1. Nernst Equation  G = -nF  cell  G o = -nF  o cell  G =  cell =  o cell - RT nF ln Q standard non-standard GoGo + RT ln Q

2. Nernst Equation Cu (s)   cell =  o cell - RT ln Q nF  o cell = 0.80 - 0.34 = 0.46 V Cu 2+ + 2e -  CuAg + + e -  Ag(s)  o red -  o ox =  o = 0.34 V  o = 0.80 V 2( ) b) reductionoxidation anode cathode Cu 2+ (aq) (1M ) 4M4M Cu (s)  Cu 2+ + 2e - a) oxidation  Ag + (aq) (1M)  Ag (s)

3. Nernst Equation Cu (s)  Cu 2+ (aq) (4M)  Ag + (aq) (1M)  Ag (s)  cell =  o cell - RT ln Q nF Cu (s)  Cu 2+ + 2e - 2Ag + + 2e -  2Ag (s) Q = = [Cu 2+ ] [Ag + ] = 4 1212 [products] m initial [reactants] n initial anode cathode 2

4. Nernst Equation Cu (s)  Cu 2+ (aq) (4M)  Ag + (aq) (1M)  Ag (s)  cell =  o cell - RT ln Q nF Cu(s)  Cu 2+ + 2e - 2Ag + + 2e -  2Ag  cell = 0.46V 2 = 0.44 V -(298) ln 4 (96,500) (8.314) anodecathode

5. Non-standard conditions G =G = = -85 kJ -(2 mol e - )(96,500 C/mol e - ) (0.44V) Cu (s)  Cu 2+ (aq) (4M)  Ag + (aq) (1M)  Ag (s) Cu(s)  Cu 2+ + 2e - 2Ag + + 2e -  2Ag  G o = = -89 kJ -(2 mol e - )(96,500 C/mol e - ) (0.46V)  G = -nF 

6. V Ni (s) Ag (s) 1 x 10 -3 M NiCl 2 1.0 M HCl AgCl (s) ξ = 0.54 VK sp AgCl = 0.54 =1.03 Ag + + e - → Ag (s) ξ o = 0.80 V Ni 2+ + 2e - → Ni (s) ξ o = -0.23 V -RT/nFln Q ln Q = 38.17 Q = 3.78 x 10 16 [Ag + ] 2 = [Ni 2+ ] [Ag + ] = 1.62 x 10 -10 [Ag + ][Cl - ] [AgCl] 1.62 x 10 -10 K sp =

7. Concentration cell Cu (s)  Cu(s)Cu 2+ oxidationreduction  cell =  o cell =  o red -  o ox = 0.34 - 0.34= 0  o cell - RT nF ln Q Cu 2+ (aq) (0.5M)  Cu 2+ (aq) (2M)  Cu (s) + 2e -  Cu(s)  Cu 2+ + 2e - 0 V

8. Concentration Cells Cu (s)  Cu 2+ (aq) (0.5M)  Cu 2+ (aq) (2M)  Cu (s) Cu(s)  Cu 2+ + 2e - Cu 2+ + 2e -  Cu(s) = [products] m initial =  o cell = [Cu 2+ ] [reactants] n initial anode cathode = 0.5= 0.25 2 Q  cell = -RT 2F ln.25 =.02 V oxidationreduction [Cu 2+ ] cathode -RT nF ln Q0

9. Concentration Cells P 680 QAQA + - h Mn 2H 2 O4H + pH = 3.0 pH = 7.0  cell =  o cell - RT ln Q nF  cell =  o cell - 0.059 log Q n ADP + P i ATP  G > 0 Q = 10 -7 / 10 -3  cell = 0.24 V  G < 0 + O 2 + 4e - +

10. 2H + + 2e -  H 2  o = 0.00 V MnO 4 - + 8H +  Mn 2+ + 4 H 2 O +5e -  o = 1.51 V  o cell =  G o = -nF  o 2MnO 4 - + 6H + + 5H 2  2Mn 2+ + 8 H 2 O Go =Go =  o red -  o ox = 1.51 - 0 = 1.51 V = - (10)(96,500)(1.51) = -1457 kJ -RT ln K= -1457 kJ -RT K = e 588

11. Balancing Redox Reactions MnO 4 -  Mn 2+ manganese is balanced MnO 4 -  Mn 2+ + balance oxygen using H 2 O 4H2O4H2O balance hydrogen using H + MnO 4 - +  Mn 2+ + 4 H 2 O 8H + balance charge using e - MnO 4 - + 8H +  Mn 2+ + 4 H 2 O +5e -

12. Oxidation States MnO 4 - + 8H +  Mn 2+ + 4 H 2 O +5e - MnO 4 - Oxidation state of oxygen = Oxidation state of hydrogen = 4 x (-2)+ 1(Mn)= -1 Mn = 7+ Mn 2+ Mn = 2+ Mn 7+ + 5 e -  Mn 2+ 0 in O 2 2- in compounds 1+ in compounds 0 in H 2

13. 2H + + 2e -  H 2  o = 0.00 V MnO 4 - + 8H +  Mn 2+ + 4 H 2 O +5e -  o = 1.51 V reduction reaction MnO 4 - +8H + +5e -  Mn 2+ +4 H 2 O oxidation reaction H 2  2H + +2e - anode Pt (s)  2MnO 4 - + 16H + +10e - + 5H 2  2Mn 2+ + 8 H 2 O + 10H + + 10e - 2MnO 4 - + 6H + + 5H 2  2Mn 2+ + 8 H 2 O cathode H 2 (g) (1atm)  H + ( aq) (1M)  Pt (s) Mn 2+ (aq) (1M)MnO 4 - (aq)( 1M),  6,H + (1M) ( )5( )2