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Electrochemistry

Electrochemistry. w max. G =. w max = H - TS. . - n. F. w =. Electrical work. n =. # moles e -. F =. charge on 1 mol e -.  =. electrical potential. Ger. Electrochemistry. electrons . reduction oxidation. G E R. ain lectrons eduction. L E O. ose

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Electrochemistry

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  1. Electrochemistry wmax G = wmax = H - TS  - n F w = Electrical work n = # moles e- F = charge on 1 mol e-  = electrical potential

  2. Ger Electrochemistry electrons reduction oxidation G E R ain lectrons eduction L E O ose lectrons xidation

  3. Electrochemistry silver Ag(s) Ag+(aq) aluminum Al(s) Al3+(aq) Ag(s)  Ag+(aq) + e- G = 77 kJ mol-1 of Al(s)  Al3+(aq) Gof = + 3e- -481 kJ mol-1 these are reactions a) oxidation b) reduction

  4. Ag(s)  Ag+(aq) + e- Gof = 77 kJ mol-1 Al(s)  Al3+(aq) + 3e- Gof = -481 kJ mol-1 Al(s)  Al3+(aq) + 3e- Gof = -481 kJ mol-1 Ag+(aq) + e-  Ag(s) 3( ) Go = -77 kJ mol-1 3( ) __________________ ________________ Go = -712 kJ/mol + Al  3Ag + Al3+ 3Ag+ spontaneous

  5. Al  Al3+ + 3e- Ag+ + e- Ag a) oxidation b) reduction reduction 3e-  Cl- K+ Al Ag anode cathode 1M Al(NO3)3 1M AgNO3 1 mol Al3+ + 1 molecule Al3+ 3 mol NO3- 3Ag++ Al  Al3+ + 3Ag

  6. Al  Al3+ + 3e- Ag+ + e- Ag a) oxidation b) reduction reduction 3e-  Cl- K+ Al Ag anode cathode 1M Al(NO3)3 1M AgNO3 1 mol Al3+ + 1 molecule Al3+ 3 mol NO3- Al(s) Al3+(1 M)  Ag+(1 M) Ag(s)  

  7. Reduction Potential Al(s) Al3+(aq) (1M) Ag+(aq) (1M) Ag(s) Ag+ + e-  Ag Al(s)  Al3+ + 3e- 3Ag+ + Al  3Ag + Al3+ Go = -712 kJ/mol -n F o Go = wmax wmax = standard reduction potential o = standard = 1M, 1 atm reduction potential = tendency to gain e-

  8. half-reactiono(V) F2 + 2e-  2F- 2.87 spontaneousreduction 0.80 Ag+ + e-  Ag Cu2+ + 2e-  Cu 0.34 Standard Hydrogen electrode (SHE) 2H+ + 2e-  H2 0.00 o = 0 Sn2+ + 2e-  Sn -0.13 Al3+ + 3e-  Al -1.66 spontaneousoxidation Li+ + e-  Li -3.05 a) oxidizing b) reducing F2 is spontaneously reduced F2 is an agent Liis spontaneously oxidized Li is a agent reducing

  9. a) b) half-reactiono(V) F2 + 2e-  2F- 2.87 Ag+ + e-  Ag 0.80 Cu2+ + 2e-  Cu 0.34 2H+ + 2e-  H2 0.00 Sn2++ 2e- Sn -0.13 Al3+ + 3e-  Al -1.66 Li+ + e-  Li -3.05 o is intensive Al half-cell and Ag half-cell o (V) Ag+ + e-  Ag reduction reaction: 3() 0.80 _____________ ____ Al  Al3+ + 3e- oxidation reaction: 1.66 ocell =  3Ag+ +Al 3Ag +Al3+ 2.46 V

  10. ocell ocell=ored-oox Ag+ (aq) + e-  Ag(s) o = 0.80 V Al3+ (aq) + 3e- Al(s) o = -1.66 V ocell= 0.80 2.46 V - (-1.66) = spontaneous voltaic or galvanic cell ocell> 0 electrolytic cell non-spontaneous ocell< 0

  11. Electrochemical work ocell = 2.46 V 3Ag+ + Al  3Ag + Al3+ o Go = wmax = - n F n = mol of e- F = faraday = 96,500 C / mol e- o = standard reduction potential V (J/C) (2.46V) Go = -(3 mol e-) (96,500 C/mol e-) = -712170 CV = -712170 J = -712 kJ

  12. Balancing redox reactions  Cr3+(aq) Cr2O72- (aq) Cr2O72-  Cr3+ e-+ 6 H++ 14 2 + H2O 7 balance Cr balance O add H2O balance H add H+ balance charge add e-

  13. Cr2O72- + 14 H++ 6e- 2 Cr3++ 7 H2O oxidation or reduction oxidation state of Cr Cr2O72- 6+ Cr3+ 3+ O = 2- Cr has been reduced Cr is an agent oxidizing

  14. ( ) Cr2O72- + 14 H+ + 6e-  2 Cr3+ + 7 H2O oxidation or reduction gain e-reduction GER need oxidationreaction 2H+ (aq)  H2(g) balance charge with e- 2H+ + 2e- H2(g) H2(g) 2H+ (aq) + 2e- 3 another reduction reaction

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