1. Spontaneity of Redox Reactions
Electrochemistry
Spontaneity of Redox Reactions

2. Chemical Change and Electrical Work
Electrochemistry:
Chemical Change and Electrical Work
Redox Reactions and Electrochemical Cells
Voltaic Cells: Using Spontaneous Reactions to Generate Electrical Energy
Cell Potential: Output of a Voltaic Cell
Free Energy and Electrical Work
Electrochemical Processes in Batteries
Corrosion: An Environmental Voltaic Cell
Electrolytic Cells: Using Electrical Energy to Drive Nonspontaneous Reactions

3. Overview of Redox Reactions
Chemistry magic (vid): Copper coin  “Silver”  “Gold”
Oxidation: Loss of electrons
Reduction: Gain of electrons.
These processes occur simultaneously.
Oxidation results in an increase in Oxidation Numbers (O.N.). while reduction results in a decrease in O.N.
Oxidizing agent takes electrons from the substance being oxidized. The oxidizing agent is therefore reduced.
Reducing agent takes electrons from the substance being oxidized. The reducing agent is therefore oxidized.

4. Example of redox terminology
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) +1 +2

5. Half-Reaction Method for Balancing Redox Reactions
The half-reaction method divides a redox reaction into its oxidation and reduction half-reactions.
- This reflects their physical separation in electrochemical cells.
This method does not require assigning oxidation numbers (O.N.s).
The half-reaction method is easier to apply to reactions in acidic or basic solutions.

6. Steps in the Half-Reaction Method
Divide the skeleton reaction into two half-reactions, each containing the oxidation and reduction reactions.
Balance the atoms and (WPE).
First balance atoms other than O and H, then O w/ H2O, then H w/ H+.
Charge is balanced by adding electrons (e-) to the left side in the reduction half-reaction and to the right side in the oxidation half-reaction.
If necessary, multiply one or both half-reactions by an integer so that
number of e- gained in reduction = number of e- lost in oxidation
Add the balanced half-reactions, and include states of matter.

7. Balancing Redox Reactions in Acidic Solution
Cr2O72-(aq) + I-(aq) → Cr3+(aq) + I2(s)
6I-(aq) + 14H+(aq) + Cr2O72-(aq) → 3I2(s) + 7H2O(l) + 2Cr3+(aq)

8. Balancing Redox Reactions in Basic Solution
An acidic solution contains H+ ions and H2O. We use H+ ions to balance H atoms.
A basic solution contains OH- ions and H2O. To balance H atoms, we proceed as if in acidic solution, and then add one OH- ion to both sides of the equation.
For every OH- ion and H+ ion that appear on the same side of the equation we form an H2O molecule.
Excess H2O molecules are canceled in the final step, when we cancel electrons and other common species.

9. Balancing a Redox Reaction in Basic Solution
MnO4-(aq) + C2O42-(aq) → MnO2(s) + CO32-(aq) [basic solution]
2MnO4- + 2H2O + 3C2O42- → 2MnO2 + 6CO H+
2MnO4-(aq) + 3C2O42-(aq) + 4OH-(aq) → 2MnO2(s) + 6CO32-(aq) + 2H2O(l)

10. Electrochemical Cells
A voltaic cell uses a spontaneous redox reaction (DG < 0) to generate electrical energy.
- The system does work on the surroundings. Discharging Battery
A electrolytic cell uses electrical energy to drive a nonspontaneous reaction (DG > 0).
- The surroundings do work on the system. Charging Battery
Both types of cell are constructed using two electrodes (electric conductor) placed in an electrolyte solution.
Anode : the electrode at which oxidation occurs.
Cathode: the electrode at which reduction occurs.

11. Voltaic cell vs. Electrolytic cells

12. Reaction between zinc metal and Cu2+

13. Spontaneous Redox Reactions
zinc metal reacts with Cu2+ ions spontaneously:
Cu2+(aq) + 2e- → Cu(s) [reduction]
Zn(s) → Zn2+(aq) + 2e- [oxidation]
Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s)
Zn: ________ed
Cu2+: _________ed.
Electrons are being transferred in CONTACT but does not do electrical work

14. Construction of a Voltaic Cell
Each half-reaction in half-cell are physically separate.
Each half-cell consists of an electrode in an electrolyte solution.
The half-cells are connected by the external circuit to allow e- transfer
Salt bridge (inert ions) balances charges by allowing ion movement.

