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Chapter 20 Electrochemistry

Chapter 20 Electrochemistry. Electrochemistry. Electrochemistry is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy .

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Chapter 20 Electrochemistry

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  1. Chapter 20 Electrochemistry

  2. Electrochemistry Electrochemistry is the branch of chemistry that deals with the interconversion of electrical energy and chemical energy. Electrochemistry processes are redox (Oxidation-Reduction) reactions in which the energy released by a spontaneous reaction is converted to electricity or in which electrical energy is used to cause a nonspontaneous reaction to occur.

  3. Oxidation-Reduction Reactions Redox Oxidation-Reduction Reactions are considered electron-transfer reactions. Also known as Redox Reactions. Remember: LEO goes GER Loss of Electrons means Oxidized & Gain of Electrons means Reduced If you say an element is Oxidized, then it is called a Reducing agent because it donates electrons to another element. If you say an element is Reduced, then it is called an Oxidizing agent because it accepts electrons from another element.

  4. Let’s take a look at the formation of calcium oxide (CaO) from calcium and oxygen: 2Ca(s) + O2(g) -> 2CaO(s) For convenience, we think of this process as 2 steps, one involving the loss of electrons and one involving the gain of electrons. This separate steps are calleda half-reaction, which explicitly shows the electrons involved in a redox reaction. Loss of electrons: 2Ca -> 2Ca2+ + 4e- Gain of electrons: O2 + 4e- -> 2O2- The sum of the half-reactions gives the overall reaction: 2Ca + O2 + 4e- -> 2Ca2+ + 2O2- + 4e- We cancel the electrons that appear on both sides to get: 2Ca + O2 -> 2Ca2+ + 2O2- Finally, the Ca2+ and the O2- ions combine to form CaO: 2Ca2+ + 2O2- -> 2CaO

  5. Rules for Assigning Oxidation States • The oxidation state of an atom in an element is 0. • In a neutral molecule, the sum of the oxidation numbers of all the atoms must equal 0. • The oxidation state of a monatomic ion is the same as its charge. • In its compound, fluorine is always assigned an oxidation sate of -1. • Oxygen is usually assigned an oxidation of -2 in its covalent compounds, such as CO, CO2, SO2. Exceptions to this rule includes peroxides (compounds containing the O22- group), where each oxygen is assigned an oxidation state of -1, as in hydrogen peroxide (H2O2), and OF2 in which oxygen is assigned a +2 oxidation state. • In its covalent compounds with nonmetals, hydrogen is assigned an oxidation state of +1. For example HCl. When hydrogen is bonded to a nonmetal in a binary compound, it is assigned an oxidation state of -1. For example LiH. • For an ion, the sum of the oxidation states must equal the charge of the ion. For example, the sum of the oxidation states must equal -2 in CO32-.

  6. Half-Reaction Method for Balancing Redox Reactions in Acidic Solutions • Write separate equations for the oxidation and reduction half-reactions and show the oxidation state of all elements. • For each half-reaction, a. Balance all of the elements except hydrogen and oxygen. b. Balance oxygen using H2O. c. Balance hydrogen using H+. d. Balance the charge using electrons. • If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions. • Add the half-reactions, and cancel identical species. • Check that the elements and charges are balanced.

  7. Half-Reaction Method for Balancing Redox Reactions in Basic Solutions • Write separate equations for the oxidation and reduction half-reactions and show the oxidation state of all elements. • For each half-reaction, a. Balance all of the elements except hydrogen and oxygen. b. Balance oxygen using H2O. c. Balance hydrogen using H+. d. Balance the charge using electrons. • If necessary, multiply one or both balanced half-reactions by an integer to equalize the number of electrons transferred in the two half-reactions. • Add the half-reactions, and cancel identical species. • To both sides of the equation obtained, add a number of OH- ions that is equal to the number of H+ ions. (We want to eliminate the H+ by forming H2O.) • Form H2O on the side containing both H+ and OH- ions, and eliminate the number of H2O molecules that appear on both sides of the equation. • Check that the elements and charges are balanced.

