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Electrochemistry and Its Applications

Chapter 19. Electrochemistry and Its Applications. Humphry Davy 1778-1829. Prepared metallic K, Na, Sr, Ca, B, Ba, Mg, Li by electrolysis. LEO the lion goes GER. Oxidation-Reduction Reactions. OXIDATION = loss of electrons. Examples:. Na  Na + + e - Al  Al 3+ + 3e -

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Electrochemistry and Its Applications

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  1. Chapter 19 Electrochemistry and Its Applications Humphry Davy 1778-1829. Prepared metallic K, Na, Sr, Ca, B, Ba, Mg, Li by electrolysis.

  2. LEO the lion goes GER Oxidation-Reduction Reactions OXIDATION = loss of electrons Examples: Na  Na+ + e- Al  Al3+ + 3e- S2- S + 2e- OXIDATION = increasing the oxidation number (more positive) Example: NO  NO2 Oxid. Nos: +2 -2 +4 -2 change = +2 +2 to +4 : N is oxidized

  3. Oxidation-Reduction Reactions REDUCTION = gain of electrons Examples: N + 3e- N3- Fe3+ +e- Fe2+ LEO the lion goes GER REDUCTION = decreasing the oxidation number (more negative) Example: MnO4- Mn2+ Oxid. Nos: +7 +2 change = -5 +7 to +2 : Mn is reduced

  4. Review of Oxidation Numbers O.N. = 0 for atom in element form (C, Ag, O2, H2, P4, etc.) = charge for any monoatomic ion (e.g., +1 for Na+, -2 for S2- , +3 for Al3+) = -2 for oxygen in compound or ion (except peroxides and when bonded to fluorine) = +1 for hydrogen when bonded to non-metals (e.g., H2O, CH4, H3N, HCl) = -1 for hydrogen when bonded to metals (e.g., NaH) = -1 for oxygen in peroxides (e.g., HOOH) Sum of O.N. = 0 for neutral compound (e.g., for Na2CO3, Na = +1, O = -2, C = +4) Sum of O.N. = ion charge for polyatomic ion (e.g., for CO3-2, O = -2, C = +4)

  5. Oxidation-Reduction Reactions = “REDOX” What we have looked at are called half-reactions. Half-reactions are fully balanced with respect to both mass (atoms) and electrons (net charge) and are either reduction or oxidation (but not both), e.g., Sn2+ + 2e- Sn (Sn is reduced) FeCl3 + e- FeCl2 + Cl- (Fe is reduced) Mn2+(aq) + 4 H2O (l)  8H+(aq) + MnO4-(aq) + 5e- (Mn is oxidized) But half-reactions do not occur by themselves in the real world; reduction cannot occur without oxidation (and vice versa). Hence, we have REDOX occurring (both oxidation and reduction) in a chemical reaction.

  6. Oxidation-Reduction Reactions Half reactions are combined to get REDOX reactions. When combining, • charges must balance (this is conservation of electrons) and • atoms must balance (this is conservation of mass). Half-reaction method of balancing REDOX equations Consider the (unbalanced) reaction: MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (1) Write out separate reduction and oxidation half-reactions; (2) Add H2O and/or H+ as needed for mass balance, and add electrons (e-) for charge balance; (3) Multiply each half reaction by a common denominator so that electrons (e-) will cancel; (4) Add the half reactions and (mathematically) simplify.

  7. Balancing Equations by the Method of Half-Reactions MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) 1. The two incomplete half reactions are: MnO4-(aq)  Mn2+(aq) C2O42-(aq)  2CO2(g) 2a. Balance 1st half-reaction: MnO4-(aq)  Mn2+(aq) 5e- + 8H+ + MnO4- Mn2+ + 4H2O 2b. Balance 2nd half-reaction: C2O42-(aq)  CO2(g) C2O42- 2CO2 + 2e-

  8. Balancing Equations by the Method of Half-Reactions 3a. Find common denominator so that electrons will cancel. x 2 5e- + 8H+ + MnO4- Mn2+ + 4H2O x 5 C2O42- 2CO2+ 2e- The common denominator is 10. 3b. Multiply equations so that electrons will cancel. 10e- +16H+ + 2MnO4- 2Mn2+ + 8H2O 5C2O42- 10CO2 + 10e- The 10 e- on each side cancel.

