2. Assign oxidation states and set up skeletal OX and RED half reactions including number of electrons gained or lost.

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1 Electrochemistry Notes for Chem 1001S These are notes and outlines of the material covered in class. They are not meant as a substitute for reading the text or attending lectures. 1. Balancing Redox reactions method differs from that in text 1. first balance elements other than H and O 2. Assign oxidation states and set up skeletal OX and RED half reactions including number of electrons gained or lost. SHORT CUT if balanced half reactions not required otherwise do 4 and 5 for each half rxn first. 3. Multiply each half reaction by integers such that electrons gained = electrons lost and add them. The full equation contains no electrons at this point. 4. In acid add H + and in base OH - to balance charge. 5. Add H 2 O to balance either H or O and then check the other to complete. There are many examples to work in Chapter 5. Work them to perfect your skills. A few tips on balancing equations 1. Always balance the charge. 2. Never have H + and OH - in the same redox equation 3. Never add H 2 or O 2 unless they are actually present and involved in the reaction. 4. ALWAYS check H,O and charge balance when finished. 5. Become familiar with the most common oxidizing and reducing agents and their oxidation numbers. 6. Memorize the O 2 /H 2 O half cell as students often mess this one up. 7. Begin by asking what was oxidized and what was reduced. 8. Make sure you haven t combined two oxidations or two reductions or have the electrons on the wrong side. ASSIGNING OX NUMBERS 1. H has ox state +1 except in H 2 (0) or metal hydrides (-1), simple cations like Na + (+1) Mg 2+ (+2) etc. 2. O has oxidation state -2 except in peroxides (-1) as in H 2 O 2 or O 2 or O 3 (0) 3. Work out other elements such that the sum of the oxidation numbers adds up to the net charge on the species + - Examples N in : NH 3 and NH 4-3 N 2 H 4-2 N 3 (-1/3) HNO 2 (+3) NO - 3 (+5) KNO 3 (+5) Cl in Cl - or KCl or CCl 4 (-1) Cl 2 O (+1) HClO or ClO - (+1) KClO 3 (+5) LEWIS STRUCURES AND OXIDATION NUMBERS If you have studied Lewis structures in Chem 1000 you may note that oxidation numbers derive from one of the three arbitrary ways of assigning electrons to atoms in molecules. 1. For assigning octet count both electrons in a bond for each atom i.e count them twice. 2. For formal charge divide the bonding electrons equally between the two atoms. 3. For ox number award both bonding electrons to the more electronegative element. (For homonuclear divide equally) The peculiar oxidation number of -1/3 in N 3 - is seen to arise because of the different ox numbers of each N in a given Lewis structure which average to -1/3. In N=N=N each terminal N has 2 lone pairs giving 6 total when

2 bonding electrons are divided equally (no difference in electronegativity of N vs. N) OX # = -1 each. The central N has no lone pairs and thus 4 electrons to call its own OX # = +1 On avg we have (-1+1-1)/3 If you write the other resonance form with one single and one triple bond. The single bonded N has 3 lone pairs and thus 7 electrons (OX #= -2), the triply bonded N has one lone pair and thus 2+3 = 5 (OX # =0) and the central N has no lone pairs and thus = 4 electrons (OX # = +1) On avg we have (-2+0+1)/3 = -1/3. E vs. G. G = -nef where n = number of electrons and F = Faraday constant = 96,485 Coulombs/mol 1 Coulomb volt = joule so E in volts X F gives Joules/mol. Note that E is an intensive quantity, not on a per mol or gram basis. E has the same value at any point in a solution. G is an extensive quantity - on a per mol basis. An analogy is heat Q vs. temperature T. Heat is extensive, temperature intensive. Almost everything we have said about G and G o applies to E and E o E is a measure of spontaneity + goes forward 0 at equilibrium - goes in reverse While for G - goes forward 0 at equilibrium + goes in reverse because G = -nef The naught refers to standard