1. Third Edition Lecture Power Points
Chemistry
Third Edition
Julia Burdge
Lecture Power Points
Chapter 19
Electrochemistry
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2. Electrochemistry Applications
Waste water treatment
Corrosion
Energy storage
Energy production
Solar to electrical energy
Production of chemicals
Sensors

3. Electrochemistry 19.1 Balancing Redox Reactions 19.2 Galvanic Cells
19.3
Standard Reduction Potentials
19.4
Spontaneity of Redox Reactions Under Standard-State Conditions
19.5
Spontaneity of Redox Reactions Under Conditions Other Than Standard State
19.6
Batteries
19.7
Electrolysis
19.8
Corrosion
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4. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Electrochemistry
What is electrochemistry?
It is a branch of chemistry that studies chemical reactions “called redox reactions” which involve electron transfer.
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Oxidation states +2 +2
Oxidation reactions
Reduction reactions
Redox reactions
Oxidizing agent
(reduced species)
Reducing agent
(oxidized species)

5. Assigning Oxidation States in Redox Reactions
19.1 F
Group 1A or 2A H O
Group 7A(Halogens)
2Na(s) + Cl2(g) NaCl(s)
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
CH CO2 + 8e‒
2O e‒ CO H2O ‒1 +1 ‒4 +4
‒ 2
‒2×2
+1×4
+1×2
CH4 is oxidized (it is a reducing agent) ‒4 +4
O2 is reduced (it is an oxidizing agent) ‒2
‒2×2

6. Balancing Redox Reactions in Acidic Media
19.1
Balancing Redox Reactions in Acidic Media
H2O(l)
Cr2O72-(aq) + C2H5OH(l) Cr3+(aq) + CO2(g)
Step 1: Write half equations (oxidation-reduction).
Cr2O72-(aq) Cr3+(aq) (reduction)
C2H5OH(l) CO2(g) (oxidation)
H+(aq)
Notice that we didn’t include H+ and H2O because they are going to be added at a later stage to balance the redox reaction.

7. Balancing Redox Reactions in Acidic Media
19.1
Balancing Redox Reactions in Acidic Media
Step 2: Balance all elements and charges for each half reaction.
For the reduction half reaction:
Cr2O Cr3+
a) Balance all elements except oxygen and hydrogen atoms.
b) Balance oxygen with H2O.
c) Balance hydrogen with H+.
d) Balance charges using electrons.
Similarly, for the oxidation half reaction:
C2H5OH + 3H2O CO2 + 12H+ + 12e-
6e- +
14H+ + 2
+ 7H2O

8. Balancing Redox Reactions in Acidic Media
19.1
Balancing Redox Reactions in Acidic Media
Step 3: Equalize the number of electrons in both balanced half reactions.
12e- + 28H+ + 2Cr2O Cr H2O
Step 4: Add up the two half reactions:
C2H5OH + 3H2O CO2 + 12H+ + 12e-
16H+ + 2Cr2O72- +C2H5OH Cr H2O + 2CO2
Step 5: Check that elements and charges in the final reaction are balanced.
The reduction half reaction after being multiplied by 2

9. Balancing Redox Reactions in Basic Media
19.1
Balancing Redox Reactions in Basic Media
Ag(s) + CN–(aq) + O2(g) Ag(CN)2–(aq) + H2O(l)
Step 1: Ag + CN– Ag(CN)2– (oxidation)
O2 H2O(l) (reduction)
Step 2:
Ag + 2CN– Ag(CN)2– + e–
O2 + 4H+ + 4e– H2O
Step 3:
4Ag + 8CN– Ag(CN)2– + 4e–
O2 + 4H+ + 4e– H2O
basic

