1. Spontaneity of Redox Reactions How is related to other thermodynamic quantities such as  G  and K. In an electrochemical cell, chemical energy is converted to electrical energy. Electrical energy in this case is the product of the emf of the cell and the total electrical charge (in coulombs) that passes through the cell: The total energy charge is determined by the number of moles of electrons(n) that pass through the circuit. By definition Where F is Faraday constant, is the electrical charge contained in 1 mole of electrons. Expts. Show 1 faraday is equivalent to 96,485.3 coulombs or 96,500 coulombs ( roughly). Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

2. 1F = 96,500 C/mol Because 1J = 1 C X 1V We can also express the units of faraday as 1 F = 96,500 J/V. mol The measured emf is the max. voltage that a cell can achieve. This value is used to calculate the max. amount of electrical energy obtained from the chemical reaction. This energy is used to do the electrical work (W ele ) hence W max =W ele = - n F E cell The –ve sign on the right hand side indicates that the electrical work done by the system on the surroundings.  G is the free energy available to do work, hence  G = W max We can now write  G = -n F E cell both n and F are positive quantities and  G is negative quantity for a spontaneous process, so E cell must be positive quantity. Under standard state conditions the above eqn. becomes  G 0 = -n F E 0 cell ………………..(1) Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

3. From our study on second law of thermodynamics we know the relation between free energy and equilibrium constant (K)  G  = - RT ln K …………(2) Combining eqn. (1) and (2) -n F E 0 cell = - RT ln K Solving for E 0 cell = …………(3) when T = 298K, eqn (3) can be simplified by substituting for R and F: E 0 cell = = the eqn can be rewritten using base 10 logarithm E 0 cell = Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

4. Thus, if any of the three quantities  G , K or E 0 cell is known, the other quantities can be calculated. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

5. Eg. Calculate the equilibrium constant for the following reaction at 25  C: Sn(s) +2Cu 2+ (aq)  Sn 2+ (aq) +2 Cu + (aq) Solution : to calculate the K first calculate E 0 cell of the Sn 2+ /Sn and Cu 2+ /Cu couples. The half cell reaction are Oxidation: Sn(s)  Sn 2+ (aq) + 2e - Reduction: 2Cu 2+ (aq) +2e -  2 Cu + (aq) From table we find E 0 Sn 2+ /Sn = -0.14 V and E 0 Cu 2+ /Cu + = 0.15 V. Thus E 0 cell = E 0 Cu 2+ /Cu + - E 0 Sn 2+ /Sn =0.15 V – (-0.14)= 0.29V E 0 cell = In the overall reaction we find n=2. Therefore lnK =22.6 K=e 22.6 K = 6.5 X 10 9 Practice: Calculate the equilibrium constant for the following reaction at 25  C : Fe 2+ (aq) +2Ag(s)  Fe(s) +2Ag + (aq) Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

6. Calculate the standard free energy change for the following reaction at 25˚ C. 2Au(s) + 3 Ca 2+ (1M)  2Au 3+ (1M) + 3Ca(s) To calculate  G 0 for the reaction we can use  G 0 = -n F E 0, first we need to calculate E 0. Calculate  G 0 for the following reaction at 25˚ C. 2 Al 3+ (aq) + 3Mg(s)  2Al(s) + 3Mg 2+ (aq) Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

7. The Effect of Concentration on Cell Emf The discussion so far were on redox reactions in their standard state conditions, but standard states are difficult to maintain and sometimes impossible. Nevertheless, there is a mathematical relationship between the emf of a cell and the concentration of reactants and products in a redox reaction under non standard state conditions. The Nernst Equation Consider a redox reaction of the type aA + bB  cC + dD from equation  G =  G˚ + RT ln Q ( from thermodynamics) Because  G =  nFE and  G˚ =  nFE˚, the equation can be expressed as  nFE =  nFE˚ + RT ln Q dividing this eqn by –nF, we get E=E˚  lnQ ( The Nernst Equation) where Q is the reaction quotient Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

8. E = E˚  ln Q At 298 K, the Nernst equation becomes E = E˚  ln Q (nat log) or E = E˚  log Q (base 10 log) At equilibrium, there is no net transfer of electrons, so E=0, Q=K, where K is the equilibrium. The Nernst equation enables us to calculate E as a function of reactant and product concentrations in a redox reaction. For example, for the Daniell cell Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) +Cu(s) E = E˚  ln Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

