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Oxidation-Reduction Reactions Electrochemistry: Oxidation-Reduction Reactions gaining to 2e- Zn(s) + Cu+2 (aq)  Zn2+(aq) + Cu(s) loss of 2e- Copper is reduced because it goes down in charge Zinc is oxidized - it goes up in charge

Messing with electrons Electrochemistry Messing with electrons What is the charge on each atom +1 +7 -8 +3 -2 = 1 +3 -3 +1 +7 -2 +3 -2 +1 +3 -1 KMnO4 Cr(OH)3 Fe(OH)2+1

Zn(s) + 2H+(aq)  Zn2+(aq) + H2(aq) 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 0 +1 +2 0 Zn(s) + 2H+(aq)  Zn2+(aq) + H2(aq) In any oxidation-reduction reaction (redox), both oxidation and reduction must occur.

Balancing Oxidation-Reduction Reactions Electrochemistry: Balancing Oxidation-Reduction Reactions Oxidation Number Method -3e’s X 4 = -12 Al(l) + MnO2  Al2O3 + Mn 4 3 2 3 0 +4 -2 +3 -2 0 +4e’s X 3 = +12

Balancing Oxidation-Reduction Reactions Electrochemistry: Balancing Oxidation-Reduction Reactions Sample problem: +2 +4 -2e’s X 5 = -10 I2O5(s) + CO(g)  I2(s) + CO2(g) 5 5 +5e’s X 2 = 10 +5

MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (acid) Half-Reaction Method : MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (acid)

Half-Reaction Method (in acid) : MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) Step 1: Divide into two half-reactions, MnO4-  Mn2+ C2O42-  CO2 Step 2: Balance main element MnO4-  Mn2+ C2O42-  2CO2

MnO4-  Mn2+ + 4H2O C2O42-  2CO2 MnO4-  Mn2+ + 4H2O 8H+ + Step 3: Balance the O atoms by adding H2O : : MnO4-  Mn2+ + 4H2O C2O42-  2CO2 Step 4: Balance the H atoms by adding H+ 8H+ + MnO4-  Mn2+ + 4H2O C2O42-  2CO2 Step 5: Balance charge with electrons +7 +2 5e- + 8H++ MnO4-  Mn2+ + 4H2O -2 C2O42-  2CO2 + 2e-

5e- + 8H++ MnO4-  Mn2+ + 4H2O C2O42-  2CO2 + 2e- Step 6: Make the electrons balance : 10 16 2 2 8 5e- + 8H++ MnO4-  Mn2+ + 4H2O 5 10 10 C2O42-  2CO2 + 2e- Step 7: Cancel and add 10e- + 16H++ 2MnO4-  2Mn2+ + 8H2O + 5C2O42-  10CO2 +10e- 16H+ + 2MnO4- + 5C2O42-  2Mn2+ + 10CO2 + 8H2O

CN- + MnO4-  CNO- + MnO2 -1 +1 H2O + CN-  CNO- + 2H+ + 2e- 3e-+ Balancing in base CN- + MnO4-  CNO- + MnO2 -1 +1 3 3 6 3 6 H2O + CN-  CNO- + 2H+ + 2e- 6 2 8 2 4 3e-+ 4H+ + MnO4-  MnO2 + 2H2O +3 2H+ + 3CN- + 2MnO4-  3CNO- + 2MnO2 + H2O + 2OH- 2OH- 2H2O 1

Cr(OH)3 + ClO  CrO4-2 + Cl2 Have at it Balancing in base 4Cr(OH)3 + 6ClO + 8OH- 4CrO4-2 + 3Cl2 + 10H2O

metals ordered from the highest to lowest 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 2H+ (1 M) + 2e-  H2 (1 atm, 25 °C) E°red = 0 V

Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) Cell EMF or Voltage From the table: Zn+2 + 2e-  Zn red= -.76V Flip Cu+2 + 2e-  Cu red= .34V Zn  Zn+2 + 2e- ox = .76V Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) E°cell = 1.10 V E ° ox + E ° red = E°cell

Voltaic or Galvanic Cells - oxidation + reduction I love M dawg Voltaic or Galvanic Cells

Sample Problem. Draw a voltaic cell using the following equation and label all parts include line notation. Cr2O72-(aq) + I-(aq)  Cr3+(aq) + I2(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 A. Cu(s) + 2H+(aq)  Cu2+(aq) + H2(g) Cu(s)  Cu2+(aq) + 2e- Eox = -.34V 2H+(aq) + 2e-  H2(g) Ered = 0.0V Ecell = -.34V B. Cl2(g) + 2I-(aq)  2Cl-(aq) + I2(s) Cl2(g) + 2e-  2Cl-(aq) Ered = 1.36V 2I-(aq)  2e- + I2(s) Eox = -.54V Ecell = .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.) G° = -nE ° n = # of moles of e-s transferred  = 96500 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.

