AgNO3, and the other halfcell has a Pt electrode immersed in 1 M Cr(NO3)2 and 1 M Cr(NO3)3 Write the half-reactions and overall reaction, and determine the cell potential under standard conditions. 14

15 20.5 Free Energy and Redox Reactions G for a redox reaction can be found by using the equation: G = nfe ; G = nfe n = #e transferred in balance equation (unitless); see the 1 2 reactions C F = Faraday Constant = ; 1 mole e E =cell voltage (V); V = J C G = nfe[=]( C mol )(V) = J mol (as written) Spontaneous reaction: G, E. Nonspontaneous reaction: G, E. For: Zn + Cu 2+ Zn 2+ + Cu E = 1.10 V G = nfe = (2) (96485 C mol ) (1.10 J C ) G = J mol G = 212 kj mol 20.6 Cell Potentials Under Nonstandard Conditions We know there is a relationship between the reaction quotient, Q, the equilibrium constant, K, and the free energy change, G. Changing the concentrations of the reactants and products so they are not 1 M will affect the standard free energy change, G. Because G determines the cell potential, E cell, the voltage for the cell will be different when the ion concentrations are not 1 M. E = E If [Zn 2+ ] = [Cu 2+ ], Q = 1, ln Q = 0, E = E If [Zn 2+ ] > [Cu 2+ ], Q > 1, ln Q, E = E x G = G + RT ln Q G = nfe ; G = nfe nfe = nfe + RT ln Q E = E RT ln Q nf V ln Q or E = E n V log Q n Zn + Cu 2+ Zn 2+ + Cu E = 1.10 V Q = [Zn2+ ] [Cu 2+ ] ; n = 2 ; E = V E ln Q n E is lower than E, less spontaneous than std. Why? Higher [Zn 2+ ] shifts eq. reverse. More product and/or less reactant makes the reaction less favorable. 15

16 Zn + Cu 2+ Zn 2+ + Cu E = 1.10 V Q = [Zn2+ ] [Cu 2+ ] ; n = 2 ; E = V E ln Q n If [Zn 2+ ] < [Cu 2+ ], Q < 1, ln Q, E = E ( x) E is higher than E, more favorable than std. Why? More reactant and/or less product makes the reaction more favorable. E cell When Ion Concentrations are Not 1 M Q. MnO 4 + 8H + + 5e Mn H 2 O E = V O 2 (g) + 4H + + 4e 2H 2 O E = V Calculate the potential of the permanganate-oxygen cell operating at ph=7.00 when P O2 = atm, [MnO 4 ] = 0.100M, [Mn 2+ ] = 0.300M? 16

17 Concentration Cells The electrode compartments contain the same substances at different concentrations. Cu e Cu red cat E = V Cu Cu e ox an E = 0.34 V Cu + Cu Cat Cu + Cu An E = 0 V As it comes to equilibrium, the concentrations become equal. In the more concentrated solution, [Cu 2+ ] decreases (cat). In the more concentrated solution, [Cu 2+ ] increases (an). Q = [Cu2+ ] anode dilute [Cu 2+ ] catode conc. E = E V V ln Q = 0 ln = V n Spontaneous. As long as the concentrations are different, E will not be 0. Application ph meters: Inside the probe we have an electrode with a known concentration [H + ]. We dip it in a solution with unknown ph. The ph meter measures the potential across an ion-selective glass bulb membrane. It converts potential to ph reading. Also potentials across cell membranes see page 853. As a cell operates, its voltage drops (reactants get used up, and products form; E decreases). When the voltage = 0, it s at equilibrium. At equilibrium: E = 0 and Q = K. K is the thermodynamic eq. constant (gases in atm, solutes in M). E = E 0 = E E = V ln Q n V n ln K V ln n E = RT ln K nf 17