10. Balancing Redox Reactions in Basic Media
19.1
Balancing Redox Reactions in Basic Media
Step 4: Ag + 8CN– + O2 + 4H Ag(CN)2– + 2H2O
But in alkaline, we have HO-
Step 5:
4Ag + 8CN– + O2 + 4H+ + 4OH– Ag(CN)2– +2H2O + 4OH–
Step 6: Eliminate H2O molecules from both sides:
4Ag + 8CN– + O2 + 2H2O Ag(CN)2– + 4OH–
Step 7: Recheck the balance of elements and charges.
4H2O

11. Oxidation half-reaction Reduction half-reaction
Galvanic Cells
19.2
A galvanic cell (also called voltaic cell) is the experimental apparatus for generating electricity through the use of a spontaneous reaction.
Salt bridge e‒ e‒
Anode (‒)
Cathode (+)
Aqueous solution
Aqueous solution
Oxidation half-reaction
Reduction half-reaction
Electrodes

12. Oxidation half-reaction Reduction half-reaction
Galvanic Cells
19.2
Electrodes
Oxidation
Reduction
Salt bridge
Oxidation half-reaction
Reduction half-reaction
An electric current flows from anode to cathode because there is a difference in electrical potential energy between the two electrodes.

13. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
Galvanic Cells
19.2
Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)
The two half-reactions can be physically separated in two beakers.
As a result, the electrons lost by Zn travel through an external wire in order to be gained by Cu2+ ions. This generates electricity.
What is electroplating?

14. Salt Bridge
19.2
A salt bridge is an essential conducting medium through which the cations and anions can move from one half-cell to the other. It maintains balance of the charge between the two half-cells.
Ion migration:
Cations – migrate toward the cathode
Anions – migrate toward the anode
Without the presence of a salt bridge, current will not flow.  galvanic cell will not work

15. Galvanic Cell Notation: How we express cell reaction?
19.2
Galvanic Cell Notation: How we express cell reaction?
Anode
Cathode
Phase boundary
Salt bridge
Phase boundary

16. Cell Potential
19.2
The cell potential, cell voltage, Electromotive force, (E) is the difference in electrical potential between the cathode and the anode half-cells.
 Ecell = Ecathode - Eanode
Cell potential is:
concentration dependent.
temperature dependent.
and nature of reactants .
The cell potential is measured using a voltmeter with units of volts (V).

17. What we learned in last class…
Oxidation & reduction
Oxidizing agent & reducing agent
Balancing Redox reactions
Galvanic cell (Design, Anode, Cathode, Salt bridge, etc)
Galvanic cell notation
Cell potential (Ecell = Ecathode - Eanode)
Standard reduction potential

18. What we hope to learn today…
Standard reduction potential
Standard Hydrogen Electrode (SHE)
Identifying cathode and anode
Predicting the occurrence of redox reactions
Electrochemistry & Thermodynamics

19. Standard Reduction Potential
19.3
The standard reduction potential (E) is the potential associated with a reduction half-reaction at an electrode when the ion concentration is 1 M and the gas pressure is 1 atm.
The standard reduction potential (E) for the standard hydrogen electrode (SHE) is assigned the value 0 V.
shown as reduction
The relative standard reduction potentials of other half-reactions are measured relative to the SHE potential.

20. Standard Hydrogen Electrode
19.3
Hydrogen/Pt electrode can be used as a reference for other electrodes.
2H+ (aq) + 2e-  H2 (g)
The platinum (Pt) electrode is used to:
provide a surface on which the oxidation of H2 can take place.
serve as an electrical conductor.
Under standard state condition at 25C , the reduction potential of H+ is defined as exactly zero.
The standard hydrogen electrode (SHE)
By definition

21. Measuring
standard reduction potential for zinc
19.3

22. Measuring The standard reduction potential of zinc is ‒ 0.76 V.
19.3
The standard reduction potential of zinc is ‒ 0.76 V.
The standard reduction potential of zinc is negative, which means, when connected with hydrogen half cell, zinc tends to be oxidized, not reduced.