  9. Balancing Equations by the Method of Half-Reactions 4. Add equations and simplify*: 16H+ +2MnO4- +5C2O42- 2Mn2++8H2O + 10CO2 5. Check to make sure both mass and charges balance. Left side: 16 H, 2 Mn, 28 O, 10 C, (+16-2-10) = +4 Right side: 16 H, 2 Mn, 28 H, 10 C, +4 *In some instances H+ and/or H2O appear on both sides of the Equation, so that simplification can be performed by subtracting H+ and/or H2O from the equation.

  10. Balancing Equations by the Method of Half-Reactions • Reactions Occurring in Basic Solution • We use OH- and H2O rather than H+ and H2O. • First, balance as usual, then add OH- to both sides so that all H+ is consumed. MnO4- (aq) + Br-(aq)  MnO2 (s) + BrO3-(aq) (basic soln) 2x [ 3e- + 4H+ + MnO4- MnO2 + 2H2O] 1x [ Br- + 3H2O  BrO3- + 6H+ + 6e- ] Add half reactions: 2H+ + 2MnO4- + Br- 2MnO2 + BrO3- + H2O Add 2OH-to both sides to remove 2H+ on left 2H+ + 2OH- + 2MnO4- + Br- 2MnO2 + BrO3- + H2O + 2OH- 2H2O + 2MnO4- + Br- 2MnO2 + BrO3- + H2O + 2OH- H2O + 2MnO4- + Br- 2MnO2 + BrO3- + 2OH-

  11. Oxidizing and Reducing Agents That which is oxidized is the reducing agent; That which is reduced is the oxidizing agent. Consider, for example, the following REDOX reaction: undergoing reduction undergoing oxidation Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Reducing agent is oxidized. Oxidizing agent is reduced. Note: Reactions that are not redox include acid-base rxs. (e.g., HCl + NaOH  H2O + NaCl); and metathesis (replacement, or ppt.) rxs. (e.g., NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl (s)).

  12. Cu Ag+ Electrolytic Cells There are essentially two types of cells: • Voltaic (or galvanic): • spontaneous REDOX rxn → electricity • (b) Electrolytic • electricity → nonspontaneous REDOX rxn • We’ll start with spontaneous processes. Consider copper (Cu) metal in silver ion (Ag+) soln Cu What happens? • soln becomes blue • (production of Cu2+) 2. White whiskers grow on Cu surface (production of metallic silver, Ag)

  13. Cu Ag+ Silver whiskers (Ag) growing on Cu metal Blue solution due to production of Cu2+ ion What are the half reactions? Ag+(aq) + e- Ag(s) Cu(s)  Cu+2(aq) + 2e- What is the overall REDOX reaction? 2Ag+(aq) + Cu(s)  2Ag(s) + Cu+2(aq) Silver ions are reduced to metal silver. Copper metal is oxidized to cupric ions. Obviously, the process is spontaneous (because we observe it to happen).

  14. e- Another setup: Voltaic Cells (external flow of electrons) Ag Cu Salt bridge Ag Same half reactions Cu2+ Ag+ Electrons flow from Cu on left to Ag on right Cu will be oxidized to produce Cu2+ {Cu(s)  Cu+2(aq) + 2e-}. Ag+ will be reduced to produce Ag {Ag+(aq) + e- Ag(s)}. This needs to be balanced by negative (-) charges. But how? Add a salt bridge (inverted U tube with solution of ions, e.g. Na+ NO3-) Through salt bridge, anions (NO3-) can now move into left chamber to balance Cu2+ being produced. Similarly, Na+ will move into right compartment to replace the Ag+ being reduced.

  15. e- (Same) Voltaic Cell + - This is a complete cell (made up of two half cells) Salt bridge anode Cu Ag cathode Ag Cu2+ Ag+ Anode Cathode Cu → Cu2+(aq) + 2e- Ag+ +(aq) + e- → Ag (oxidation) (reduction) Overall redox reaction: Cu (s) + 2 Ag+(aq)→ Cu2+(aq) + 2 Ag (s) The individual metals are the anode and cathode. Here, Cu is the anode. Ag is the cathode Oxidationalways takes place at the anode. Reductionalways takes place at the cathode. Electrons always flow in the external circuit from the anode (marked “-” to the cathode (marked “+”).