concentrations (1 molar ) or pressures (1 atm) just as with G. At any other conditions E is obtained from the Nernst equation which can be derived as follows: G = G o + RT lnq -nef = -ne o F + RTlnQ E = E o RT/nF ln Q which at 298 and using logs becomes E = E o /n log Q At equilibrium E = 0 and thus E o = /n log K by analogy to G o = log K at 298 Standard Half Cell Potentials and their use. E o 1/2 are half cell potentials measured relative to the standard hydrogen electrode (SHE). The potential for 2H + + 2e - = H 2 (g) at 1 M H + and 1 atm H 2 = 0 This definition also sets G o f of H + (aq) = 0 and forms the basis for the free energies of ions in solution. An electrochemical cell consists of two half reactions one where oxidation occurs (ANODE) and the other where reduction occurs (CATHODE). The cell potential is obtained as follows. TEXT METHOD : E (cell) = E cathode E anode where both E s are reduction potentials given in tables. PREFERRED METHOD E (cell) = E OX + E RED where E OX = the oxidation potential obtained by reversing the reaction and changing the sign of E for the half cell where oxidation occurs. EXAMPLE : Zn + Cu 2+ Cu + Zn 2+ given Zn e - Zn E o = V Cu e - Cu E o = V A. Text method : note that Zn is oxidized and Cu reduced in the reaction E (cell) = 0.34 (-0.76) = V B. Preferred method : reverse the Zn half reaction to Zn Zn e - E (ox) = V Now add the two half reaction and the E s = = V

3 Why I prefer method B. When combining reactions either using G or E one ADDS. If a reaction is reversed one must change the sign of E or G for the reverse direction. In the preferred method one can see that the electrons drop out for a net reaction as they must. The text method is a rule which makes two mistakes that cancel out. You use a reduction potential for an oxidation half reaction and then you subtract instead of add potentials. This is an E-chem for dummies approach. LATIMER DIAGRAMS or Reduction Potential Diagams. ( see Chap 22.2 for example) Reduction potential data can be displayed as in Table 20.1 listing potentials in order from high to low. (About as sensible as phone book listings in order of phone number). Such tables give only a few potentials and one must search for the one you want. They provide no sense of the redox chemistry of a given element. Latimer diagrams (download a copy from the 1001S website) display reduction potentials by element showing the range of stable oxidation states, the relative stability of them, and the proper standard states in acid and base solution. (Continuing the phone book analogy, not only do you get the number by looking up the name, but you also get the next door neighbor s as well). For example Mn(IV) is shown as the insoluble MnO 2 (s), weak acids are protonated in 1 M H + (HNO 2 ) and not in 1 M base (NO 2 - ) etc. Skip step E o 1/2. Note that when combining half cell potentials to generate another half cell, they are not additive. The correct net half cell potential is the weighted average of the components. Thus for Cr e - Cr o (s) E o 1/2 = {E o 1/2 ( 1 e - + Cr 3+ Cr 2+ ) + 2 E o 1/2 ( 2e - + Cr 2+ Cr o ) }/3 E o 1/2 = { V + 2 (-0.91V)}/3 = V Think of E as the driving force per electron. We can prove this is the case by combing G s which are additive. G = -nef. G o ½ = -{n 1 E 1 F + n 2 E 2 F } = - (1)(-0.41)F + - (2)(-0.91)F = F E o 1/2 = - G o ½ /nf = - (+2.23) F/3F = V You can check your skill at this using any of the skipped step values shown in the Latimer diagrams.. PREDICTING CHEMICAL REACTIONS USING E o 1/2 s Combining any oxidation half reaction with any reduction half reaction gives a net chemical reaction. If the net E >0 the reaction is spontaneous. Because E is an intensive quantity you do not need to concern yourself with numbers of electrons or balancing to decide on spontaneity noting that E o s refer to standard conditions of 1M all reactants, products and H + if that is present in the net balanced reaction. Specific conditions can influence the outcome. Also the E o does not consider kinetics which is typically slow when gases like H 2 and O 2 are involved. Kinetic factors lead to an overpotential for H 2 and O 2 reactions at electrode surfaces. There are four reactions which must be examined to decide on thermodynamic stability of a chemical species in aqueous solution in air. These largely determine what form you find elements on earth. 