23. 19.3
Measuring
standard reduction potential for copper

24. Measuring The standard reduction potential of copper is + 0.34 V.
19.3
Measuring
The standard reduction potential of copper is V.
The standard reduction potential of copper is positive, which means, when connected with hydrogen half cell, copper tends to be reduced, not oxidized.

25. Comparison of E Values
19.3
Comparison of E Values
Cu2+ (aq) + 2e‒  Cu (s) E = V 2H+ (aq) + 2e‒  H2 (g) E = 0 V Zn2+ (aq) + 2e‒  Zn (s) E = ‒ 0.76 V
Cathode
Easier to be reduced
Easier to be oxidized
Anode

26. of a Zinc-Copper Cell
19.3

27. Predicting the Occurrence of Redox Reactions

28. Standard reduction potentials (E) of some half-reactions at 25C
19.3
Increasing ease of reduction
Increasing strength as a reducing agent
Increasing ease of oxidation
Increasing strength as an oxidizing agent
Diagonal Rule

29. Predicting the Occurrence of Redox Reactions
19.3
Predicting the Occurrence of Redox Reactions
Example:
Determine whether redox reaction will occur or not at 25C when lead metal (Pb) is added to:
(a) 1.0 M solution of NiCl2.
(b) 1.0 M solution of HCl.
Also, write cell notation, calculate cell potential, write overall reaction?
Pb2+ (aq) + 2e  Pb(s) V
Cl2 (gas) + 2e  2Cl-(aq) V
Ni2+(aq) + 2e  Ni (s) V
2H+(aq) + 2e  H2(gas) V
Cl2 (gas) + 2e  2Cl-(aq) V
2H+(aq) + 2e  H2(gas) V
Pb2+ (aq) + 2e  Pb(s) V
Ni2+(aq) + 2e  Ni (s) V

30. Cell notation: Pb(s) | Pb2+(1 M) || 2H+(1 M) | H2(g)
Solution
The proposed reaction will occur only if
Thus, the redox reaction will NOT occur.
The proposed reaction will occur only if
Thus, the redox reaction will occur
Cell notation: Pb(s) | Pb2+(1 M) || 2H+(1 M) | H2(g)
Cell potential: E°cathode - E°anode = 0 - ( V) = 0.13 V
Overall reaction: Pb(s) + 2H Pb2+ + H2(g)

31. Calculating the Standard Cell Potential ( )
19.3
Calculating the Standard Cell Potential ( )
Example:
Write the overall cell reaction and calculate the standard cell potential of the following cell?
Zn(s) | Zn2+(1 M) | || | Ag+(1 M) | Ag(s)
EZn2+/Zn(s) = V
EAg+/Ag(s) = V

32. Balanced equation of the overall cell reaction
Zn (s)  2e‒ + Zn2+ (aq)
Ag+ (aq) + e‒  Ag (s) 2
Zn (s) + 2Ag+ (aq)  Zn2+ (aq) + 2Ag (s)

33. Cell Potential Zn (s) + 2Ag+ (aq)  Zn2+ (aq) + 2Ag (s)
Note: Standard reduction potential is an intensive property (like temperature and density), and it is not an extensive property (like mass and volume). In other words, it does NOT depend on the amounts of substances involved in the cell.

34. Example - Redox reaction will occur or not - E0cell - Cell notation
Determine the following for a Cadmium – Lead cell .
- Redox reaction will occur or not
- E0cell
- Cell notation
- Balanced equation of the overall cell reaction
ECd2+/Cd(s) = V
EPb2+/Pb(s) = V
Note: Cadmium – Lead cell means - Cd electrode in 1.0 M Cd(NO3)2 and Lead electrode in 1.0 M Pb(NO3)2)?

