Electrochemistry: Oxidation-Reduction Reactions Zn(s) + Cu +2 (aq)  Zn 2+ (aq) + Cu(s) loss of 2e - gaining to 2e - Zinc is oxidized - it goes up in charge and is the reducing agent reduced Copper is reduced because it goes down in charge and is the oxidizing agent

Electrochemistry Messing with electrons What is the charge on each atom KMnO 4 Cr(OH) 3 Fe(OH) =

Electrochemistry: Oxidation-Reduction Reactions By writing the oxidation number of each element under the reaction equation, we can easily see the oxidation state changes that occur Zn(s) + 2H + (aq)  Zn 2+ (aq) + H 2 (aq) In any oxidation-reduction reaction (redox), both oxidation and reduction must occur.

Electrochemistry: Balancing Oxidation-Reduction Reactions Oxidation Number Method Al(l) + MnO 2  Al 2 O 3 + Mn e’s +4e’s X 4 = -12 X 3 =

Electrochemistry: Balancing Oxidation-Reduction Reactions Sample problem: I 2 O 5 (s) + CO(g)  I 2 (s) + CO 2 (g) -2e’s +5e’s X 2 = 10 X 5 =

: MnO 4 - (aq) + C 2 O 4 2- (aq)  Mn 2+ (aq) + CO 2 (g) (acid) Half-Reaction Method

: Half-Reaction Method (in acid) MnO 4 -  Mn 2+ C 2 O 4 2-  CO 2 MnO 4 - (aq) + C 2 O 4 2- (aq)  Mn 2+ (aq) + CO 2 (g) Step 1: Divide into two half-reactions, Step 2: Balance main element MnO 4 -  Mn 2+ C 2 O 4 2-  2CO 2

:: Step 3: Balance the O atoms by adding H 2 O MnO 4 -  Mn 2+ C 2 O 4 2-  2CO 2 Step 5: Balance charge with electrons 8H + + MnO 4 -  Mn H 2 O C 2 O 4 2-  2CO 2 + 4H 2 O 8H e e - Step 4: Balance the H atoms by adding H + MnO 4 -  Mn H 2 O C 2 O 4 2-  2CO 2

: Step 6: Make the electrons balance 5e - + 8H + + MnO 4 -  Mn H 2 O C 2 O 4 2-  2CO 2 + 2e - 10e H + + 2MnO 4 -  2Mn H 2 O 5C 2 O 4 2-  10CO 2 +10e - Step 7: Cancel and add 16H + + 2MnO C 2 O 4 2-  2Mn CO 2 + 8H 2 O

Balancing in base CN - + MnO 4 -  CNO - + MnO 2 CN -  CNO - MnO 4 -  MnO 2 H 2 O + + 2H 2 O + 2H + 4H e - 3e H + + 3CN - + 2MnO 4 -  3CNO - + 2MnO 2 + H 2 O 2OH - + 2OH - 2H 2 O 1

Balancing in base Cr(OH) 3 + ClO  CrO Cl 2 Have at it 4Cr(OH) 3 + 6ClO + 8OH -  4CrO Cl H 2 O

Note that the activity series is simply the oxidation half- reactions of the metals ordered from the highest to lowest

Using the hydrogen electrode as a reference Using the hydrogen electrode as a reference 2H + (1 M) + 2e - 2H + (1 M) + 2e -  H 2 (1 atm, 25 °C) E° red = 0 V

Cell EMF or Voltage Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s) E° cell = 1.10 V E ° ox + E ° red = E° cell From the table: Zn e -  Zn  red = -.76V Cu e -  Cu  red =.34V Flip Zn  Zn e -  ox =.76V

Voltaic or Galvanic Cells - oxidation + reduction

Sample Problem. Draw a voltaic cell using the following equation and label all parts Cr 2 O 7 2- (aq) + I - (aq)  Cr 3+ (aq) + I 2(s)

Spontaneity and Extent of Redox Reactions A positive emf indicates a spontaneous process, and a negative emf indicates a nonspontaneous one. Sample Problem: Using the standard electrode potentials, determine whether the following reaction are spontaneous  Cu 2+ (aq) + H 2(g) A. Cu (s) + 2H + (aq)  Cu 2+ (aq) + H 2(g)  2Cl - (aq) + I 2(s) B. Cl 2(g) + 2I - (aq)  2Cl - (aq) + I 2(s)  Cu 2+ (aq) + 2e - E ox  = -.34V Cu (s)  Cu 2+ (aq) + 2e - E ox  = -.34V  H 2(g) E red  = 0.0V 2H + (aq) + 2e -  H 2(g) E red  = 0.0V E cell  = -.34V  2Cl - (aq) E red  = 1.36V Cl 2(g) + 2e -  2Cl - (aq) E red  = 1.36V  2e - + I 2(s) E ox  = -.54V 2I - (aq)  2e - + I 2(s) E ox  = -.54V E cell  =.82V