  16. Voltaic Cells (a porous barrier can be used instead of salt bridge)

  17. e- V Voltage of Cells (Same Cu-Ag Cell) + - cathode Ag Salt bridge Cu Ag anode [Ag+]=1.0 M [Cu2+]=1.0 M Cu2+ Ag+ Anode Cathode Cu(s) → Cu2+(aq) + 2e- Ag+ +(aq) + e- → Ag (s) (oxidation) (reduction) Cu (s) + 2 Ag+(aq)→ Cu2+(aq) + 2 Ag (s) If the solutions of Cu2+ and Ag+ were “standard” (i.e., 1.0 M), and we placed a voltmeter in the electron flow line, we would get a reading of +0.46 volt (at 25oC). This is the standard cell potential or Eocell (This is also called electromotive force (emf) Why is this reaction spontaneous? Why is the voltage 0.46 volt?

  18. 0.46 V This electromotive series presents data showing the tendency of substances to gain or lose electrons. Choose any two entries; the top one will act as the cathode and will be reduced; the bottom one will act as the anode and will be oxidized. The standard voltage will be the algebraic difference between the two respective potentials.

  19. The Ag|Ag+ half-cell has a higher reduction potential than the Cu|Cu2+ half-cell. This means that the Ag/Ag+ half-cell will more readily undergo reduction when compared to the Cu/Cu2+ half-cell, and the Cu|Cu2+ half-cell will undergo oxidation. The Ag|Ag+standard reduction potential is Eored = +0.80 v. The Cu|Cu2+standard reduction potential is Eored =+0.34 v. The overall cell potential is given by: Eocell= Eored (higher) – Eored (lower) The overall cell potential is always positive if it is spontaneous (which it will be if the higher reaction in the table is the cathode and the lower reaction in the table is the anode). Standard Reduction Potentials (SRPs) are all based on and compared to the hydrogen (H2|H+) half-cell which is assigned Eored =0 volts by convention. Caution: spontaneous processes have positive voltages (E°, or V°), but negative free energies (ΔG).

  20. Cell EMF – different electrodes Standard Reduction Potentials (SRPs) In this voltaic cell, hydrogen is the cathode and zinc is the anode. The measured voltage is 0.76 V.

  21. Hydrogen is the cathode and the zinc is the anode. The algebraic difference between the standard potentials is 0.76 V. Note: EMF (electromotive force), voltage, and cell potential are all synonyms 0.76 V

  22. F2 is the strongest oxidizing agent; F- is the weakest reducing agent. Li is the strongest reducing agent; Li+ is the weakest oxidizing agent.

  23. Using the Standard Reduction Potentials (RSP) to Predict Spontaneity Some metals are easily oxidized whereas others are not. e.g., Fe is oxidized by Ni2+ but Ni is not oxidized by Fe2+. The reaction: Fe + Ni2+ Fe2+ + Ni occurs, but Ni + Fe2+ Ni2+ + Fedoes not. Remember, the SRP table is an Activity series (see Ch 5), a list of metals arranged in order of ease of oxidation.* The lower a metal is on the SRP table, the more active that metal is, i.e., the more easily it is oxidized. Any metal can be oxidized by the ions of elements above it. *Caution! The Activity series of Ch 5 is in reverse order of the SRP table of this chapter.

  24. Let’s repeat the exercise using specific values of V: The voltage of the reaction Ni + Fe2+ Ni2+ + Fe is calculated by adding the reactions: Ni Ni2+ (V = +0.28) Fe2+ Fe (V = -0.44) The summed voltage is V = –0.16 and the reaction is nonspontaneous. The voltage of the reaction Fe + Ni2+ Fe2+ + Ni is calculated by adding the reactions: Fe Fe2+ (V = +0.44) Ni2+ Ni (V = -0.28) The summed voltage is V = +0.16 and the reaction is spontaneous.

  25. e- Nernst Equation – for nonstandard solutions – explains why batteries “run down” 0.0592 n log Q E = Eo - As Ag+ is depleted, the Ag+(aq)  Ag(s) equilibrium shifts to the left (le Châtlier’s principle) As Cu2+ builds up, the Cu(s)  Cu2+(aq) equilibrium shifts to the left (le Châtlier’s principle) + - Salt bridge Also, if the Cu(s) is completely consumed, the reaction cannot proceed. Cu Ag Ag Cu2+ Ag+ Anode Cathode Cu → Cu2+(aq) + 2e- Ag+ +(aq) + e- → Ag (oxidation) (reduction)

