1. Electrochemical cells

2. Electrochemical Cells
_______
__________
_______
__________
spontaneous
redox reaction

3. Electrochemical Cells
Cell Diagram
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
[Cu2+] = 1 M & [Zn2+] = 1 M
Zn (s) | Zn2+ (1 M) || Cu2+ (1 M) | Cu (s)
anode
cathode

4. Standard Electrode Potentials
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
Anode (oxidation):
Zn (s) Zn2+ (1 M) + 2e-
Cathode (reduction):
2e- + 2H+ (1 M) H2 (1 atm)

5. Standard Electrode Potentials
(E0) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm.
Any time you see º, think “standard state conditions”
Reduction Reaction
2e- + 2H+ (1 M) H2 (1 atm)
E0 = 0 V
Standard hydrogen electrode (SHE)

6. Standard Electrode Potentials
E0 = 0.76 V
cell
Standard emf (E0 )
cell
E0 = Ecathode - Eanode
cell
E0 = Ereduct’n - Eoxid
cell
Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s)
E0 = EH /H - EZn /Zn
cell + 2+ 2
0.76 V = 0 - EZn /Zn 2+
EZn /Zn = V 2+
Zn2+ (1 M) + 2e Zn E0 = ____________

7. Standard Electrode Potentials
E0 = 0.34 V
cell
E0 = Ecathode - Eanode
cell
Ecell = ECu /Cu – EH /H 2+ + 2
0.34 = ECu /Cu - 0 2+
ECu /Cu = 0.34 V 2+
Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s)
Anode (oxidation):
H2 (1 atm) H+ (1 M) + 2e-
Cathode (reduction):
2e- + Cu2+ (1 M) Cu (s)

8. E0 is for the reaction as written
The more positive E0 the greater the tendency for the substance to be reduced
The more negative E0 the greater the tendency for the substance to be oxidized
Under standard-state conditions, any species on the left of a given half-reaction will react spontaneously with a species that appears on the right of any half-reaction located below it in the table (the diagonal rule)

9. The half-cell reactions are reversible
The sign of E0 changes when the reaction is reversed
Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

10. Can Sn reduce Zn2+ under standard-state conditions?
How do we find the answer?
Look up the Eº values in Table.
Zn2+(aq) + 2e- —> Zn(s) (Is this oxidation or reduction?)
Which reactions in the table will reduce Zn2+(aq)?

11. Cd is the stronger oxidizer
What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution?
Cd2+ (aq) + 2e Cd (s) E0 = V
Cd is the stronger oxidizer
Cd will oxidize Cr
Cr3+ (aq) + 3e Cr (s) E0 = V
Anode (oxidation):
Cr (s) Cr3+ (1 M) + 3e-
x 2
Cathode (reduction):
2e- + Cd2+ (1 M) Cd (s)
x 3
E0 = Ecathode - Eanode
cell
E0 = – (-0.74)
cell
E0 = _________
cell

12. Spontaneity of Redox Reactions
DG = -nFEcell
n = number of moles of electrons in reaction
F = 96,500 J
V • mol
DG0 = -nFEcell
= 96,500 C/mol
DG0 = -RT ln K
= -nFEcell
Ecell = RT nF
ln K
(8.314 J/K•mol)(298 K)
n (96,500 J/V•mol)
ln K = = V n
ln K
Ecell = V n
log K
Ecell

13. Spontaneity of Redox Reactions
If you know one, you can calculate the other…
If you know K, you can calculate Eº and Gº
If you know Eº, you can calculate Gº

14. Spontaneity of Redox Reactions
Relationships among G º, K, and Eºcell

15. What is the equilibrium constant for the following reaction at 250C
What is the equilibrium constant for the following reaction at 250C? Fe2+ (aq) + 2Ag (s) Fe (s) + 2Ag+ (aq) = V n
ln K
Ecell
Oxidation:
2Ag Ag+ + 2e-
n = ___
Reduction:
2e- + Fe Fe
E0 = EFe /Fe – EAg /Ag 2+ +
E0 = – (0.80)
E0 = ___________ V
x n E0
cell
exp
K = V
x 2
-1.24 V
= exp
K = ________________

