Presentation is loading. Please wait.

Presentation is loading. Please wait.

Electrochemistry G = wmax w = - Pext V w = - n F  Mechanical work

Similar presentations


Presentation on theme: "Electrochemistry G = wmax w = - Pext V w = - n F  Mechanical work"— Presentation transcript:

1 Electrochemistry G = wmax w = - Pext V w = - n F  Mechanical work
Electrical work n = # moles e- F = charge on 1 mol e-  = electrical potential

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

3 Electrochemistry silver Ag(s) Ag+(aq) aluminum Al(s) Al3+(aq) Ag(s) 
77 kJ mol-1 of Al(s)  Al3+(aq) + 3e- Gof = -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 = kJ mol-1 3( ) Ag+(aq) + e-  Ag(s) Go = kJ mol-1 3( ) __________________ ________________ 3Ag+ + Al  3Ag + Al3+ Go = -712 kJ/mol spontaneous

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

7 Reduction Potential Al(s) Al3+(aq) (1M) Ag+(aq) (1M) Ag(s)
Al(s)  Al3+ + 3e- Ag+ + e-  Ag 3Ag+ + Al  3Ag + Al3+ Go = -712 kJ/mol wmax = -n F o Go = 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 Ag+ + e-  Ag 0.80
spontaneous reduction Ag+ + e-  Ag 0.80 Cu2+ + 2e-  Cu 0.34 Standard Hydrogen electrode 2H+ + 2e-  H o = 0 (SHE) Sn2+ + 2e-  Sn Al3+ + 3e-  Al spontaneous oxidation Li+ + e-  Li a) oxidizing b) reducing F2 is spontaneously reduced F2 is an agent Li is spontaneously oxidized Li is a agent reducing

9 o is intensive _____________ ____ half-reaction o (V)
F2 + 2e-  2F Ag+ + e-  Ag 0.80 Cu2+ + 2e-  Cu 0.34 2H+ + 2e-  H Sn2+ + 2e- Sn Al3+ + 3e-  Al Li+ + e-  Li o is intensive a) b) Al half-cell and Ag half-cell o (V) reduction reaction: 3( ) Ag+ + e-  Ag 0.80 oxidation reaction: _____________ ____ Al  Al3+ + 3e- 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 = V ocell = 0.80 - (-1.66) = 2.46 V spontaneous ocell > 0 voltaic or galvanic cell ocell < 0 electrolytic cell non-spontaneous

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

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

13 Cr2O H+ + 6 e-  2 Cr H2O oxidation or reduction oxidation state of Cr Cr2O72- 6+ Cr3+ 3+ O = 2- Cr has been reduced Cr is an agent oxidizing

14 oxidation or reduction
Cr2O H+ + 6 e-  2 Cr H2O oxidation or reduction GER gain e- reduction need oxidation reaction 2H+ (aq)  H2(g) balance charge with e- H2(g)  2H+ (aq) + 2e- 2H+ + 2e- H2(g) 3 ( ) another reduction reaction


Download ppt "Electrochemistry G = wmax w = - Pext V w = - n F  Mechanical work"

Similar presentations


Ads by Google