15. Overall (cell) reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
A voltaic cell based on the zinc-copper reaction
Oxidation half-reaction
Zn(s) → Zn2+(aq) + 2e-
Reduction half-reaction
Cu2+(aq) + 2e- → Cu(s)
Overall (cell) reaction
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

16. Reactions in the Voltaic Cell
Oxidation (loss of e-): at the anode, the source of e-.
Zn(s) → Zn2+(aq) + 2e-
Over time, Zn(s) anode↓ [Zn2+] ↑
Reduction (gain of e-): at the cathode, e- are used up.
Cu2+(aq) + 2e- → Cu(s)
Over time, [Cu2+] ↓ Cu(s) cathode↑

17. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Zinc-copper reaction in Voltaic cell
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Oxidation half-reaction
Zn(s) → Zn2+(aq) + 2e-
Zn anode weighs less as Zn is oxidized to Zn2+.
Reduction half-reaction
Cu2+(aq) + 2e- → Cu(s)
Cu cathode gains mass as Cu2+ ions are reduced to Cu.

18. Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Electrodes
Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
Anode, the negative (“-”) electrode in a voltaic cell, produces e- from oxidation of Zn(s)..
Electrons flow through the external wire from the anode to the cathode
Cathode, the positive (“+”) electrode in a voltaic cell, receives electrons for reduction of Cu2+ ion

19. Salt Bridge Without salt bridge, charge accumulates:
“+” charge ↑ at Anode from more Zn2+
“-” charge ↑ at Cathode from loss of Cu2+
Salt bridge maintains electrical neutrality, allowing excess Zn2+ ions to enter from the anode, and excess negative ions to enter from the cathode.
A salt bridge contains nonreacting cations and anions, often K+ and NO3-, dissolved in a gel.

20. Active and Inactive Electrodes
Active electrode involved in the half-reaction (as a reactant or product in the overall reaction). Example: Zinc-Copper voltaic cell
Inactive electrode provides a surface for the reaction and completes the circuit. It does not participate in the overall reaction.
- Inactive electrodes are necessary when none of the reaction components can be used as an electrode.
Inactive electrodes are usually unreactive substances such as graphite or platinum.

21. Inactive electrode in Action
Oxidation half-reaction
2I-(aq) → I2(s) + 2e-
Reduction half-reaction
MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l)
Overall (cell) reaction
2MnO4-(aq) + 16H+(aq) + 10I-(aq) → 2Mn2+(aq) + 5I2(s) + 8H2O(l)

22. Notation for a Voltaic Cell (ABC)
The components of each half-cell are written in the same order as in their half-reactions.
Anode components: on the left.
Cathode components: on the right.
Zn(s)│Zn2+(aq)║Cu2+(aq) │Cu(s)
“|”: phase boundary between components of a half-cell.
“||”: half-cells are physically separated (by salt Bridge)
Concentrations may be given in parentheses . (If not stated, concentrations are assumed as 1 M.)

23. Notation for a Voltaic Cell
graphite I-(aq)│I2(s)║MnO4-(aq), H+(aq), Mn2+(aq) │graphite
The inert electrode is specified.
A comma is used to show components that are in the same phase.

24. Example: Draw the Voltaic Cell Diagram
Cr(s)│Cr3+(aq)║Ag+(aq)│Ag(s)

25. Why Redox Reaction between two half cells?
Zn as stronger reducing agent than Cu
When the switch is OFF, electrons accumulate on zinc electrode.
When the switch is ON, e- flow (Zn  Cu) to equalize the difference in electrical potential
The spontaneous reaction occurs because of the different abilities of these metals to give up their electrons.

26. Cell Potential
Cell potential (Ecell, Voltage) depends on the different abilities of these metals to give up their electrons
Ecell can be measured using voltammeter (one of many functions in a multimeter).
Ecell > 0 for a spontaneous process.