9. If the ratio [Zn 2+ /Cu 2+ ] is less than 1, ln [Zn 2+ /Cu 2+ ] is a negative number, so the second term on the right- hand side of the above equation is positive. Under this condition E is greater than the standard emf E˚. If the ratio is greater than 1, E is smaller than E˚. Predict whether the following reaction would proceed spontaneously as written at 298 K: Co(s) + Fe 2+( aq)  Co 2+ (aq) + Fe(s) given that [Co 2+ ]= 0.15 M and [Fe 2+ ]= 0.68 M. Reasoning and solution : We can use the value of E for the redox process to determine the spontaneity of the reaction. To do so we need to know E˚ and ln Q in the Nernst equation. The half reaction reactions are Oxidation: Co(s)  Co 2+ (aq) + 2e - Reduction: Fe 2+ (aq) + 2e -  Fe(s) The values of E˚ Co 2+ /Co = - 0.28V and E˚ Fe 2+ /Fe = -0.44V. There fore the standard emf is E˚ = E˚ Fe 2+ /Fe - E˚ Co 2+ /Co = -0.44V-(-0.28V) = -0.16V Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

10. Lets find the value of E E = E˚ - = - 0.16 V - = - 0.16+ 0.019V = - 0.14V Because E is negative, the reaction is not spontaneous in the direction written. Practice ques: Will the following reaction occur spontaneously at 25˚C, given that the [Fe 2+ ]=0.06M and [Cd 2+ ]= 0.010M? Cd(s) + Fe 2+ (aq) → Cd 2+ (aq) + Fe(s) Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

11. Now suppose you want to know what ratio of [Co 2+ ]/[Fe 2+ ] the reaction will would be spontaneous we can use the equation E = E˚ - We set E=0, which corresponds to the equilibrium situation 0 = - 0.16 V - = -12.5 = e -12.5 = K =4 X 10 -6 Thus for the reaction to be spontaneous, the ratio [Co 2+ ]/[Fe 2+ ] must be smaller 4 X 10 -6. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

12. Consider the electrochemical cell shown below. In a certain experiment, the emf(E) of the cell is found to be 0.54V. Suppose that the [Zn 2+ ]= 1.0M and p H 2 =1 atm. Calculate [H + ]. Write the redox reaction to find n, we know the standard reduction potential of E˚ Zn 2+ /Zn = 0.76V, substitute the values. ( ans= 2X10 -4 M) It may be useful to realize here that an electrochemical cell which involves the measurement of H + ions can be used to measure pH. This principle is used in pH meter. Home work: What is the emf of a cell consisting of a Cd 2+ / Cd half cell and a Pt/H + /H 2 half cell if [Cd 2+ ] = 0.20M, [H + ]= 0.16M and p H 2 =0.80atm ? This can be solved using the Nernst equation. Note that the conc. of gas expressed in atm. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

13. Concentration cells Electrode potentials depend on ion concentration, it is possible to construct a cell from two half cell of the SAME material but differing in ion conc. Such a cell is called concentration cell. Two zinc electrodes in two different aq. Solution of zinc sulphate. Eg 1.0M and 0.1M There fore reduction occur in 1.0M compartment and oxidation occur in 0.1M compartment. The cell diagram Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

14. And the half-reactions are Oxidation: Zn(s) → Zn 2+ (0.10M) +2e - Reduction: Zn 2+ (1.0M) +2e - → Zn(s) Overall: : Zn 2+ (1.0M) → Zn 2+ (0.10M) The emf of the cell is The of this cell is zero ( same electrode, same ions) so the emf E of the cell is The potential is small which eventually becomes zero when the conc. Becomes same. A biological cell can be compared to conc. Cell. Membrane potential develop due to unequal conc of same ions on different sides of membrane. Nerve impulses, heart beat, communication in living cells. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

15. Batteries An electrochemical cell or a series of combined electrochemical cells that can be used as a source of direct electric current at a constant voltage. The operation of battery is similar in principle to the elctrochemical cells (galvanic, daniell, voltaic), a battery has the advantage of being completely self contained. ( doesn’t require salt bridge or other components). Types of batteries The dry cell battery or Leclanche’ cell The mercury battery The lead storage battery The solid –state lithium battery Fuel cells Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

16. Corrosion: Deterioration of metals by an electrochemical process. Corrosion causes enormous damage-buildings, bridges, ships, automobiles etc. Examples-Rusting of iron, tarnish on silver, green patina on copper or brass etc. What happens in rusting? In the formation of rust on iron, oxygen and water must be present. The reaction involved are quite complex and not completely understood, the main steps are believed to be as follows. The metal surface acts as the anode where oxidation occurs Fe (s) → Fe 2+ (aq) + 2e - E  =- 0.45 V The electrons given up by iron reduce atmospheric oxygen to water at cathode which is another region of the same metal surface: O 2 (g) + 4H + (aq) + 4e - → 2H 2 O(l) E  = 1.23V (note that the E  of the reduction half reaction will be less than 1.23V because the H+ is not 1 M but even at pH 7 the reduction half reaction is 0.81 V, which means the cell potential is positive and the reaction is spontaneous.) The overall reaction is 2Fe (s) + O 2 (g)+ 4H + (aq)→ 2Fe 2+ (aq) +2H 2 O(l) E  = E  cat - E  and = 1.23V-(-0.44V)= 1.67 V Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

