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Electrochemistry Mechanical work  G = Electrical work w = n = w = w max # moles e - F =charge on 1 mol e -  = electrical potential - P ext VV - nF.

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Presentation on theme: "Electrochemistry Mechanical work  G = Electrical work w = n = w = w max # moles e - F =charge on 1 mol e -  = electrical potential - P ext VV - nF."— Presentation transcript:

1 Electrochemistry Mechanical work  G = Electrical work w = n = w = w max # moles e - F =charge on 1 mol e -  = electrical potential - P ext VV - nF 

2 Electrochemistry electrons ose lectrons xidation ain lectrons eduction reduction oxidation LEOLEO GERGER Ger

3 Electrochemistry silver Ag(s) aluminum Al(s) Ag(s)  Al(s)   G =  G o f = Ag + (aq) Al 3+ (aq) 77 kJ mol -1 -481 kJ mol -1 Ag + (aq)+ e - ofof Al 3+ (aq) + 3e -

4 Ag(s)  Ag + (aq) + e - Al(s)  Al 3+ (aq) + 3e -  G o f = 77 kJ mol -1  G o f = -481 kJ mol -1 Al(s)  Al 3+ (aq) + 3e -  G o f = -481 kJ mol -1 Ag + (aq) + e -  Ag(s)  G o = -77 kJ mol -1 3( ) __________________ ________________ 3Ag +  G o = spontaneous + Al  3Ag +Al 3+ -712 kJ/mol

5 1 molecule Al 3+ 3Ag + + Al  Al 3+ + 3Ag 1M AgNO 3 1M Al(NO 3 ) 3 Ag Al Ag + + e -  AgAl  Al 3+ + 3e - 3e -  oxidation reduction K+K+ Cl - anodecathode 1 mol Al 3+ +3 mol NO 3 -

6 1M AgNO 3 1M Al(NO 3 ) 3 Ag Al K+K+ Cl - anodecathode Ag + +e -  Ag Al  Al 3+ + 3e - oxidation reduction Al(s)  Al 3+ (1 M)  Ag + (1 M)  Ag(s)

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

8 half-reaction  o (V)  o = 0 Standard Hydrogen electrode spontaneous reduction spontaneous oxidation (SHE) F 2 + 2e -  2F - 2.87 Ag + + e -  Ag 0.80 Cu 2+ + 2e -  Cu 0.34 2H + + 2e -  H 2 0.00 Sn 2+ + 2e -  Sn -0.13 Li + + e -  Li -3.05 Al 3+ + 3e -  Al -1.66

9 half-reaction  o (V) F 2 + 2e -  2F - 2.87 Ag + + e -  Ag0.80 Cu 2+ + 2e -  Cu0.34 2H + + 2e -  H 2 0.00 Sn 2+ + 2e -  Sn -0.13 Al 3+ + 3e -  Al -1.66 Li + + e -  Li -3.05 Al half-cell and Ag half-cell reduction reaction: oxidation reaction:

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

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

12  G o = w max = - nF oo = -RT ln K Equilibrium constant, K ln K= -712 kJ - RT K = e 287 = - nF  o - 8.314x10 -3 x 298 = 287


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