27. Standard cell potential E°cell
The standard cell potential (E°cell) is measured at a specified temperature with no current flowing and all components in their standard states (298 K, partial pressure = 1 atm and solution = 1 M).

28. Voltages of Some Voltaic Cells
Voltage (V)
Common alkaline flashlight battery
1.5
Lead-acid car battery (6 cells ≈ 12 V)
2.1
Calculator battery (mercury)
1.3
Lithium-ion laptop battery
3.7
Electric eel (~5000 cells in 6-ft eel
= 750 V)
0.15
Nerve of giant squid
(across cell membrane)
0.070

29. Standard Hydrogen Electrode (SHE)
The standard hydrogen electrode has a standard electrode potential defined as zero (E°reference = 0.00 V).
Component: a Pt electrode, H2(g, 1 atm) bubbling through it. The Pt electrode is in 1 M strong acid.
2H+(aq; 1 M) + 2e H2(g; 1 atm) E°ref = 0.00V
Half-cell potentials are measured relative to SHE

30. Standard Reduction Potential
The standard reduction potential are measured depends on the difference between the abilities of the two electrodes to act as reducing agents.
E°cell = E°cathode (reduction) - E°anode (oxidation)

31. Determining an unknown E°half-cell with the standard reference (hydrogen) electrode
Oxidation half-reaction
Zn(s) → Zn2+(aq) + 2e−
Reduction half-reaction
2H3O+(aq) + 2e- → H2(g) + 2H2O(l)
Overall (cell) reaction
Zn(s) + 2H3O+(aq) → Zn2+(aq) + H2(g) + 2H2O(l)

32. Example: Find Unknown E°half-cell from E°cell
A voltaic cell houses the reaction between aqueous bromine and zinc metal:
Br2(aq) + Zn(s) → Zn2+(aq) + 2Br-(aq) E°cell = 1.83 V.
Calculate E°bromine, given that E°zInc = V
E°cell = E°cathode (reduction) - E°anode (oxidation)
E°bromine = 1.07 V

33. Selected Standard Electrode Potentials (298 K)
Half-Reaction
E°(V)
F2(g) + 2e− F−(aq)
+2.87
Cl2(g) + 2e− Cl−(aq)
+1.36
MnO2(g) + 4H+(aq) + 2e− Mn2+(aq) + 2H2O(l)
+1.23
NO3-(aq) + 4H+(aq) + 3e− NO(g) + 2H2O(l)
+0.96
Ag+(aq) + e− Ag(s)
+0.80
Fe3+(g) + e− Fe2+(aq)
+0.77
O2(g) + 2H2O(l) + 4e− OH−(aq)
+0.40
Cu2+(aq) + 2e− Cu(s)
+0.34
2H+(aq) + 2e− H2(g)
0.00
N2(g) + 5H+(aq) + 4e− N2H5+(aq)
−0.23
Fe2+(aq) + 2e− Fe(s)
−0.44
2H2O(l) + 2e− H2(g) + 2OH−(aq)
−0.83
Na+(aq) + e− Na(s)
−2.71
Li+(aq) + e− Li(s)
−3.05

34. Comparing E°half-cell values
E°half-cell : the ability of the reactant to act as an oxidizing agent.
F2(g) + 2e− F−(aq)
+2.87 V
Positive E°half-cell: the more readily the ___________ will act as an oxidizing agent. Fluorine gas as the strongest oxidizing agent.
Li+(aq) + e− Li(s)
−3.05 V
Negative E°half-cell: the more readily the ____________ will act as a reducing agent. Lithium metal as the strongest reducing agent.

