2. Using Half Cell Potentials and Latimer Diagrams. 100 measured half cell potentials generate 10,000 full reactions

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1 Electrochemistry 1. Balancing Redox Reactions 2. Using Half Cell Potentials and Latimer Diagrams 100 measured half cell potentials generate 10,000 full reactions 3. E as a Thermodynamic state function G = -n E F 4. Applying the Nernst Equation - non standard conditions 5. Electrochemical Cells Voltaic and Electrolytic 6.Batteries 7. Applications to Analytical and Industrial Chemistry

2 Oxidation State Assignments rules on p-82 Assign O as -2 except for peroxides and O 2 Assign H as +1 except for hydrides and H 2 Assign halides -1 except in X 2 and oxyhalide species HClO, HClO 2 ClO 4- have Cl in +1, +3,and +7 ox states Sum of ox numbers adds up to net charge In Lewis structures there are 3 ways to divide up the bonding electrons. For octet - count them twice- once for each atom For formal charge - divide them equally For OX number - award both bonding electrons to more electronegative element.

3 LATIMER DIAGRAMS SHOW REDOX CHEMISTRY OF EACH ELEMENT

4 / HO 2 H-O-O. [:S-SO 3 ] 2- [:N=N=N:] or +2ea /3ea

5 Balancing Redox Reactions (review) * Assign oxidation states and write skeletal OX and RED half reactions Ox leo RED ger Multiply to eliminate electrons on adding OX and RED Balance Charge with H + in acid or OH - in base Balance H or O with H 2 O, check with O or H. * If only full rxn is needed it is easier to combine ox and red first Work examples from Chapter 5. `47-58 * Make sure skeletal atoms are balanced red 6 e - + Cr 2 O Cr 3+ X 1 not fully balanced ox Fe 2+ Fe e - X 6 6 Fe 2+ + Cr 2 O Cr Fe H Fe 2+ + Cr 2 O Cr Fe H 2 O

6 Zn + NO 3- NH 3 + Zn(OH) 4 2- ox 4 OH - + Zn Zn(OH) e - x 4 red 6 H 2 O + NO e - NH OH - 7 OH H 2 O + NO Zn NH Zn(OH) 4 2- Disproportionation. Cl 2 is both oxidized and reduced Cl 2 OCl - + Cl - ox 2 OH - + ½ Cl 2 OCl - + 1e - + H 2 O red ½ Cl e - Cl - 2 OH - + Cl 2 OCl - + Cl - + H 2 O

34 + 0.76= 1.10 V E 0 cell = -0.34 + 0.80 = + 0.

7 Activity series Na,K Mg Al Zn Fe Sn Pb H 2 Cu Ag Au Zn + Cu 2+ Cu + Zn 2+ Cu + 2 Ag + Cu Ag E o cell = = 1.10 V E 0 cell = = V anode is Zn, reverse sign of -0.76V anode is Cu, reverse sign of +0.34

8 SHE H + (1 M) / H 2 (1 atm) Pt E 1/2 = 0.00 V H 2 + Cu 2+ 2 H + + Cu Zn + 2H + H 2 + Zn 2+ E o ½ redn of Cu2+ = V E o ½ ox of Zn = V E o 1/2 red of Zn2+ = V

9 A CAUTION ON SIGNS and COMBINING HALF RXNS. TEXT Method E cell = E cathode - E anode or E rxn = E 1/2 redn - E 1/2 redn potential for the oxidn half rxn how confusing is that? Recommended Method Write both half reactions, reverse one of them and change the sign of E 1/2. Then multiply as needed and add making sure the electrons drop out. red 6 e - + Cr 2 O Cr 3+ X 1 E o 1/2 = V ox Fe 2+ Fe e - X 6 E o 1/2 = V 6 Fe 2+ + Cr 2 O Cr Fe V 14 H Fe 2+ + Cr 2 O Cr Fe H 2 O *

11 E o s are the potential at standard conditions GASES 1 atm solutes 1 M solids or liquids unit activity In ACID - the species which exists at 1 M acid is shown HClO 4, HClO 3 are strong acids and do not exist in 1 M H + HClO 2 and HClO are present as the acids IN BASE all are present as the conjugate bases. ph dependence of E o : species differing by H or O will require H + or OH - to balance the rxn and are ph dependent

12 ACID BASE Redn of ClO 4-, ClO 3- ClO 2, and Cl 2 are ph independent in base H 2 O + ClO e - ClO OH - is balanced

13 G = - n E F G is an extensive quantity kj/mol E is an intensive quantity Volts ( Note T vs. heat Q : 1 ml of 50 o water vs ml of 50 o water) forward E>0, G< 0 ; at equil E = 0, G = 0; reverse E< 0, G>0 Nernst Equation G = G o + RT ln Q -n E F = -n E o F + RT ln Q E = E o - RT/nF ln Q at T = 298 and using log RT/F = E = E o /n log Q The naught has exactly the same meaning as it did for G E is the potential at the conditions of concentration given by the reaction quotient Q. E o is the potential for all species in their standard states