1. Oxidation by water 2. Reduction by water 3. Oxidation by air (O 2 ) 4. Disproportionation These will usually be ph dependent so be careful to note whether you are in acid or base. A half cell involving a difference in the number of H s or O s between redox pairs will be ph dependent! Cl - /Cl 2 NO ClO 3 - /ClO 4 - YES

4 In 1 M H + 1. Water acts as an oxidation agent by having the H in it get reduced. 2H + + 2e - H 2 + 2e - E o 1/2 = 0.0 V Only species with oxidation potentials greater than 0.0 V will liberate H 2 from 1 M H +. Eg. Mn o 2. Water acts as a reducing agent by getting oxidized 2 H 2 O O 2 + 4e - 4H + E o 1/2 = V Only species with reduction potentials greater than V will liberate O 2 from water. Eg. MnO 4-3. Oxygen oxidizes by getting reduced to water O 2 + 4H + + 4e - 2 H 2 O E o 1/2 = V ( Note 3 = reverse of 2. All species with oxidation potentials greater than V will oxidize in air. Thus Fe 2+ and Sn 2+ do while Co 2+ and Ni 2+ don t. 4. Disproportionation The species oxidizes and reduces itself. Such species are unstable unless kinetics are slow and usually are not commonly found in solution for very long. Examples : H 2 O 2, MnO 4 Cl 2 in base but not in acid, Cu +, Mn 3+ Example 1. Mn e - Mn 2+ E (red) = +1.5 V Mn 3+ MnO 2 (s) + 1e - E (OX) = -1.0 V 2 Mn 3+ Mn 2+ + MnO 2 E = V spontaneous balanced 2 Mn H 2 O Mn 2+ + MnO 2 (s) + 4 H + Example 2. MnO e - MnO 2 E (red) = V MnO 4 MnO e - E (ox) = V - 3 MnO 4 2 MnO 4 + MnO 2 (s) E = V spontaneous balanced 3 MnO H MnO 4 + MnO 2 (s) + 2 H 2 O Example 3 in base ½ Cl 2 (g) + 1 e - Cl - E = V ½ Cl 2 (g) OCl e - E = V Cl 2 (g) OCl - + Cl - E = V spontaneous in 1 M OH - Balanced 2 OH - + Cl 2 (g) OCl - + Cl - + H 2 O balanced in base Note in 1 M acid the balanced reaction is Cl 2 (g) + H 2 O HOCl + Cl - + H + and E o = = V (not spontaneous at stnd conditions but see below) NERNST EXERCISES The Nernst equation is used to correct for non-standard conditions in a half reaction and in an electrochemical cell. For equilibrium conditions (when e-flow ceases) set E = 0 and E o = log K at 298 K. EX1. E 1/2 for the SHE at ph = 7. Write the balanced reaction 2 H e - H 2 (g) Write the Nernst eqn E = E o /n log Q Put in the correct value of n and Q and E o E = 0.0 V /2 log { P H2 / [H + ] 2 } Plug in 1 atm P H2 and 10-7 M H + E = log (1/[10-7 ] = V Ex2 Obtain the value of E 1/2 for O 2 /H 2 O in 1M OH - using the value for 1 M H +. Ex 3. If Cl 2 gas is bubbled through neutral water the water takes up 0.09 moles of the gas per litre of solution. The saturated solution becomes acidic. Obtain the concentrations of all relevant species. Cl 2 disproportionates. The balanced reaction in acid is Cl 2 (g) + H 2 O HOCl + Cl - + H +

5 And the E o = V ( using the 1 M acid reference state) Let X = the concentration of Cl -. From stoichiometry [H + ] = [OCl - ] = [Cl - ] = X Nernst E = E o log {[OCl - ][Cl - ][H + ] / P Cl2 } = 0 at equilibrium E o = = log {X 3 /1atm } X = 0.03 At equilibrium we have 0.03 M HOCl, Cl -, and H +. Of the 0.09 moles of Cl 2 gas taken up by a litre of water 0.06 are present as Cl 2 (aq) and.03 each as Cl - and OCl -. NOTE : You must be consistent in the choice of E o, reference states, n, and Q in your Nernst expression. Example A : For 2H + + 2e - H 2 (g) E o is 0.0, n = 2 and Q = P H2 / [H + ] 2 For H e - 1/2 H 2 (g) E o is 0.0, n = 1 and Q = (P H2 ) 1/2 / [H + ] Both give the same result Example B: At ph = 4 using the standard base reference potentials you balance the half cell as 2e - + H 2 O H 2 (g) + 2 OH - E = /2 log { P H2 [OH - ] 2 } and plug in 1 atm of hydrogen and OH - = to give E 1/2 = V OR you could use the standard 1 M H + reference and balance the half cell below and plug in [H + ] = 10-4 M 2H + + 2e - H 2 (g) E = /2 log{p H2 / [H + ] 2 } = V Both give the same result EXAMPLE 4. What would be the concentration of Cu + remaining in solution at equilibrium following disproportionation of a 0.2 M solution of cuprous chloride. 