35. E0cell Cell notation Pb2+ + 2e- Pb(s) E = ‒ 0.13 V
Cd e Cd(s) E = ‒ 0.40 V
(We have Pb2+ and Cd(s), so the reaction will occur)
E0cell
Cell notation
Cd(s) | Cd2+ (1M) || Pb2+ (1M) | Pb(s)

36. Balanced equation of the overall cell reaction
Pb e Pb(s) Reduction
Cd(s) Cd e Oxidation
Cd(s) Pb2+  Cd2+ + Pb(s)

37. What we discussed in last lecture…
Standard reduction potential
Standard Hydrogen Electrode (SHE)
Identifying cathode and anode
Predicting the occurrence of redox reactions

38. Today… Electrochemistry & Thermodynamics (ΔG or ΔGo)
Electrochemistry & Equilibrium constant

39. Relating Electrochemistry to Thermodynamics
ΔGo = ‒ nFEocell
ΔG = ‒ nFEcell
ΔGo : Standard free energy change
n: Number of moles of electrons that pass through the circuit.
F: Faraday’s constant (the electric charge contained in 1 mol of electrons) 1 F = 96,500 J / V ∙ mol e-
Eocell : Standard cell potential.
When Ecell is +ve, the value of ΔG is ‒ve (the process is spontaneous).
When Ecell is ‒ve, the value of ΔG is +ve (the process is nonspontaneous).
G = negative, spontaneous
Standard state cond.
Room temp. 25 0C
1 M concentration
1 atm. pressure

40. Relationship between and ΔG
19.4
Relationship between and ΔG
ΔGo = ‒ nFEocell
Example:
Give the balanced equation and determine the standard free energy change (ΔG) for the following reaction at 25C :
Al(s) + Mn2+(aq)(1.0 M)  Al3+(aq) (1.0 M) Mn(s)
The half-cell reactions:
reduction: Mn2+(aq) e-  Mn(s)
oxidation: Al (s)  Al3+(aq) + 3e-
ΔGo = ‒ nFEocell
Eocell = Eocathode – Eoanode = – 1.18 V – (– 1.66 V) =
ΔGo = ‒ nFEocell = ‒ (6 e-)(9.65 × 104 J/V∙mol e-)(0.48 V)
= ‒ 2.78 × 105 J/mol = ‒ 2.78 × 102 kJ/mol
(product favored) 2 3 2 3 3 6 3 2 2 6

41. Relationship between and K
19.4
Relationship between and K
ΔGo = ‒ RT lnK  (from Chapter 18) ΔGo = ‒ nFEocell Combining the above two equations: ‒ RT lnK = ‒ nFEocell Solving for Eocell :
At 25C:

42. Relationship between and Kc
19.4
Relationship between and Kc
Example:
Calculate the equilibrium constant for the following reaction at 25C :
Al (s) + Mn2+(aq)(1.0 M)  Al3+(aq) (1.0 M) Mn (s)
Remember that product will be favored (very large Kc value)! 2 3 2 3
Eocell = – 1.18 V – (– 1.66 V) =
Very large or very small K, which can be difficult to determine, is easily calculated using electrochemistry!!

43. Nonstandard States
19.5
The spontaneity of a chemical reaction at a condition other than the standard states (a nonstandard state means that concentrations are NOT 1 M) can be predicted using Nernst Equation.
Since ΔGo = ‒ nFEo and ΔG = ‒ nFE
then:
ΔG = ΔGo RT lnQ  (from Chapter 18)
‒ nFE = ‒ nFEo RT lnQ
dividing the whole equation by ‒ nF ; and setting T = 298 K : or

44. 19.5
Nonstandard States
Electrons flow spontaneously from the anode to cathode. Product concentration increases and the reactant concentration decreases.
So as time proceeds Q increases and E decreases.
When equilibrium is reached ( Q = K ), there is no net flow of electrons (E = 0) and we get:
Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
Q =
[Zn2+]
[Cu2+]
Section 19.4