14 3 criteria for thermodynamic equilibrium. K = Q E = 0 G = 0 If K > Q E >0 or G <0 reaction goes forward If K < Q E < 0 or G > 0 reaction goes in reverse Which you use is your choice. When a system is at equilibrium E = 0 = E /n log K eq E o = /n log K eq in volts and G o = -RT ln K in KJ/mol at T = x J K -1 mol -1 x 298 = 5.7 kj/mol Keq = 10 +n Eo/ = 10 - Go/5.7

15 SKIP STEP POTENTIAL 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. 1e - + Cr 3+ 2e - + Cr 2+ Cr o 3e - E o 1/2 G0 1/2 = -n Eo F Cr V F V + 2(0.91)F + Cr 3+ Cr o (s) V F E o 1/2 = G0 /nf = F /3F = V E 1/2 is the weighted average of the E 1/2 s weighted by the number of electrons

16 HF or F -?? PICK YOUR REFERENCE STATE 1) ½ F 2 (g) + 1 e - F - E o acid = E o base = V 2) ½ F 2 (g) + 1 e - + H + HF E o acid = V 3) HF H + + F - pk a = 3.14 the third can be computed from the other two, at 1 atm F 2 gas E 1 = log [F - ] E 2 = log [HF]/[H + ] but E 1 = E 2 so 0.19 = log [F - ][H + ]/[HF] 0.19 / = pka = 3.2 eq 1) is ph independent, eq 2) depends on ph If we want the E at ph = pka with 1 M total HF + F - we can use either. E 1 = log (1/2) = E 2 = log ( 1/2/ 7.2 e-4) =

17 Adding Half RXNS to give Full REACTIONS A bonus of E being an intensive quantity is that you don t need to consider the number of electrons in each half IF you produce a proper FULL RXNS ( which will not have e s in it!!!) In the oxidation of iron in air G o = -ne o F O H e - 2 H 2 O E o 1/2 = V F 4Fe 2+ 4 Fe e - E o 1/2 = V F 4 Fe 2+ + O H + 4 Fe H 2 O V F E o = - (-1.84)F / 4F = 0.46 V you get this by simply combining the E o 1/2 s without regard for balancing or number of e s. E is the driving force per electron - G is the driving force per mole of rxn

18 G o f for ions. Taking SHE as a reference, we assign Go f for H + (aq) = 0 Then for Zn (s) + 2H + Zn 2+ + H 2 1 Coul Volt = Joule G o = - n E o F = -2 ( V) 94,485 C/mol = 146 kj/mol = G o f H 2 + Go f Zn G o f H+ - G o f Zn (s) 0 + G o f Zn = -146 kj/mol ( see App D) From G o f in Appendix D obtain E o 1/2 for Fe e - Fe 0 G o = G o f (Feo ) - G o f (Fe2+ ) = 0 - (-78.90) = kj/mol E o = - G o /nf = x 10 3 J/mol / (96,485 C/mol x 2) = V a Joule = Coulomb Volt or explicitly include reference electrode to give same result. Fe 2+ + H 2 Fe H + G o = G o f (Fe2+ )

19 1) species oxidizes water to O 2 E redn > V x4 Co e - Co 2+ E o 1/2 = 1.82 V 2 H 2 O 4 H e - + O 2 E o 1/2 = V 4 Co H 2 O 4 H Co 2+ + O V 2) species reduces water to H 2 E ox > 0 2 H e - H 2 Eo 1/2 = 0.00 V Mn Mn e - E o 1/2 = V V 3) species is oxidized by O 2 E ox > V O H e - 2 H 2 O E o 1/2 = V Fe 2+ Fe e - E o 1/2 = V V reactions involving gases may have overpotential ( kinetics slow)

20 4. disproportionation MnO e - MnO 2 (s) Eo 1/2 = V x 2 MnO 2-4 MnO e - E o 1/2 = V 4 H MnO MnO MnO 2 (s) + 2 H 2 O V Cu e - Cu o E o 1/2 = V Cu + Cu e - E o 1/2 = V 2 Cu + Cu 2+ + Cu o V EX 1. How much Cu + remains at equilibrium if 0.1 mol of CuCl is dissolved in a litre of water? 2 Cu + Cu 2+ + Cu o since E >>0 assume reaction goes all the way to right. this would give 0.05 M Cu 2+ and 0.05 mol of Cu o metal plated out E= 0 = log ([Cu 2+ ]/[Cu + ] = log (0.05) - 2 log [Cu + ] [Cu + ] = 1.4 x 10-4 M