16. Calculate DG0 for the following reaction at 250C.
2Al3+(aq) + 3Mg(s) Al(s) + 3Mg+2(aq)
Oxidation:
2Mg Mg2+ + 6e-
n = ?
Reduction:
6e- + 3Al Al
E0 = Ecathode - Eanode
cell
DG0 = -nFEcell

17. Calculate DG0 for the following reaction at 250C.
2Al3+(aq) + 3Mg(s) Al(s) + 3Mg+2(aq)
Oxidation:
2Mg Mg2+ + 6e-
E0 = __ V
Reduction:
6e- + 3Al Al
E0 = __ V
n = __
E0 = Ecathode - Eanode
cell
= ___ V
DG0 = -nFEcell
= ___ X (96,500 J/V mol) X ___ V
DG0 = _______ kJ/mol

18. The Effect of Concentration on Cell Emf
DG = DG0 + RT ln Q
DG = -nFE
DG0 = -nFE
-nFE = -nFE0 + RT ln Q
E = E0 -
ln Q RT nF
_____________ equation
At 298K - V n
ln Q E
E = - V n
log Q E
E =

19. The Nernst equation enables us to calculate E as a function of [reactants] and [products] in a redox reaction.

20. Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
Will the following reaction occur spontaneously at 250C if [Fe2+] = 0.60 M and [Cd2+] = M?
Fe2+ (aq) + Cd (s) Fe (s) + Cd2+ (aq)
Oxidation:
Cd Cd2+ + 2e-
n = ___
Reduction:
2e- + Fe Fe
E0 = EFe /Fe – ECd /Cd 2+
E0 = – (-0.40) - V n
ln Q E
E =
E0 = V - V 2 ln
-0.04 V
E =
0.010
0.60
E = ____________
E ___ 0
________________

21. Batteries Dry cell Leclanché cell Anode: Zn (s) Zn2+ (aq) + 2e-
Cathode:
2NH4 (aq) + 2MnO2 (s) + 2e Mn2O3 (s) + 2NH3 (aq) + H2O (l) +
Zn (s) + 2NH4 (aq) + 2MnO2 (s) Zn2+ (aq) + 2NH3 (aq) + H2O (l) + Mn2O3 (s)

22. Batteries Mercury Battery Anode:
Zn(Hg) + 2OH- (aq) ZnO (s) + H2O (l) + 2e-
Cathode:
HgO (s) + H2O (l) + 2e Hg (l) + 2OH- (aq)
Zn(Hg) + HgO (s) ZnO (s) + Hg (l)

23. Batteries Lead storage battery Anode:
Pb (s) + SO2- (aq) PbSO4 (s) + 2e- 4
Cathode:
PbO2 (s) + 4H+ (aq) + SO2- (aq) + 2e PbSO4 (s) + 2H2O (l) 4
Pb (s) + PbO2 (s) + 4H+ (aq) + 2SO2- (aq) PbSO4 (s) + 2H2O (l) 4

24. Solid State Lithium Battery
Batteries
Solid State Lithium Battery

25. Batteries
A ______ ______ is an electrochemical cell that requires a continuous supply of reactants to keep functioning
Anode:
2H2 (g) + 4OH- (aq) H2O (l) + 4e-
Cathode:
O2 (g) + 2H2O (l) + 4e OH- (aq)
2H2 (g) + O2 (g) H2O (l)

26. Corrosion

27. Cathodic Protection of an Iron Storage Tank

28. _______________ is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.

29. Electrolysis of Water

30. Electrolysis and Mass Changes
charge (C) = current (A) x time (s)
1 mole e- = 96,500 C
So what is the charge on a single electron?