18 Q. Calculate the equilibrium constant for: Cu Br Cu (s) + Br 2 (l) using standard reduction potentials Batteries and Fuel Cells Batteries are galvanic cells (portable, self-contained). When connected in series (i.e. cathode of one connected to anode of next), the total voltage is the sum of voltages. Cathode (+) and Anode ( ). Primary cell cannot be recharged. Secondary cell can be recharged. Lead Batteries in Cars Plates of Pb (s) and PbO 2 (s) in H 2 SO 4 (aq). PbO 2 (s) + HSO 4 + 3H + + 2e PbSO 4 (s) + 2H 2 O (l) Pb (s) + HSO 4 PbSO 4 (s) + H + + 2e cat-red an-ox PbO 2 (s) + Pb (s) + 2HSO 4 + 2H + 2PbSO 4 (s) + 2H 2 O (l) 6 of these connected in series yields 12 V Battery. Anode Pb, Cathode PbO 2. Overall: V Both solids, there is no need to separate compartments. Both get coated with PbSO 4 sticks to electrodes so battery can be recharged. As this battery is discharged, H 2 SO 4 (aq) is used up (we can measure the density of the solution to determine how discharged the battery is). 18

19 Alkaline Battery Used in Flashlights, etc. Cathode: MnO 2 (s) + C gr Anode: Zn + KOH aq (conc. In gel) Salt bridge: porous fabric sealed Cat: MnO 2 (s) + 2H 2 O + 2e 2MnO(OH) (s) + 2OH An: Zn + 2OH Zn(OH) 2 (s) + 2e E = 1.55 V (room T) NiCad Battery Ni-Cd, Ni-metal-hydride, Lithium-ion batteries. Electrolyte is concentrated KOH solution Anode = Cd Cd (s) + 2OH (aq) Cd(OH ) 2 (s) + 2e E = 0.81 V Cathode = Ni coated with NiO 2 NiO 2 is reduced NiO 2 (s) + 2H 2 O (l) + 2e Ni(OH ) 2 (s) + 2OH (aq) E = 0.49 V Cell voltage = 1.30 V Rechargeable, long life, light however recharging incorrectly can lead to battery breakdown Ni-MH Battery Electrolyte is concentrated KOH solution Anode = metal alloy with dissolved hydrogen oxidation of H from H 0 to H + M H (s) + OH (aq) M (s) + H 2 O (l) + e E = 0.89 V Cathode = Ni coated with NiO 2 NiO 2 is reduced NiO 2 (s) + 2H 2 O (l) + 2e Ni(OH) 2 (s) + 2OH E = 0.49 V Cell voltage = 1.30 V Rechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad 19

20 Lithium Ion Battery Electrolyte is concentrated KOH solution Anode = graphite impregnated with Li ions Cathode = Li - transition metal oxide reduction of transition metal Work on Li ion migration from anode to cathode causing a corresponding migration of electrons from anode to cathode. Rechargeable, long life, very light, more environmentally friendly, greater energy density. Fuel Cells Like batteries in which reactants are constantly being added so it never runs down! Anode and cathode both Pt coated metal Electrolyte is OH solution Anode reaction: 2H 2 + 4OH 4H 2 O (l) + 4e Cathode Reaction: O 2 + 4H 2 O (l) + 4e 4OH Twice as efficient as an internal combustion engine (more efficient = less energy wasted as heat) 20.8 Corrosion Corrosion is the spontaneous oxidation of a metal by chemicals in the environment (mainly O2). Since many materials we use are active metals, corrosion can be a big problem: a) strength and malleability are lost. b) In many cases the product of corrosion does not adhere to the metal and it sheds off, leaving the surface exposed to more corrosion. O2 is very easy to reduce in moist conditions O 2 (g) + 2H 2 O (l) + 4e 4OH E = 0.40 V O2 is even easier to reduce under acidic conditions O 2 (g) + 4H + + 4e 2H 2 O E = 1.23 V The reduction potential of most metals happens to be below O 2 (g) ; therefore, their oxidation by oxygen is spontaneous. The Corrosion Process At the anodic regions, Fe(s) is oxidized to Fe 2+ ; the electrons travel through the metal to a cathodic region where O2 is reduced. The Fe 2+ ions migrate through the moisture to the cathodic region where they are further oxidized to Fe 3+ which combines with the oxygen and water to form rust. 20