  8. Redox Reactions Cu(s) + 2NO3-(aq) + 4H+ -> 2NO2(g) + Cu2+(aq) + 2H2O(l) 2Ag+(aq) + Cu(s) -> 2Ag(s) + Cu2+(aq) 2NO(g) + O2(g) -> 2NO2(g) 2Fe(s) + 3Cl2(g) -> 2FeCl3(s)

  9. Electrochemical Cells If we place a piece of Zinc metal into a CuSO4 solution, Zn is oxidized to Zn2+ ions while Cu2+ ions are reduced to metallic copper: Zn(s) + Cu2+(aq)g Zn2+(aq) + Cu(s) The electrons are transferred directly from the reducing agent (Zn) to the oxidizing agent (Cu2+) in solution. However, if we physically separate the oxidizing agent from the reducing agent, the transfer of electrons can take place via an external conducting medium (a metal wire). As the reaction progresses, it sets up a constant flow of electrons and hence generates electricity (it produces electrical work such as driving an electric motor). An electrochemical cell is the experimental apparatus for generating electricity through the use of a spontaneous redox reaction. (An electrochemical cell is sometimes referred to as a galvanic cell or voltaic cell, after the scientists Luigi Galvani and Allessandro Volta, who constructed early versions of this device.)

  10. External Wire that electrons travel on. Anode – Where oxidation (loss of electrons) occurs. Cathode – Where reduction (gain of electrons) occurs. Salt Bridge – Contains inert electrolyte that allows cations to move toward cathode and anions toward anode so build up of (+) and (–) charges in the solutions does not occur and stop the reaction.

  11. This Particular Galvanic / Voltaic Cell is Known as the Daniell Cell When the concentrations of the Cu2+ and Zn2+ ions are both 1.0 M, we find that the voltage or emf of the Daniell cell is 1.10 V at 25 oC

  12. Galvanic/Voltaic Cells An electric current flows from the anode to the cathode because there is a difference in electrical potential energy between the electrodes. This is similar to the flow of gas from a high-pressure region to a low-pressure region. Experimentally, the difference in electrical potential between the anode and the cathode is measured by a voltmeter and the reading (in volts) is called cell voltage. Two other terms, electromotive force or emf (E) and cell potential are also used to denote cell voltage. The conventional notation for representing electrochemical cells is the cell diagram. For the previous cell we used, if we assume the concentrations of Zn2+ and Cu2+ ions are 1 M, the cell diagram is: Zn(s) Zn2+ (1 M) Cu2+ (1 M) Cu(s) The single vertical line represents a phase boundary. For example, the zinc electrode is a solid and the Zn2+ ions (from ZnSO4) are in solution. Thus we draw a line between Zn and Zn2+ to show the phase boundary. The double vertical line denotes the salt bridge. You must always write the anode to the left of the double lines, and the cathode on the right of the double lines to show the movement of electrons from anode to cathode.

  13. Standard Reduction Potentials When the concentrations of the Cu2+ and Zn2+ ions are both 1.0 M, we find that the voltage or emf of the Daniell cell is 1.10 V at 25 oC. This voltage must be related directly to the redox reactions, but how? The measured emf of the cell is the sum of the electrical potentials at the Zn and Cu electrodes. If we know one of these electrode potentials, we could obtain the other from subtraction (from 1.10 V). It is impossible to measure the potential of just a single electrode, but if we set the potential value of a particular electrode to zero, we can use it to determine the relative potentials of other electrodes. The hydrogen electrode, shown in next slide, serves as the reference for this purpose.

  14. The hydrogen electrode, shown in the previous slide, serves as the reference for setting the potential value of a particular electrode to zero so that we can determine the relative potentials of other electrodes. Hydrogen gas is bubbled into a hydrochloric acid solution at 25 oC. The platinum electrode has two functions. First, it provides a surface on which the dissociation of hydrogen molecules can take place: H2g 2H+ + 2e- Second, it serves as an electrical conductor to the external circuit. Under standard state conditions (when the pressure of H2 is 1 atm and the concentration of HCl solution is 1 M), the potential for the reduction of H+ at 25 oC is taken to be exactly zero: 2H+ (1 M) + 2e-g H2 (1 atm) Eo = 0 V We can use hydrogen electrode potential = 0 to figure out the other potentials of other kinds of electrodes.

  15. We can use the hydrogen potential = 0 to figure out the previous Daniell Cell voltage = 1.1 Volts, by breaking it up into two separate parts. For the Daniell Cell: Zn(s) + Cu2+(aq)g Zn2+(aq) + Cu(s) If we use hydrogen as the middle point, the cell diagram is: Zn(s) Zn2+ (1 M) H+ (1 M) H2(1 atm) Pt (s) (The platinum electrode provides the surface on which the reduction can take place.) The half reactions are: Anode (oxidation): Zn(s) g Zn2+(1 M) + 2e- Cathode (reduction): 2H+(1 M) + 2e-g H2(1 atm) Overall: Zn(s) + 2H+(1 M) g Zn2+(1 M) + H2(1 atm) The standard emf of any cell, Eocell, is: Eocell = Eocathode - Eoanode