35. Selected Standard Electrode Potentials (298 K)
Half-Reaction
E°(V)
F2(g) + 2e− F−(aq)
+2.87
strength of oxidizing agent
strength of reducing agent
Cl2(g) + 2e− Cl−(aq)
+1.36
MnO2(g) + 4H+(aq) + 2e− Mn2+(aq) + 2H2O(l)
+1.23
NO3-(aq) + 4H+(aq) + 3e− NO(g) + 2H2O(l)
+0.96
Ag+(aq) + e− Ag(s)
+0.80
Fe3+(g) + e− Fe2+(aq)
+0.77
O2(g) + 2H2O(l) + 4e− OH−(aq)
+0.40
Cu2+(aq) + 2e− Cu(s)
+0.34
2H+(aq) + 2e− H2(g)
0.00
N2(g) + 5H+(aq) + 4e− N2H5+(aq)
−0.23
Fe2+(aq) + 2e− Fe(s)
−0.44
2H2O(l) + 2e− H2(g) + 2OH−(aq)
−0.83
Na+(aq) + e− Na(s)
−2.71
Li+(aq) + e− Li(s)
−3.05

36. Predict Spontaneous Redox Reactions
Each half-reaction contains both reducing agent and oxidizing agent.
Br2(l) + 2e− Br−(aq) V
Cu2+(aq) + 2e− Cu(s) V
A spontaneous redox reaction (E°cell > 0) will occur between an oxidizing agent and any reducing agent that has (more, less) positive value for E°.
What would be the spontaneous redox reaction from the above two half reactions? Br Cu  Br + Cu
The oxidizing agent is the reactant from the half-reaction with the (more, less) positive E°half-cell.

37. Example: Write a spontaneous redox reaction:
Sn2+(aq) + 2e- → Sn(s) E°tin = V
Ag+(aq) + e- → Ag(s) E°silver = 0.80 V
E°cell = E°cathode (reduction) - E°anode (oxidation)
> 0
Which half reaction should be the cathode reaction?
The half reaction with (higher, lower) cell potential should be the cathode reaction
E°cell = E°silver – E°tin = 0.80 – (-0.14) = 0.94 V
2Ag+(aq) + Sn(s) → 2Ag(s) + Sn2+(aq)

38. Which halogen can oxidize gold metal to form gold(III) ion?
Example:
Which halogen can oxidize gold metal to form gold(III) ion?
Strategy:
A redox reaction would be spontaneous if combination of the half-reactions leads to positive standard cell potential (E°cell = E°cat - E°an > 0)
Collect the reduction potentials:
Au3+(aq) + 3e- → Au(s) E° = 1.50 V
F2(g) + 2e- → 2F-(aq) E° = 2.87 V
Cl2(g) + 2e- → 2Cl-(aq) E° = 1.36 V
Br2(g) + 2e- → 2Br-(aq) E° = 1.07 V
I2(g) + 2e- → 2I-(aq) E° = 0.53V
Write down the suggested reaction:
Which half reaction should be the cathode rxn?
X2 + Au(s) → Au3+(aq) + X-
Only fluorine can oxidize gold metal

39. Which metal can reduce iron(II) ion in aqueous solution?
Strategy:
A redox reaction would be spontaneous if combination of the half-reactions leads to positive standard cell potential (E°cell = E°cat - E°an ____ 0)
Common metals:
Cu2+(aq) + 2e- → Cu(s) E° = +.34 V
Fe2+(aq) + 2e- → Fe(s) E° = V
Zn2+(aq) + 2e- → Zn(s) E° = V
Al3+(aq) + 3e- → Al(s) E° = V
Write down the suggested reaction:
Which half reaction should be the cathode reaction rxn?
Fe2+(aq) + M → Mn+(aq) + Fe(s)

40. Voltaic Cell in Mercury Dental Fillings
Biting down with a filled tooth on a scrap of aluminum foil will cause pain. The foil acts as an active anode (E°aluminum = V), saliva as the electrolyte, and the filling as an inactive cathode as O2 is reduced to H2O.

41. Free Energy and Electrical Work
For a spontaneous redox reaction, DG < 0 and Ecell > 0.
DG = -nFEcell
n = mol of e- transferred
F is the Faraday constant
= 9.65x104 J/V·mol e-
Under standard conditions, DG° = -nFE°cell and
E°cell =
ln K RT nF
E°cell =
log K V n
for T = K or

42. The interrelationship of DG°, E°cell, and K
Reaction Parameters at the Standard State
DG° K
E°cell
Reaction at standard-state conditions
< 0
> 1
> 0
spontaneous 1
at equilibrium
< 1
nonspontaneous
DG°
E°cell K
ΔG° = -nFE°cell
ΔG = -RT ln K
E°cell =
ln K RT nF