45. Conditions other than Standard State
19.5
Conditions other than Standard State
Example
Predict whether the following reaction will occur spontaneously as written at 298 K:
assuming [Co2+] = 0.15 M and [Fe2+] = 0.68 M.
ECo2+/Co(s) = V
EFe2+/Fe(s) = V

47. SUMMARY ΔG = ‒ nFEcell ΔGo = ‒ nFEocell Standard state
Non-standard state
ΔG = ‒ nFEcell
ΔGo = ‒ nFEocell
At 25 0C

48. 19.5
Concentration Cells
Concentration cells are galvanic cells with two half-cells composed of the same material but are different in ion concentrations.
Concentration cells produces some potential due to the difference in concentrations.
0.1 M
1.0 M

49. Concentration Cells Oxidation: Zn(s)  Zn2+(aq) (0.10 M) + 2e-
19.5
Concentration Cells
Oxidation: Zn(s)  Zn2+(aq) (0.10 M) + 2e-
Reduction: Zn2+(aq) (1.0 M) + 2e-  Zn(s)
Overall: Zn2+(aq) (1.0 M)  Zn2+(aq) (0.10 M)
The cell potential is:
E of the concentration cell above is V. In general, concentration cells have small potentials.
As the process continues, E gradually decreases. When [Zn2+]dilute becomes equal to [Zn2+]conc. then E becomes zero.
equal to zero

50. 19.7
Electrolysis
Electrolysis is the process that uses electric energy to force a nonspontaneous chemical reaction to take place.
One main application of electrolysis is to recharge some batteries by reversing the ordinary electro-chemical process.
An electrolytic cell is the cell used to carry out electrolysis.
Cell Type
Chemical reaction
Electric energy
Galvanic
Spontaneous
Produced
Electrolytic
Nonspontaneous
consumed

51. Nonspontaneous flow of electrons
19.7
Electrolysis
Electrolysis of molten NaCl.
This process can be used to produce sodium metal and chlorine gas from NaCl (m.p. = 800C).
Molten NaCl is used as a medium that allows the flow of electrons.
However, the Ecell for this process is ‒ 4 V. The battery serves to push the electrons in the direction they would not flow spontaneously.
Nonspontaneous flow of electrons
E = V
E = ‒ 2.71
Ecell = ‒ 4.07 V

52. Electrolysis Downs cell.
19.7
Electrolysis
Downs cell.
It is used for large-scale electrolysis of NaCl.

53. Electrolysis Electrolysis of water.
19.7
Electrolysis
Electrolysis of water.
At normal conditions, water will not spontaneously decompose into hydrogen and oxygen gases.
H2O(l)  H2(g) + O2(g) ΔG = kJ/mol
highly nonspontaneous
E = V
E = V
Ecell = ‒ 1.23 V

54. Electrolysis Electrolysis of water.
19.7
Electrolysis
Electrolysis of water.
Pure water does not have sufficient ions, so 0.1 M H2SO4 solution is used to conduct electric current and establish the circuit.
There is no net consumption of H2SO4 in the over all reaction.

55. 19.7 Electrolysis of an Aqueous Sodium Chloride Solution
Thus, the half-cell reactions in the electrolysis of aqueous sodium chloride are
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56. Quantitative Applications of Electrolysis
19.7
Quantitative Applications of Electrolysis
The amount (mass) of products formed during an electrolysis process can be calculated by measuring the current (in amperes) that passes through an electrolytic cell in a given period of time.
Then:
1 C = 1 Amp × 1 s
and:
96500 C = 1 mol e‒
Thus, from electrochemical equations, the number of moles (and hence the mass) of products can be calculated.

57. 19.7 Quantitative Applications of Electrolysis
Consider an electrolytic cell in which molten CaCl2 is separated into its constituent elements, Ca and Cl2.
Suppose a current of A is passed through the cell for 1.50 h.
How much product will be formed at each electrode?
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58. 19.7 Electrolysis Quantitative Applications of Electrolysis 58
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