  16. The emf for this cell is = 0.76 V at 25 oC The emf for this cell is = 0.34 V at 25 oC

  17. The standard emf of any cell, Eocell, is: Eocell = Eocathode – Eoanode Therefore, for the first half reaction of Zn and H2: Eocell = EoH+/H2 – EoZn2+/Zn 0.76 V = 0 - EoZn2+/Zn Therefore: (EoZn2+/Zn = -0.76 V) We can also find thestandard electrode potential for copper with Hydrogen, where the cell diagram is: Pt (s) H2(1 atm) H+ (1 M) Cu2+ (1 M) Cu(s) The half reactions are: Anode (oxidation): H2 (1 atm)g 2H+(1 M) + 2e- Cathode (reduction): Cu2+(1 M) + 2e-g Cu(s) Overall: H2 (1 atm) + Cu2+(1 M) g 2H+(1 M) + Cu(s)

  18. Therefore, for the first half reaction of Cu and H2: Eocell = EoCu2+/Cu – EoH+/H2 0.34 V = EoCu2+/Cu – 0 Therefore: EoCu2+/Cu = 0.34 V For the Daniell Cell (Cu and Zn) we can now write that the combination of both will be: Eocell = Eocathode – Eoanode Eocell = EoCu2+/Cu - EoZn2+/Zn Eocell = 0.34 V – (-0.76 V) Eocell = 1.10 V This is an example of how we use hydrogen as a reference point to find thestandard electrode potential of any galvantic cell.

  19. Free Energy and Chemical Equilibrium (This is from Ch 19.7) If we start a reaction in solution with all the reactants in their standard states, 1 M concentrations, then as soon as the reaction starts, the standard-state conditions will no longer exist. Under these conditions, (not standard-state), we must use DG rather than DGo. The relationship between the two is: DG = DGo + RT ln Q (R is the gas law constant, 8.314 J / K x mol, T is the absolute temperature of the reaction, and Q is the reaction quotient derived from the equilibrium expression.)

  20. Free Energy and Chemical Equilibrium Let’s look at 2 different types of cases for DG = DGo + RT ln Q: • If DGo is a large (-) value, the RT ln Q term will not become (+) enough to match the DGo term until a significant amount of product has formed. • If DGo isa large (+) value, the RT ln Q term will be more negative than DGo is(+) only as long as very little product formation has occurred and the concentration of the reactant is high relative to that of the product. At equilibrium, by definition, DG = 0 and Q = K. Therefore, we get: 0 = DGo + RT ln K DGo = -RT ln K

  21. Spontaneity of Redox Reactions To relate cell potential, Eocell to thermodynamic quantities such as DGo and K, we can change chemical energy into electrical energy by the following equation: DGo = -nFEocell Where n = the moles of electrons that pass through the circuit, F = faraday = 96,500 J / V x mol and Eocell is the cell potential and is spontaneous when it is (+). Remember that from previous slides: DGo = -RT ln K Where R is the gas law constant 8.314 J/K x mol, T is the temperature in Kelvin and K is the equilibrium constant of the reaction. Therefore: nFEocell = RT ln K For a spontaneous reaction, K > 1 , Eocell =(+) and DGo = (-) For a reaction at equilibrium, K = 1 , Eocell =(0) and DGo = (0) For a non-spontaneous reaction, K < 1 , Eocell =(-) and DGo = (+)

  22. The Effect of Concentration on Cell emf So far we have been talking about redox reactions with reactants and products in their standard states. But this is difficult to find. There does exist a relationship between the emf of a cell and the concentration of reactants and products when they are not in their standard states. The Nernst Equation Consider a redox reaction: aA + bB g cC + dD DG = DGo + RT ln Q Because DG = -nFE and DGo= -nFEo, we can get: -nFE = -nFEo + RT ln Q Which is the Nernst Equation: E = Eo - (RT/nF) ln Q This can also be written as:E = Eo - (0.0257 V/n) ln Q

  23. Corrosion Corrosion is the term usually applied to the deterioration of metals by an electrochemical process. The most familiar example is the formation of rust from iron. Oxygen gas, water and acid in the atmosphere must be present for the formation of rust. Iron serves as the anode: Fe(s) g Fe2+(aq) + 2e- The electron given up by the iron is used to reduce the oxygen in the atmosphere into water (along with the presence of acid in the atmosphere). O2(g) + 4H+(aq) + 4e-g 2H2O(l) The overall redox rxn is: 2Fe(s) + O2(g) + 4H+(aq) g 2Fe2+(aq) + 2H2O(l)

  24. The Fe2+ ions formed at the anode are further oxidized by oxygen into Rust: 4Fe2+(aq) + O2(g) + (4+2x)H2O(l)g 2Fe2O3.xH2O(s) + 8H+(aq)