2 Cu + Cu o (s) + Cu 2+ E o = = V = log [Cu 2+ ]/[Cu + ] 2 Since E o is positive reaction goes largely to right giving (0.1-x/2 ) M [Cu 2+ ] and x M [Cu + ] K = /.0592 = 1.6 x 10 6 = (0.1)/ x 2 x = 2.4 x 10-4 M ELECTROCHEMICAL CELLS There are two kinds of electrochemical cells. The voltaic (AKA galvanic cell or battery or fuel cell) is a cell in which electrons spontaneously move from anode ( LEO =lose electrons oxidation) to cathode (GER = gain electrons reduction). These cells output the free energy of chemical reactions into useful electrical work. Electrolytic cells are ones where an external voltage is applied to drive a nonspontaneous reaction uphill. Electroplating of metals, production of Chlorine or electrolysis of water to give H 2 + O 2 are examples. These cells input energy to do work on chemicals producing higher energy chemicals. The basic principles are the same for both. The anode is always the electrode at which oxidation occurs. The only difference is the sign of the electrodes and the direction of electron flow. In a voltaic cell the anode is negative and electrons flow downhill from anode to + cathode. In an electrolytic cell the anode is positive and electrons are forced to flow uphill from + anode to cathode. A shorthand notation for cells is < anode(red form\anode)(ox form)// cathode(ox form)/ cathode(red form> with the spontaneous direction of e-flow being left to right.

6 Thus < Zn/Zn 2+ (0.01 M) // Cu 2+ (0.1 M)/Cu> describes a cell combining the zinc and copper half cells at nonstandard concentrations. E(cell) = E /2 log [Cu 2+ ]/[Zn 2+ ] = /2 log (0.1)/0.01) = 1.31 V Note that the reaction Zn + Cu 2+ Zn 2+ + Cu becomes less favorable (less positive E) compared to 1 M. ELECTRON FLOW, STOICHIOMETRY AND VOLTAGE CHANGE. The Faraday constant corresponds to the charge of a mole of electrons in Coulombs. You can treat electrons just as you would a stoichiometric reagent in a chemical reaction. Note that 1 amp = 1 Coulomb per second. Example 1. How long must a 10 amp current pass through a solution of Cu 2+ before 6.35 grams of copper is plated out? 6.35 grams of Cu = 0.1 mole, we need 2 electrons for each copper. Thus we need 0.2 F 0.2 (96,485) = 1.93 x 10 5 C. = amp X time = 10 t t = 1.93 x 10 4 seconds. Example 2. Let us consider the < Zn/Zn 2+ (0.01 M) // Cu 2+ (0.1 M)/Cu> above with a KNO 3 salt bridge. We will assume a litre of each of the two solutions. What do we have after passing 2 amp current for 20 minutes? 2 amp x 20 minutes x 60 s/min = 2.4 x 10 3 C or 2.4 x 10 3 C / 96,485C/mol = mol of electrons In the cell reaction Zn + Cu 2+ Zn 2+ + Cu initial change final and E(cell) = /2 log {[.0224]/[0.0876] = 1.08 As you see even large changes have only a small effect on the voltage until one of the concentrations gets near zero. Assuming an excess of the zinc metal, the cell would go dead when essentially all the Cu 2+ was used up. This would require 0.2 moles of electrons or x 10 4 C or a 2 amp current for 9648 seconds or minutes or 2.68 hrs. At equilibrium E = 0 and the remaining conc of Cu 2+ would be : log K = n E o / = 2(1.04 V)/ = 35.1 K = = [Zn 2+ ]/ [Cu 2+ ] = 0.11/ [Cu 2+ ] thus [Cu 2+ ] = M ( You see why we say redox rxns tend to go all the way. ) One thing to note is that the cell voltage changes very little until one of the reagents is almost used up. APPLICATIONS We will briefly discuss applications to analytical chemistry, batteries, and the chemical industry. Some of these are discussed in Chapters Batteries Pb acid battery, Ni/Cd, dry cell Analytical Chemistry; 1. Titration of halides using the Ag/Ag+ electrode Determining Ksp for AgCl or Kf for Ag(NH 3 ) 2 Electrolytic cells: Electrolysis of water, brine, Cl 2 production, Al, Na, Cu refining.