17. Notice the reaction occurs in acidic medium : the H + ions are supplied in part by the reaction of atmospheric CO 2 with H 2 O to form H 2 CO 3. the Fe 2+ ions is further oxidized by oxygen. Fe 2+ (aq) + O 2 (g) + (4 + 2x) H 2 O(l)→ 2Fe 2 O 3. xH 2 O + 8H + (aq) The hydrated form of iron(III) oxide is called rust. As the amount of water associated with iron oxide varies, we represent the formula as 2Fe 2 O 3. xH 2 O Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

18. The electrical circuit is completed by the migration of electrons and ions causing rusting rapidly. The fast rusting of automobiles in cold placed is due to salts (NaCl or CaCl 2 ) spread on the roads to melt ice. Aluminum has greater tendency to oxidize than Iron (compare reduction potential). We might expect to see lot of corroded aluminum like soda cans, our cooking pots etc. How ever this doesn’t happen because Al forms insoluble Al 2 O 3 layer when it is exposed in air. This layer is not porous ( like iron) and protects the underlying metal. Copper forms a layer of green substance called patina (CuCO 3 ) which protects the underlying metal Silver forms silver sulfide when it comes in contact with food stuff. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

19. Methods adopted to protect metals from corrosion. Most methods are targeted to prevent rust formation. Painting metal surface with paint. But sooner the paint is scratched the iron is exposed and rusting starts. Metals such as chromium, tin or zinc afford a more durable surface coating. The steel used in making automobiles, for example, is coated by dipping it in a bath of molten zinc, a process known as galvanizing. As the potentials indicate Fe 2+ (aq) + 2e -  Fe(s) E˚= -0.45V Zn 2+ (1.0M) +2e - → Zn(s) E˚= -0.76 V Zinc is oxidized more easily than iron, and therefore when the metal corrodes, zinc is oxidizes instead of iron. An incipient oxidation of iron would reverse the process immediately since Zn can reduce Fe 2+. As long as zinc and iron are in contact, Zn protects iron from oxidation even if zinc layer is scratched. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

20. Cathodic protection is The technique of protecting a metal from corrosion by connecting it to second metal that is more easily oxidized. It is unnecessary to cover the entire surface of the metal with second metal, as in galvanizing. All it requires is an electrical contact with the second metal. An under steel pipeline can be protected by connecting it through an insulated wire to a stake of magnesium, which acts as a sacrificial anode and corrodes instead of the iron. In effect the arrangement is a galvanic cell in which the easily oxidized magnesium acts as the anode and the pipeline behaves as the cathode and the moist soil acts as the electrolyte. Anode: Mg(s) Mg 2+ (aq) + 2e - E˚= 2.37V* Cathode: O 2 (g) + 4 H + (aq) +4e - 2H 2 O(l) E˚ =1.23V* *written as oxidation potential PROBLEM Mg is often attached to the hull of ships to protect to steel from rusting. Write a balanced equation for the corrosion reactions that occur (a) in the presence of Mg and (b) in the absence of Mg. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

21. Electrolysis Process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur. An electrolytic cell is an apparatus used for carrying out electrolysis. The same principles underlie electrolysis and the processes that take place in a electrochemical cell. Electrolysis of molten sodium chloride. Sodium chloride is an ionic compound which can be electrolyzed to form sodium metal and chlorine. The cell used for large scale electrolysis of NaCl is called a Down’s cell. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

22. The reactions at the electrodes are Anode(oxidation) : 2Cl - (l) → Cl 2 (g)+ 2 e - Cathode(reduction: 2Na + (l) +2e - → 2Na(l) Overall; 2Na + (l) + 2Cl - (l) → 2Na(l)+ Cl 2 (g) This process is the major source of pure sodium metal and chlorine gas. Theoretical aspects show that the E˚ value for the overall process is about -4V, which means the reaction is non spontaneous process. There fore a minium voltage of 4V must be supplied by the battery to carry out the reaction. Inpractice however a higher voltage is required because of inefficiencies in the electrolytic process because of over voltage. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