EMF and Free Energy Change Any redox reaction involves free energy change (  G) which also may be used as a measure of spontaneity or work (max or min.) n = # of moles of e- s transferred  = C the charge of one mole of e -s or 96,500 J/V-mol e - Note that because n and  are both positive values, a positive value in E leads to a negative value of  G.  G° = -n  E °

Use the standard electrode potentials to calculate the standard free energy change,  G°, for the following reaction 2Br - (aq) + F 2(g)  Br 2(l) + 2F - (aq) 2Br - (aq)  Br 2(l) + 2e E° ox = F 2(g) + 2e -  2F - (aq) E° red = Br - (aq) + F 2(g)  Br 2(l) + 2F - (aq) E° cell = 1.81 V  G = -n  E  G = -(2 mol e - )(96,500 J/volt-mol e - )( 1.81 V)  G° = x 10 5 J = -349 kJ

EMF and Equilibrium Constant “Remember in Chapter 19, we related  G° to the equilibrium constant, K?”  G° = -RT lnK  G ° = -n  E ° -n  E° = -RT lnK E ° = logK n Simplify

Calculate the K for the following reaction O 2 (g) + 4H + (aq) + 4Fe 2+ (aq)  4Fe 3+ (aq) + 2H 2 O (l) O 2 (g) + 4H + (aq) + 4e -  2H 2 O (l) E° red = 1.23 V 4Fe 2+ (aq)  4Fe 3+ (aq) + 4e - E° ox = V O 2 (g) + 4H + (aq) + 4Fe 2+ (aq)  4Fe 3+ (aq) + 2H 2 O (l) E° cell = 0.46 V log K = n E ° V 4(0.46V) V = = 31.1 K = 1.36 x 10 31

Calculate the equilibrium constant for the reaction IO 3 - (aq) + 5Cu (s) + 12H + (aq)  I 2 (s) + 5Cu 2+ (aq) + 6H 2 O (l) 2IO 3 - (aq) + 12H + (aq) + 10e -  I 2 (s) + 6H 2 O (l)  red = 1.20V 5Cu (s)  5Cu 2+ (aq) + 10e -  ox = -.34V  cell =.86V  =.0591 log k n.86V =.0591 log k 10 Log K = K =

 G =  G° + RT lnQ  G° = -RT lnK Standard Conditions Non- Standard Conditions  G =  G° RT log Q -n  E = -n  E ° RT log Q E = E ° RT log Q n  Nernst Equation E = E ° log Q n When T=298K

Zn (s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu (s) E °= 1.10 V E = 1.10 ° log 2 [Zn 2+ ] [Cu 2+ ] If [Zn 2+ ]=0.05M, and [Cu 2+ ] = 0.50 M E = 1.10 ° log 2 [0.05] [.50] = 1.13V 2e - + Cu 2+ (aq)  Cu (s) Zn (s)  Zn 2+ (aq) + 2e - Therefore

Calculate the emf generated by the the following reaction Cr 2 O H + + 6I -  2Cr I 2 + 7H 2 O [Cr 2 O 7 2- ] = 2.0 M [H + ] =.05 M [I-] =.25 M [Cr 3+ ] [Cr 3+ ] = 1.0 x M  red = 1.33V  ox = -.54V  cell =.79V  =.79V log (1 x ) 2 6(2.0)(.05) 14 (.25) 6  =.68V

Concentration Cells What happens when both cells are identical except for concentration. Nature will try to equalize the two cells. E= Eº log Q n E = log.01 =.591V.1

 + 2e - anode: Pb (s) + SO 4 2- (aq)  PbSO 4 (s) + 2e - E °= V  cathode:PbO 2(s) + SO 4 2- (aq) + H + (aq) + 2e -  PbSO 4(s) + 2H 2 O (l) E °= V  Pb (s) + PbO 2(s) + 4H + +2SO 4(aq)  2PbSO 4(s) + 2H 2 O (l) E °= V Note that one advantage of the lead storage battery is that it can be recharged because the PbSO 4 produced during discharge adheres to the electrodes Lead Storage Battery