21 Conditions Rust is: Fe 2 O 3 nh 2 O (the exact composition depends on the conditions). Moisture must be present as both: a reactant and as a medium for the flow of ions. Electrolytes and H + promote rusting by enhancing the current flow. The cracks and pits that develop weaken the metal. Corrosion Prevention One way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment. a) paint b) some metals, such as Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corroding (there is a problem when the cover cracks). Another method to protect one metal is to attach it to a more reactive metal (Zn, Mg) that is cheap (sacrificial electrode). Once used, the electrode needs to be replaced Electrolysis In all electrochemical cells, oxidation occurs at the anode and reduction occurs at the cathode. Electrons are drawn off the anode (by the + terminal of the battery) and forced toward the canode (by the terminal of the battery). The electricity makes the nonspontaneous reaction happen. Note: the anode is and the cathode is. It can be done in one container. Electrolysis of molten salts Think of liquid (molten) sodium chloride: NaCl (l) contains Na + and Cl (but they are not aqueous!) Half reactions are: Na + + e Na (s) (reduction, cathode) Cl 1 2 Cl 2 (g) + e (oxidation, anode) The sodium metal forms at the cathode and chlorine gas at the anode. This is how Na (s) is produced (it is so reactive that it is not found in nature by itself). 21

22 Aqueous electrolysis water is present and can also react. To decide what will happen, consider all possible reductions and oxidations, and then choose the easiest of each. [Sometimes there are other complications like the speed of reactions, the activation energies and concentrations but this is a pretty good prediction.] Electrolysis of NaCl (aq) Present: Na + (aq), Cl (aq), H 2 O Possible reductions: Na + + e Na E = 2.71 V 2H 2 O + 2e H 2 + 2OH E = 0.83 V (reduction of water is easiest because it has a less negative potential.) Possible oxidations: 2Cl Cl 2 + 2e E = V 2H 2 O O 2 + 4H + + 4e E = 1.23 V (values are close, but oxidation of water is easiest. Reaction with Cl is faster) In actuality: if [Cl ] is high, Cl gets oxidized. If [Cl ] is low, H 2 O gets oxidized. Note: possible ½ reactions include major species as reactants. For the above example, don t consider ½ reactions such as: Cl 2 2Cl + 2e, 4H + + O 2 + 4e 2H 2 O, 2H 2 O + O 2 4OH, etc because they don t have major species as reactants. Q. Predict the ½ reactions you would expect at the anode and cathode in the electrolysis of aqueous CoF 2 using copper electrodes. What would you expect to see at each electrode? 22

23 Electroplating One metal can be deposited (plated) on another. The cathode is the object to be plated. In the electroplating of silver we find: At anode: Ag Ag + + e At cathode: Ag + + e Ag A layer of Ag is plated on iron. Electrochemical Stoichiometry Current is measured in amperes (A): 1 V = 1 J C 96,485 C ; F = 1 mol e 1A = 1 C sec If the current is 1 A, 1 C of charge passes a given point in 1 sec. In problems: We need to know the # mol e transferred in the reaction. Look at the ½ reaction. Q. How many grams of O 2 are liberated by the electrolysis of H 2 O after 185 sec with a current of 56.5 ma? 23

24 Q. A constant electric current deposits 546 mg Zn in 150. min from an aqueous ZnCl 2 solution. What is the current? Electrical work W max = nfe ; because G = W max If spontaneous rxn, W max system can do work. If nonspon, W max - the work needed to force the rxn to happen. W = nfe external Power watt: 1 W = 1 J sec 1 watt sec = 1 J sec = 1 J sec 1000 J sec 1 kw hr = (1000 W)(1 hr) = ( ) (1 hr) (3600 ) = J sec 1 hr Q. Calculate # kilowatt-hours (kwh) of electricity needed to produce kg Al by electrolysis of Al 3+ if the applied voltage is 4.50 V. 24

25 More Problems Q. Zn e Zn E = V Ag + + e Ag E = V If [Zn 2+ ] = 1.00 M, what [Ag + ] will give a voltage of V? Q. Ag (s) Ag + (saturated AgI) Ag + (0.100M) Ag (s) Cell voltage = V What is K sp for AgI (s)? 25