43. Use Oxidation state change to determine n (#e- transferred)
6I-(aq) + 14H+(aq) + Cr2O72-(aq) → 3I2(s) + 7H2O(l) + 2Cr3+(aq)
2MnO4-(aq) + 3C2O42-(aq) + 4OH-(aq) → 2MnO2(s) + 6CO32-(aq) + 2H2O(l)

44. Calculating K and DG° from E°cell
Lead can displace silver from solution, and silver occurs in trace amounts in some ores of lead.
Pb(s) + 2Ag+(aq) → Pb2+(aq) + 2Ag(s)
As a consequence, silver is a valuable byproduct in the industrial extraction of lead from its ore. Calculate K and DG° at K for this reaction.
Ag+(aq) + e- → Ag(s) E° = V
Pb2+(aq) + 2e- → Pb(s) E° = V
n = 2
K = 2.6x1031
DG° = -1.8x102 kJ/mol rxn

45. Cell Potential and Concentration
Fresh alkaline battery has 1.6 V. Its voltage decreases as we use it. Why?
Nernst Equation
Ecell = E°cell -
ln Q RT nF
When Q < 1, [reactant] > [product], ln Q < 0, so Ecell > E°cell
When Q = 1, [reactant] = [product], ln Q = 0, so Ecell = E°cell
When Q > 1, [reactant] < [product], ln Q > 0, so Ecell < E°cell
Ecell = E°cell -
log Q V n
We can simplify the equation as before for T = K:

46. Nernst Equation to Predict Spontaneous Reaction

47. Ecell depends on log Q for the zinc-copper cell
If the reaction starts with [Zn2+] < [Cu2+] (Q < 1), Ecell is higher than the standard cell potential.
As the reaction proceeds, [Zn2+] ↑ vs. [Cu2+] ↓, so Ecell ↓
Eventually the system reaches equilibrium and the cell can no longer do work.

48. The relation between Ecell and log Q for the zinc-copper cell.
A summary of the changes in Ecell as any voltaic cell operates.

49. Concentration Cells
The same half-reaction in both cell compartments, but with different concentrations of electrolyte:
Cu(s) → Cu2+(aq; 0.10 M) + 2e [anode; oxidation]
Cu2+(aq; 1.0 M) → Cu(s) [cathode; reduction]
Cu2+(aq; 1.0 M) → Cu2+(aq; 0.10 M)
E°cell = 0, Q  0, thus Ecell  0
As long as the concentrations of the solutions are different, the cell potential is > 0 and the cell can do work.

50. Overall (cell) reaction
A Cu/Cu2+ concentration cell
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Ecell > 0 as long as the half-cell concentrations are different.
The cell is no longer able to do work once the concentrations are equal.
Oxidation half-reaction
Cu(s) → Cu2+(aq, 0.1 M) + 2e-
Reduction half-reaction
Cu2+(aq, 1.0 M) + 2e- → Cu(s)
Overall (cell) reaction
Cu2+(aq,1.0 M) → Cu2+(aq, 0.1 M)

51. Application of Concentration Cell: pH meter
The glass electrode monitors the [H+] of the solution relative to its own fixed internal [H+].
An older style of pH meter includes two electrodes.
Modern pH meters use a combination electrode.

52. Some Ions Measured with Ion-Specific Electrodes
Species Detected
Typical Sample
NH3/NH4+
Industrial wastewater, seawater
CO2/HCO3-
Blood, groundwater F-
Drinking water, urine, soil, industrial stack gases
Br-
Grain, plant tissue I-
Milk, pharmaceuticals
NO3-
Soil, fertilizer, drinking water K+
Blood serum, soil, wine H+
Laboratory solutions, soil, natural waters

53. Electrochemical Processes in Batteries
A battery consists of self-contained voltaic cells arranged in series, so their individual voltages are added.
A primary battery cannot be recharged. The battery is “dead” when the cell reaction has reached equilibrium.
A secondary battery is rechargeable. Once it has run down, electrical energy is supplied to reverse the cell reaction and form more reactant.