Use the standard electrode potentials to calculate the standard free energy change, G°, for the following reaction 2Br-(aq) + F2(g) Br2(l) + 2F-(aq) 2Br-(aq)  Br2(l) + 2e E°ox = -1.06 F2(g) + 2e-  2F-(aq) E°red = 2.87 2Br-(aq) + F2(g) Br2(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° = -3.49 x 105 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 Simplify E ° = 0.0591 logK n

Calculate the K for the following reaction O2 (g) + 4H+ (aq) + 4Fe2+ (aq)  4Fe3+ (aq) + 2H2O (l) O2 (g) + 4H+ (aq) + 4e-  2H2O (l) E°red = 1.23 V 4Fe2+ (aq)  4Fe3+ (aq) + 4e- E°ox = -0.77 V O2 (g) + 4H+ (aq) + 4Fe2+ (aq)  4Fe3+ (aq) + 2H2O (l) E°cell = 0.46 V n E ° log K = 4(0.46V) = = 31.1 0.0591V 0.0591V K = 1.36 x 1031

 = .0591 log k n .86V = .0591 log k 10 Log K = 145.5 K = 10145.5 Calculate the equilibrium constant for the reaction IO3- (aq) + 5Cu (s) + 12H+ (aq) I2 (s) + 5Cu2+ (aq) + 6H2O (l) 2IO3- (aq) + 12H+ (aq) + 10e-  I2 (s) + 6H2O (l) red= 1.20V 5Cu (s)  5Cu2+ (aq) + 10e- ox= -.34V  = .0591 log k n cell= .86V .86V = .0591 log k 10 Log K = 145.5 K = 10145.5

Lead Storage Battery anode: Pb (s) + SO42- (aq)  PbSO4 (s) + 2e- E °= 0.356 V cathode:PbO2(s)+ SO42-(aq)+ H+(aq)+ 2e-  PbSO4(s)+ 2H2O(l) E °= 1.685 V Pb(s)+ PbO2(s)+ 4H+ +2SO4(aq) 2PbSO4(s)+ 2H2O(l) E °= 2.041 V Note that one advantage of the lead storage battery is that it can be recharged because the PbSO4 produced during discharge adheres to the electrodes

Dry Cell alkaline anode: Zn (s)  Zn 2+ (aq) + 2e- cathode:2NH4+(aq)+2MnO2(s)+ 2e-  Mn2O3(s) + 2NH3(aq) + H2O (l)

Fuel Cells anode: 2H2(g)+ 4OH-(aq) 4H2O(l) + 4e- cathode: 4e- + O2(g) + H2O(l)  4OH-(aq) 2H2(g) + O2(g)  2H2O(l)

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

2H2O + 2e-  H2 + 2OH- E°red = -0.83 Na+ + e-  Na(s) E°red = -2.71 Electrolysis of Aqueous Solutions Sodium cannot be prepared by electrolysis of aqueous solutions of NaCl, because water is more easily reduced than Na+: Possible Cathode reactions (Note: Not in table on AP exam) 2H2O + 2e-  H2 + 2OH- E°red = -0.83 Na+ + e-  Na(s) E°red = -2.71 The possible Anode reactions are: 2Cl-  Cl2 E°ox = -1.36 2H2O  4H+ + O2 + 4e- E°ox = -1.23 Therefore 2Cl-(aq)  Cl2(g) E°ox = -1.36 2H2O + 2e-  H2 + 2OH- E°red = -0.83 E°cell > -2.19

Electrolysis of With Active Electrodes Possibilities Ni(s)  Ni2+(aq) + 2e- E°ox = 0.28 2H2O  4H+ + O2 + 4e- E°ox = -1.23 Ni(s)  Ni2+(aq) + 2e- anode: cathode: Ni2+(aq) + 2e-  Ni(s) Electroplating creates a silver lining

= 3.36 g of Al Quantitative Aspects of Electrolysis Calculate the mass of aluminum produced in 1.00 hr. by the electrolysis of molten AlCl3 if the current is 10.0 A. (C = amperes x seconds) (10.0A) (1.00 hr) (3600 sec) (1C) (1) (1molAl) (27.0 g Al) (1 hr) (1 A-s) (96,500C) (3) (1 mol Al) = 3.36 g of Al

30.2g Mg Quantitative Aspects of Electrolysis The half-reaction for the formation of magnesium metal upon electrolysis of molten MgCl2 is Mg2+ + 2e-  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) (A•s)(96,500C) (2) (mole Mg) 30.2g Mg

Electrical Work since G= wmax and G = - nE then wmax= - nE The unit employed by electric utilities is kilowatt-hour (kWh =3.6 x 106J)

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 O2 is present. E°red = 1.23 V E°red = 0.44 V Note that as the pH increases, the reduction of O2 becomes less favorable

The corrosion of Iron:Protecting the surface with tin 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. Fe(s)  Fe2+ + 2e- E°ox = 0.44 V Sn(s)  Sn2+ + 2e- E°ox = 0.14 V

Galvanization The corrosion of Iron:Cathodic Protection Fe(s)  Fe2+(aq) + 2e- E°ox = 0.44 V Zn(s)  Zn2+(aq) + 2e- E°ox = 0.76V Zinc is more positive and goes first and has an oxide coat that seals. sacrificial anode Galvanization

The corrosion of Iron:Cathodic Protection