21 EX 2. Cl 2 water. When Cl 2 gas is bubbled thru water at 1 atm it absorbs 0.09 moles of gas per litre. What is present in chlorine water? What is the solubility of Cl 2 in water? disproportionation Cl 2 + H 2 O HOCl + Cl - + H + E o = = V 1 atm x x x we are not at standard conditions, we are at equilibrium, let x = E = 0 = log [HOCl][H + ][Cl - ] / P Cl = log x 3 x = M soln contains 0.03 M H +, 0.03 M Cl - and 0.03 M HOCl it also contains = 0.06 M Cl 2 (aq) = solubility of Cl 2 the ph = 1.52 and pka for HOCl = 7.54, we can neglect [OCl - ] = K a [HOCl]/[H + ] = 2.8 x 10-8 M note especially the importance of standard states and proper equilibria

The cell shown provides the Ksp of AgI E = 0.80-0.059 log (0.1/ [Ag + ] = 0.417 [Ag + ] = 9.1 x 10-9 M Ksp = [Ag + ][I - ] = x 2 = 8.

22 The cell shown provides the Ksp of AgI E = log (0.1/ [Ag + ] = [Ag + ] = 9.1 x 10-9 M Ksp = [Ag + ][I - ] = x 2 = 8.5 x What if we omit the AgI? concentration cells E o 1/2 for AgI (s) + 1 e- Ag o + I - is neither of these but requires saturated AgI in 1 M I - where [Ag + ] = M

23 The standard AgCl/Ag potential is listed as V This refers to the half reaction AgCl (s) + 1e - Ag o + Cl - Using the standard Ag + /Ag potential of 0.80 V and applying the Nernst eqn for non standard conditions E ½ = log (1 / [Ag + ]) The standard AgCl/Ag potential refers to 1 M [Cl - ] from Ksp = 1.8 x = [Ag + ][Cl - ] = [Ag + ][1] in saturated AgCl in 1 M Cl - [Ag + ] = Ksp E ½ = log (1/[Ksp]) = 0.22 V pksp = 0.58/ = 9.8 Ksp = = 1.6 x The AgCl/Ag electrode measures [Cl - ] E 1/2 = log [Cl - ]

24 Titration of Halides with Ag + /Ag electrode Titrate 100 ml of 0.03 M Cl -, 0.02 M Br - and 0.01 M I - using 0.1 M AgNO 3. Given Ksp AgCl 10-10, AgBr 10-13, AgI for simplicity we will neglect volume change, and use approx Ksp s etc. one drop ppts AgI : [Ag + ] = Ksp/ [I - ] = E = log[ag + ] = -0.04V 9.90 ml ppts 99% of I - [Ag + ] = Ksp/ [10-4 ] = E = V at this point we have yet to exceed Ksp of AgBr or AgCl Q= [Ag + ][Br - ] = 0.02 x first ppt of AgBr; [Ag + ] = /0.02 = 5 x ; E = log (5x10-12 ) = V 99% of Br - is ppt after 10 ml ml added Ag + ; E = log(5x10-10 ) = V we have yet to exceed Ksp of AgCl Q = (5 x )(0.03) < first ppt of AgCl appears when [Ag + ] = 3.3 x 10-9 or E = log (3.3 x 10-9 ) E = 0.29 V and when 99% has ppt ( ml) ; E = log(3.3 x 10-7 ) E = 0.41 V and at eq pt (30 ml) Ag + = 10-5 and E = 0.5 V Note that a single drop of AgNO 3 gives E = 0.04 V when AgBr starts to ppt and 0.05 V when AgCl begins to ppt and 0.09 V when complete.

25 Non-standard conditions vs. E o predictions Ag + + e - Ag o (s) V I - ½ I 2 + e V Ag + + I - Ag o (s) + ½ I 2 (s) V spontaneous BUT we know Ag + + I - AgI (s) 1/Ksp = in saturated AgI solution [Ag + ] = [I - ] = 10-8 M E = E o log 1 / [Ag + ][I - ] not spontaneous E = log = = V In an electrochemical cell with Ag + and I - in separate beakers the redox reaction is spontaneous with e s flowing from anode to cathode. < I - (1M) / I 2 (s) // Ag + (1M) /Ag> E o = V

26 Using G o f / standard states A) AgI (s) Ag o (s) + ½ I 2 (s) G o = kj/mol no go! B) Ag + + I - Ag o (s) + ½ I 2 (s) G o = kj/mol GOES or G o = - n E o F = F = C) Ag + + I - AgI (s) G o = kj/mol GOES or G o = log 1/Ksp = for Ksp G o = -5.7 log Ksp = + 91 kj/mol, the reverse of the above rxn NONSTANDARD CONDITIONS AT 10-8 M each ion for A) G = log 1 NO for B G = log = + 66 kj/mol NO for C G = log = 0.0 EQ