  26. Batteries Batteries are the most practical applications of voltaic cell. All batteries have self contained anode/cathode compartments. All operate using the same principles already discussed. The Classic “dry” (LeClanché) cell. zinc anode Overall reaction: Zn + 2 MnO2 + 2NH4+→ Mushy paste Of MnO2 and NH4Cl Zn2++2MnO(OH)+ 2NH3 E~1.5 v. carbon cathode

  27. Batteries Alkaline Battery (similar to dry cell but more efficient) Anode: (Zn cap) Zn(s)  Zn2+ (aq) + 2e- Cathode: MnO2, NH4Cl and C paste: 2NH4+(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + 2H2O(l)

  28. cathode Lead-Acid Battery anode PbO2 Sulfuric acid, H2SO4 (provides H+ ions) Eocell = 2.04 V Pb At cathode: PbO2(s) + HSO4-(aq) +3H+ +2e-→ PbSO4(s) + 2H2O (l) At anode: Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2e- Overall: PbO2(s) + Pb(s) + 2HSO4-(aq) + 2H+(aq) → 2 PbSO4(s) + 2H2O(l)

  29. Batteries Lead-Acid Battery Six cells in series give a total voltage of ~12 volts in an automobile battery.

  30. Batteries Fuel Cells • Direct production of electricity from fuels occurs in a fuel cell. • On Apollo moon flights, the H2 - O2 fuel cell was the primary source of electricity. • Cathode: reduction of oxygen: 2H2O(l) + O2(g) + 4e- 4OH-(aq) • Anode: 2H2(g) + 4OH-(aq)  4H2O(l) + 4e- • Total reaction: 2H2(g) + O2(g)  2H2O(l)

  31. Batteries Fuel Cells

  32. Corrosion Corrosion is the entropy monster’s greatest weapon. It is the evil side of REDOX. It costs 100s of billions of dollars yearly to prevent and correct. Basically, it is the spontaneous process (oxidation) of iron: Fe → Fe3+ + 3e- nice shiny metal (steel) ugly brownish-red powder ……….RUST! Rusting cannot occur by itself. Can’t have only the OX in REDOX; So, what gets reduced? Usually H2O or O2

  33. Corrosion Common type of “rusting” redox: (Eo ~ 0.8 V) O2(g) + 4H+(aq) + 2Fe(s) → 2Fe2+(aq) + 2H2O(l) Easy, but even more favorable in acid conditions. There are similar equations also involving water. Stopping Corrosion • Galvanize it (coat with Zn). Fe has higher SRP than Zn. Coupled with Zn, Fe is the cathode (cathodic protection) (look for a “matte” appearance of Zn). • Use “sacrificial metal” such as Mg – (this is also cathodic protection). • Cover it (paint). • Create rust-resistant alloys, e.g., stainless steel (Fe/Ni/Cr), or nickel steels (Fe/Ni).

  34. Corrosion Preventing the Corrosion of Iron Also used on ships to prevent corrosion

  35. Electrolysis Electrolysis occurs in electrolytic cells, i.e., in cells where: electricity → chemical reaction takes place. This is the opposite of a voltaic cell. external power In electrolysis, current is forced into the cathode by external power (like a battery) cathode anode • Widely used for electroplating (silver, gold, copper) • Also for production of certain metals from ores and salts (e.g., Na, Al)

  36. Electrolysis Electrolysis of water At cathode (V = 0.00): Why is the volume of H2 twice that of O2? O2 [2H+(aq) + 2e-→ H2(g)] x 2 H2 At anode (V = -1.23): 2 H2O →O2(g) + 4H+(aq) + 4e- Overall: 2 H2O → 2H2(g) + O2(g) --not spontaneous Eo = -1.23 v. cathode anode external power needed

  37. 1.23 V Note: EMF (electromotive force), voltage, and cell potential are all synonyms

  38. Electrolysis – Electrolysis of Molten Salts Decomposition of molten NaCl Cathode: 2Na+(l) + 2e- 2Na(l) Anode: 2Cl-(l)  Cl2(g) + 2e- Industrially, electrolysis is used to produce metals like aluminum (Hall-Héroult process, where Al2O3 is electrolyzed in molten cryolite, Na3AlF6, with a carbon electrode to give an overall reaction of 2Al2O3 + 3C  4Al + 3CO2)

  39. Electrolysis with Active Electrodes – Gold plating – protects against corrosion external power source external power source Au Au Au+(aq) Au+(aq) cathode: Au+(aq) +e-→Au anode: Au→Au+(aq) +e-

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