23. Electrolysis of water. Study on your own ELECTROLYSIS of of Aqueous Sodium Chloride This is the most complicated electrolysis. In the electrolysis of aqueous NaCl, the solution contains several species that could be oxidized or reduced. The possible oxidations reaction at anode are 2Cl - (aq) → Cl 2 (g) + 2e - 2H 2 O(l) → O 2 (g) + 4H + (aq) +4e - Comparing their standard electrode potential Cl 2 (g) + 2e - → 2Cl - (aq) E= 1.36V O 2 (g) + 4H + (aq) +4e - → 2H 2 O(l)E= 1.23V The E are not very different but the values suggest that H 2 O should be preferentially oxidized at the anode. But in the actual experiment the gas liberated is Cl 2 and not O 2. In studying electrolytic process we sometimes find that the voltage required for a reaction is higher than the electrode potential indicates. Overvoltage is the difference between the electrode potential and the actual voltage required to cause electrolysis. The over voltage for O2 is quite high. Therefore under normal operating conditions Cl2 is formed instead of O2. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

24. The reductions that might occur are 2H + (aq) + 2e - → H 2 (g) E= 0.00V 2H 2 O(l)+2e - → H 2 (g) + 2OH - (aq) E= -0.83V Na+ (aq) + e- → Na(s)E= -2.71V Reaction 3 is ruled out because it has a very negative potential. Reaction 1 is preferred over reaction 2 but at pH 7 both are equally probable. Reaction 2 is used to describe the cathode reaction because the concentration of H + is too low ( 1 X10 -7 M) to make reaction 1 reasonable Thus the half cell reactions are Anode (oxidation):2Cl - (aq) → Cl 2 (g) + 2e - Cathode (reduction) 2H 2 O(l)+2e - → H 2 (g) + 2OH - (aq) Over all: 2H 2 O(l)+ 2Cl - (aq) → H 2 (g) + Cl 2 (g) +2OH - (aq) The over all reaction shows that as the Cl - ion concentration decreases OH - ion increases. In addition to H 2 and Cl 2 gas produced, a useful by product NaOH is obtained by the aqueous solution at the end of electrolysis Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

25. Note that from the analysis of electrolysis : cat ions are likely to be reduced at the cathode and the anions likely to be oxidized at anode and in aqueous solutions water itself may be oxidized or reduced. The out come depends on the nature of other species. Quantitative aspects of Electrolysis Developed primarily by Faraday- he observed that the mass of the 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 E.g. In the electrolysis of molten NaCl, The cathode reaction tells us that one Na+ ion accepts an electron from the electrode. To reduce one mole of Na+ ions, we must supply Avogadro’s number (6.02 X 1023) of electrons to the cathode. On the other hand, the stoichiometry of the anode reaction shows that the oxidation of two Cl- ion yields one chlorine molecule. Therefore, the formation of 1 mole of Cl2 results in the transfer of 2 mols of electrons from Cl- ions to the anode. Similarly, it takes 2 moles of electrons to reduce 1 mole of Mg2+ ions and three moles of electrons to reduce 1 mole of Al3+ ions. Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

26. In an electrolysis experiment, we 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 1s That is, a coulomb is a quantity of electrical charge passing any point in 1 second when the current is 1 ampere. Steps involved in calculating the quantities of subs produced in electrolysis. For example: Consider electrolysis for molten CaCl 2 in an electrolytic cell. Suppose a current of 0.452 A is passed through the cell for 1.50 hr. How much product will be formed at cathode and anode. Solution write the half reactions. Anode ( oxidation): 2Cl - (l) → Cl 2 (g) + 2 e - Cathode ( reduction): Ca 2+ (l) + 2 e - → Ca (l) Over all : Ca 2+ (l) + 2Cl - (l) → Ca (l)+ Cl 2 (g) Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk

27. The quantities of calcium metal and chlorine gas formed depends depends on current X time, or charge; C= ? C= 0.425A X 1.5 hr X 3600s/hr X1 C /1A.s= 2.44 X 10 3 C 1 mol e - =96,500 C and 2 mol of e - are required to reduce 1mol of Ca 2+ ions, the mass of Ca metal formed at the cathode is calculated as follows No. of mols of electron = 2.44 X 10 3 C / 96,500 C mol -1 2 mols of e - reduce 1 mole of Ca 2+ Therefore, 2.44 X 10 3 C / 96,500 C mol -1 will reduce =1/2 X 2.44 X 10 3 C / 96,500 C mol -1 =0.01264 mols Grams of Ca formed = 0.01264 mols X40.08g/mol =0.5067g Chang, R.(2002). Chemistry 7th ed.Mc graw Hill. bkk