27 Stoichiometry in electrochemical cells Faraday constant F = 96,485 coulombs mol -1 a Faraday is a mole of electrons, 1 amp = C s -1 1 coulomb-volt = joule 1. How much Cu will be plated out if 10 amps flow for 10 minutes in the cell <Zn\Zn 2+ (1M) \\ Cu 2+ (1M)\Cu >? mol e s = 10 C s -1 x 600 s / 96,485 C mol -1 = mol we need 2 mol e s for each Cu so mol Cu = mol 2. If we have 100 ml of solution (1 M in each ion) obtain the final conc of the ions and the cell potential after 10 min. Zn + Cu 2+ 2e Zn 2+ + Cu mol can be used, ml cancels below mol e s will plate 0.062/2 mol Cu = x = 1.97 g E = /2 log ( 0.131/0.069) =1.09 V a spontaneous reaction will always produce a decrease in E over time.

28 <Zn\Zn 2+ (0.1 M) \\ Ag + (0.1 M) \Ag> Start with 1 L in each beaker, a KNO 3 salt bridge. After 10 amps flow for 6 min what do we have? E = E o /2 log ([Zn 2 +] / [Ag + ] 2 ) OX Zn Zn e V RED x 2 Ag e - Ag V Zn + 2 Ag + Zn Ag V 10 amps x 360 s = 3600 C 3600 C/ 96,485 C/mol = mol e s For mol of e s we gain mol Zn 2+ and lose mol Ag +. From salt bridge anode gets mol NO 3-, cathode K + in millimoles [ Zn 2+ ] = ( )mmol / 1000 ml = M [Ag + ] = ( )mmol / 1000 ml = M E = E o /2 log [Zn 2+ ]/[Ag + ] 2 = 1.55 V the e s will run until Ag + is used up after 100 mmol flow = 965 seconds.

29 Batteries derive electrical energy from chemistry spontaneous e flow - to +, forced uphill e flow + to -. anode is where OX occurs anode Pb + H 2 SO 4 PbSO 4 (S) + 2 e H + ~ V cathode PbO 2 (s) + H 2 SO e H + PbSO 4 (s) + 2 H 2 O ~ 1.46 V NET Pb + PbO H 2 SO 4 2 PbSO H 2 O 2.0 V

30 Ni/Cd battery < Cd\Cd(OH) 2 \\Ni(OH) 2 \ NiO 2 > anode Cd (s) Cd(OH) 2 (s) + 2 e V cathode NiO 2 (s) + 2e - + H 2 O Ni(OH) 2 (s) V Cd + NiO 2 Cd(OH) 2 + Ni(OH) V Were you wondering wny the Cd half reaction has E o = V? Cd e - Cd E o 1/2 = V Cd(OH) 2 has a Ksp = 2.5 x in 1 M base Cd 2+ = Ksp/[OH] 2 = 2.5 x E redn = /2 log {1/[Cd 2+ ]} = V E ox then is V in base. Dry cell battery and Alkaline battery (non-rechargeable) Zn (s) + 2 Cl NH MnO 2 (s) + H 2 O Mn 2 O 3 (s) + Zn(NH 3 ) 2 Cl 2 (s) + 2 OH - LeClanche Zn (s) + 2 MnO 2 (s) + 2OH - + H 2 O Mn 2 O 3 (s) + Zn(OH) 4 2-

31 Lithium Ion Battery -new tech E 0 = 3.7 V electrolyte is nonaqueous media that allows Li + transport between graphite and a metal oxide Anode: C gr Li x C gr + X Li + + x e - Li + /Li Cathode: Li 1-X CoO 2 + X Li + + x e- LiCoO 2 CoIV /Co III Li + moves from graphite to holes in LiCoO 2 /electrons move to maintain neutrality

32 Electro-refining of Copper Crude Cu bars are used as the anode and a pure Cu wire as the cathode A 0.1 V potential is applied Metals more active than Cu dissolve Al, Fe, Ni, Co, Zn but do not plate out at cathode Metals less active drop off as anode sludge Ag, Au, Pt - these pay the cost of the process! Cu Cu e - Cu e - Cu net reaction just involves transport of Cu from anode to cathode

33 Cl 2 production in a diaphragm cell anode oxid 2 Cl - Cl 2 (g) + 2 e - cathode redn 2H 2 O + 2 e - 2 OH - + H 2 (g) E o 1/2 = V E o 1/2 = V E cell = V

Cl 2 Production via Chlor-Alkali Process Anode 2 Cl - Cl 2 + 2e - -1.36 V Cathode Na + + 1 e - Na(Hg) - 1.

34 Cl 2 Production via Chlor-Alkali Process Anode 2 Cl - Cl 2 + 2e V Cathode Na e - Na(Hg) V Mercury serves as a liquid electrode, the amalgam stabilizes Na o but even with 99% recovery of the Hg there are environmental consequences.

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