54. Alkaline battery
Anode (oxidation): Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2e-
Cathode (reduction): MnO2(s) + 2H2O(l) + 2e- → Mn(OH)2(s) + 2OH-(aq)
Overall (cell) reaction:
Zn(s) + MnO2(s) + H2O(l) → ZnO(s) + Mn(OH)2(s) Ecell = 1.5 V

55. Silver button battery
Anode (oxidation): Zn(s) + 2OH-(aq) → ZnO(s) + H2O(l) + 2e-
Cathode (reduction): Ag2O(s) + H2O(l) + 2e- → 2Ag(s) + 2OH-(aq)
Overall (cell) reaction: Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)
Ecell = 1.6 V

56. Lead-acid battery Anode (oxidation):
Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2e-
Cathode (reduction):
PbO2(s) + 3H+(aq) + HSO4-(aq) + 2e- → PbSO4(s) + 2H2O(l)
Overall (cell) reaction (discharge):
PbO2(s) + Pb(s) + H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
Ecell = 2.1 V

57. Nickel-metal hydride battery
Anode (oxidation): MH(s) + OH-(aq) → M(s) + H2O(l) + e-
Cathode (reduction): NiO(OH)(s) + H2O(l) + e- → Ni(OH)2(s) + OH-(aq)
Overall (cell) reaction:
MH(s) + NiO(OH)(s) → M(s) + Ni(OH)2(s) Ecell = 1.4 V

58. Lithium-ion battery Anode (oxidation): LixC6(s) → xLi+ + xe- + C6(s)
Cathode (reduction):
Li1-xMn2O4(s) + xLi+ + xe- → LiMn2O4(s)
Overall (cell) reaction:
LixC6(s) + Li1-xMn2O4(s) → LiMn2O4(s)
Ecell = 3.7 V
The secondary (rechargeable) lithium-ion battery in laptop computers, cell phones, and camcorders.

59. Fuel Cells
In a fuel cell, also called a flow cell, reactants enter the cell and products leave, generating electricity through controlled combustion.
Reaction rates are lower in fuel cells than in other batteries, so an electrocatalyst is used to decrease the activation energy.

60. Hydrogen fuel cell Anode (oxidation): 2H2(g) → 4H+(aq) + 4e-
Cathode (reduction): O2(g) + 4H+(aq) + 4e- → 2H2O(g)
Overall (cell) reaction: H2(g) + O2(g) → 2H2O(g) Ecell = 1.2 V

61. Corrosion: an Environmental Voltaic Cell
Corrosion : metals are oxidized to their oxides and sulfides.
The rusting of iron is a common form of corrosion.
- Rust is not a direct product of the reaction between Fe and O2, but arises through a complex electrochemical process.
- Rusting requires moisture, and occurs more quickly at low pH, in ionic solutions, and when the iron is in contact with a less active metal.

62. The Rusting of Iron
Fe(s) → Fe2+(aq) + 2e- [anodic region; oxidation]
O2(g) + 4H+(aq) + 4e- → 2H2O(l) [cathodic region; reduction]
The loss of iron metal in conjunction with reduction of oxygen:
2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l) [overall]
The rusting process:
2Fe2+(aq) + ½O2(g) + (2 + n)H2O(l) → Fe2O3·nH2O(s) + 4H+(aq)
Overall reaction:
H+ ions are consumed in the first step, so lowering the pH increases the overall rate of the process.
H+ ions act as a catalyst, since they are regenerated in the second part of the process.

63. The corrosion of iron

64. Metal-metal contact affects the corrosion of iron
Fe in contact with Cu corrodes faster.
Fe in contact with Zn does not corrode. The process is known as cathodic protection.

65. Sacrificial anodes to prevent iron corrosion
In cathodic protection, an active metal, such as zinc, magnesium, or aluminum, acts as the anode and is sacrificed instead of the iron.

66. Electrical energy  Chemical Energy
Electrolytic Cell:
Electrical energy  Chemical Energy
An electrolytic cell uses electrical energy from an external source to drive a nonspontaneous redox reaction.
Cu(s) → Cu2+(aq) + 2e [anode; oxidation]
Sn2+(aq) + 2e- → Sn(s) [cathode; reduction]
Cu(s) + Sn2+(aq) → Cu2+(aq) + Sn(s) E°cell = V and ΔG° = 93 kJ
An external source supplies the cathode with electrons (negative), and removes then from the anode (positive).
Internally, electrons flow from cathode to anode.