  25. Other metals such as copper and silver also corrode.

  26. Electrolysis In contrast to the spontaneous redox reactions, which result in the coversion of chemical energy into electrical energy, electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. An electrolytic cell is an apparatus for carrying out electrolysis. The same principles underlie electrolysis and the processes that take place in electrochemical cells. Electrolysis of Molten Sodium Chloride In its molten state, sodium chloride, an ionic compound, can be electrolyzed to form sodium metal and chlorine. The diagram on the next slide is known as the Downs cell, which is used for large-scale electrolysis of NaCl. The electrolytic cell contains a pair of electrodes connected to the battery. The battery serves as an “electron pump”,driving electrons to the cathode, where reduction occurs, and withdrawing electrons from the anode, where oxidation occurs.

  27. Theoretical estimates show that the Eo value for the overall process is about –4V, which means that this is a nonspontaneous process. Therefore, a minimum of 4 V must be supplied by the battery to carry out the reaction. • Is a practical example of the electrolytic Downs cell – you can actually see the way liquid Na and gaseous Cl2 is harnessed. • Is a simplified example of the electrolytic Downs cell.

  28. Electrolysis of Water We all know that water does not spontaneously decompose into Hydrogen and Oxygen gas under normal standard room conditions. This is because the standard free-energy change for the reaction is large and positive. But, this reaction can be induced in a cell like the one shown on the next slide. This electrolytic cell consists of a pair of electrodes made of a nonreactive metal, such as platinum, immersed in water. When the electrodes are connected to the battery, nothing happens because there is not enough ions in pure water to carry much of an electric current. The reaction occurs readily in a 0.1 M H2SO4 solution because there are a sufficient number of ions to conduct electricity. Immediately, gas bubbles begin to appear at both electrodes. The process at the anode is: 2H2O(l)g O2(g) + 4H+(aq) + 4e- The process at the cathode is: H+(aq) + e-g ½ H2(g) The overall reaction is given by: Anode (oxidation): 2H2O(l)g O2(g) + 4H+(aq) + 4e- Cathode (reduction): 4(H+(aq) + e-g ½ H2(g)) . Overall: 2H2O(l)g 2H2(g) + O2(g)

  29. Quantitative Aspect of Electrolysis The quantitative treatment of electrolysis was developed primarily by Faraday. He observed that the mass of product formed (or reactant consumed) at an electrode is proportional to both the amount of electricity transferred at the electrode and the molar mass of the substance in question. For example, in the electrolysis of molten NaCl, the cathode reaction tells us that one Na atom is produced when one Na+ ion accepts an electron from the electrode. To reduce 1 mole of Na+ ions, we must supply Avogadro’s number of electrons to the cathode. On the other hand, the stoichiometry of the anode reaction shows that oxidation of two Cl- ions yields one chlorine molecule. Therefore, the formation of 1 mole of Cl2 results in the transfer of 2 moles of electrons from the Cl- ions to the anode. In an electrolysis experiment, we generally measure the current (in amperes, A) that passes through an electrolytic cell in a given period of time. The relationship between charge (in coulombs, C) and current is: 1C = 1A x 1 second A coulomb is the quantity of electrical charge passing any point in the circuit in 1 second when the current is 1 ampere.

  30. Calculating the quantity of a substance produced in electrolysis Example problem: Suppose a current of 0.452 A is passed through the cell for 1.50 hrs. How much product will be formed at the anode and at the cathode in a molten CaCl2 electrolytic cell? 1. First step is to determine which species will be oxidized at the anode and which species will be reduced at the cathode: Anode (oxidation): 2Cl-(l)g Cl2(g) + 2e- Cathode (reduction): Ca2+(l) + 2e-gCa(l) Overall: Ca2+(l) + 2Cl-(l)gCa(l) + Cl2(g) • Find the charge (the number of electrons) that passes through the electrolytic cell. The quantity of calcium metal and chlorine gas formed will depend on the charge = current x time: ? C = 0.452 A x 1.50 hrs x 3600 s x 1 C = 2.44 x 103 C 1hr 1A x s • The mass can be calculated from the Coulomb to electron to mole to gram connection. 1 mole of e-s = 96,500 C and 2 mole of e-s are required to reduce 1 mole of Ca2+ ions, the mass of Ca metal formed would then be: ? g Ca = 2.44 x 103 C x 1 mol e-s x 1 mol Ca x 40.08 g Ca = 0.507 g Ca 96,500 C 2 mol e-s 1 mol Ca ? g Cl2 = 2.44 x 103 C x 1 mol e-s x 1 mol Cl2 x 70.90 g Cl2 = 0.896 g Cl2 96,500 C 2 mol e-s 1 mol Cl2

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