Dry Cell alkaline  2+ (aq) + 2e - anode: Zn (s)  Zn 2+ (aq) + 2e -  cathode:2NH 4 + (aq) +2MnO 2(s) + 2e -  Mn 2 O 3(s) + 2NH 3(aq) + H 2 O (l)

Fuel Cells  anode: 2H 2(g) + 4OH - (aq)  4H 2 O (l) + 4e -  4OH - cathode: 4e - + O 2(g) + H 2 O (l)  4OH - (aq)  2H 2 O (l) 2H 2(g) + O 2(g)  2H 2 O (l)

Electrolytic Cells: Electrolytic Cells: Electrolysis Electrolysis is driven by an outside energy source voltage source acts like an electron like an electron pump pump voltage source acts like an electron like an electron pump pump ( - red) These electrodes are inert These electrodes are inert Notice  (+ ox.) Eº = ( - )

Electrolysis of Aqueous Solutions Sodium cannot be prepared by electrolysis of aqueous solutions of NaCl, because water is more easily reduced than Na + : 2H 2 O + 2e -  H 2 + 2OH - E° red = Na + + e -  Na (s) E° red = The possible Anode reactions are: 2Cl -  Cl 2 E° ox = H 2 O  4H + + O 2 + 4e - E° ox = Cl - (aq)  Cl 2(g) E° ox = H 2 O + 2e -  H 2 + 2OH - E° red = E° cell > E° cell > E°cell > Possible Cathode reactions Therefore (Note: Not in table on AP exam)

Electrolysis of With Active Electrodes  Ni 2+ (aq) + 2e - E° ox = 0.28 Ni (s)  Ni 2+ (aq) + 2e - E° ox = 0.28  4H + + O 2 + 4e - E° ox = H 2 O  4H + + O 2 + 4e - E° ox =  Ni 2+ (aq) + 2e - Ni (s)  Ni 2+ (aq) + 2e - Ni 2+ (aq) + 2e -  Ni (s) anode: cathode: Electroplating creates a silver lining Possibilities

Quantitative Aspects of Electrolysis Calculate the mass of aluminum produced in 1.00 hr. by the electrolysis of molten AlCl 3 if the current is 10.0 A. (C = amperes x seconds) = 3.36 g of Al (10.0A) (1.00 hr) (3600 sec) (1 hr) (1 A-s) (1C) (96,500C) (1)(1) (3  ) (1molAl) (27.0 g Al)

Quantitative Aspects of Electrolysis The half-reaction for the formation of magnesium metal upon electrolysis of molten MgCl 2 is  Mg. Calculate the mass of magnesium formed upon passage of 60.0 A for a period of 4000 s Mg e -  Mg. Calculate the mass of magnesium formed upon passage of 60.0 A for a period of 4000 s (60.0A)(4000s)(1C)(1  ) (mole Mg)(24.3g) (As)(96,500C) (2  ) (mole Mg) 30.2g Mg

Electrical Work since  G= w max and  G = - n  E then w max = - n  E The unit employed by electric utilities is kilowatt-hour (kWh =3.6 x 10 6 J)

Corrosion reactions are redox reactions in which a metal is attacked by some substance in its environment and converted to an unwanted compound. All metals except gold and platinum are thermodynamically capable of undergoing oxidation in air at room temperature

The rusting of iron is known to require oxygen; iron does not rust in water unless O 2 is present. E° red = 0.44 V E° red = 1.23 V Note that as the pH increases, the reduction of O 2 becomes less favorable

The corrosion of Iron:Protecting the surface with tin The corrosion of Iron:Protecting the surface with tin  Fe e - Fe(s)  Fe e - E° ox = 0.44 V  Sn e - Sn(s)  Sn e - E° ox = 0.14 V Tin has lower E° ox so less likely to oxidize until the surface is broken then it accelerates as it makes a voltaic cell with iron.

The corrosion of Iron:Cathodic Protection  Fe 2+ (aq) + 2e - Fe(s)  Fe 2+ (aq) + 2e - E° ox = 0.44 V  Zn 2+ (aq) + 2e - Zn(s)  Zn 2+ (aq) + 2e - E° ox = 0.76V Galvanization Galvanization sacrificial anode Zinc is more positive and goes first and has an oxide coat that seals.

The corrosion of Iron:Cathodic Protection