67. Voltaic Cell (spontaneous) vs. Electrolytic cell (forced)
Cu(s) → Cu2+(aq) + 2e-
Sn2+(aq) + 2e- → Sn(s)
Sn2+(aq) + Cu(s) → Sn(s) + Cu2+(aq)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
voltaic cell
Sn(s) → Sn2+(aq) + 2e-
Cu2+(aq) + 2e- → Cu(s)
Cu2+(aq) + Sn(s) → Cu(s) + Sn2+(aq)

68. ELECTROLYTIC (recharge)
Lead-acid battery: Discharge (Voltaic) vs. Recharge (electrolytic)
VOLTAIC (discharge)
Switch
ELECTROLYTIC (recharge)

69. Overall (cell) reaction
Electrolysis of water
Overall (cell) reaction
2H2O(l) → 2H2(g) + O2(g)
Oxidation half-reaction
2H2O(l) → 4H+(aq) + O2(g) + 4e-
Reduction half-reaction
2H2O(l) + 4e- → 2H2(g) + 2OH-(aq)

70. Overall (cell) reaction
Electroplating: Coating metal with another corrosion-resistant metal
Overall (cell) reaction
2H2O(l) → 2H2(g) + O2(g)
Oxidation half-reaction
Cu(s, anode) → Cu2+(aq) + 2e-
Reduction half-reaction
Cu2+(aq) + 2e- → Cu(s, cathode)

71. Comparison of Voltaic and Electrolytic Cells
Cell Type DG
Ecell
Electrode
Name
Process
Sign
Voltaic
< 0
> 0
Anode
Cathode
Oxidation
Reduction - +
Electrolytic

72. Products of Electrolysis
Electrolysis is the splitting (lysing) of a substance by the input of electrical energy.
During electrolysis of a pure, molten salt, the cation will be reduced and the anion will be oxidized.
During electrolysis of a mixture of molten salts
- the more easily oxidized species (stronger reducing agent) reacts at the anode, and
- the more easily reduced species (stronger oxidizing agent) reacts at the cathode.

73. Electrolysis of Aqueous Salt Solution: Industrial production of chlorine gas
The overall reaction of electrolysis of NaCl solution:
2NaCl(aq) + 2H2O(l)  2NaOH(aq) + Cl2(g) + H2(g)

74. Electrolysis of Molten Salt: Industrial Production of Sodium metal
Molten salt is conductor to electricity as ions are mobile
When molten salt is electrolyzed
Cation is reduced
Anion is oxidized.
2NaCl(l)  2Na(l) + Cl2(g)

75. Stoichiometry in Electrolytic Cell
Quantity of electrons in Coulombs (C): 1 mole = 96,500 C
Electric current (I) in Ampere (A): 1 A = 1 C/sec
In an electrolytic cell, copper(II) ion is reduced to form copper metal.
Cu2+(aq) + 2e- → Cu(s)
If the current is 3.00 A, how many grams of copper metal will form after 10.0 minutes of electrolysis?
1.80E+3 C, 9.32E-3 mol Cu, g

76. Writing Spontaneous Redox Reactions
Each half-reaction contains both a reducing agent and an oxidizing agent.
Br2(l) + 2e− Br−(aq)
The stronger oxidizing and reducing agents react spontaneously to form the weaker ones.
A spontaneous redox reaction (E°cell > 0) will occur between an oxidizing agent and any reducing agent that lies below it in the emf series (i.e., one that has a less positive value for E°).
The oxidizing agent is the reactant from the half-reaction with the more positive E°half-cell.

77. Summary of the Electrolysis of Aqueous Salt Solutions
Cations of less active metals (Au, Ag, Cu, Cr, Pt, Cd) are reduced to the metal.
Cations of more active metals are not reduced. H2O is reduced instead.
Anions that are oxidized, because of overvoltage from O2 formation, include the halides, except for F-.
Anions that are not oxidized include F- and common oxoanions. H2O is oxidized instead.