1. Electrochemistry

2. Oxidation – Reduction Reactions
Consider the reaction of Copper wire and AgNO3(aq)
AgNO3(aq)
Cu(s)
Ag(s)

3. Oxidation – Reduction Reactions
If you leave the reaction a long time the solution goes blue!
The blue is due to Cu2+(aq)

4. Oxidation-Reduction Reactions
So when we mix Ag+(aq) with Cu(s) we get Ag(s) and Cu2+(aq)
Ag+(aq) + 1e-  Ag(s)
Cu(s)  Cu2+(aq) + 2e-
The electrons gained by Ag+ must come from the Cu
Can’t have reduction without oxidation (redox)
Each Cu can reduce 2 Ag+
2Ag+(aq) + 2e-  2Ag(s)
2Ag+(aq) + 2e- + Cu(s) 2Ag(s)+ Cu2+(aq) + 2e-
gain electrons = reduction
lose electrons = oxidation

5. Redox Cu/Ag E Cu
electron flow
Ag+
Ag+

6. Redox Cu/Ag E Cu2+ ΔE = e.Ecell Ag Ag e = charge on an electron
Ecell = Voltage in a electrochemical cell
If we could separate the two reactions we could use the energy gained by the e to do work
Redox Cu/Ag E
Cu2+
ΔE = e.Ecell Ag Ag

7. The maximum amount of energy available to do work is a definition of the free energy ΔG
Redox Cu/Ag
cell voltage
Free energy for redox rxn E
Cu2+
# electrons in redox rxn
F = Faraday’s constant
= C/mol Ag Ag

8. ΔEcell: the cell potential
In a redox reaction electrons are transferred to a more stable state
Most of the free energy of the reaction is due to these electrons
Can we access this energy?
Yes, by conducting the two half reactions in separate cells
2Ag+(aq) + 2e-  2Ag(s)
Cu(s)  Cu2+(aq) + 2e-
This is called the Voltaic electrochemical cell
The cell potential can be measured in such a cell with a voltmeter

9. Where oxidation happens
Voltaic Cell Cu/Ag
Ecell= V
DGcell= -2F x V
cations to the cathode
electrons to the cathode
Anions to the anode
Where oxidation happens
Where reduction
happens
Cu(s)  Cu2+(aq) + 2e-
2Ag+(aq) + 2e-  2Ag(s)

10. Electric Current Flowing Directly Between Atoms

11. Electrochemistry and Cells
electrochemistry is the study of redox reactions that produce or require an electric current
the conversion between chemical energy and electrical energy is carried out in an electrochemical cell
spontaneous redox reactions take place in a voltaic cell (galvanic cell)
nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy

12. Where oxidation happens
Voltaic Cell as a Battery: Electric Current Flowing Indirectly Between Atoms
Electrons do work here
electrons to the cathode
Anions to the anode
cations to the cathode
Where oxidation happens
Where reduction
happens

13. Electrodes Anode (donates electrons to the cathode)
electrode where oxidation occurs
In a Galvanic cell it is the –ve terminal and in an electrolytic cell it is the +ve terminal
anions attracted to it
connected to positive end of battery in electrolytic cell
loses weight in electrolytic cell
Cathode (attracts electrons from the anode)
electrode where reduction occurs
In a Galvanic cell it is the +ve terminal and in an electrolytic cell it is the -ve terminal
cations attracted to it
connected to negative end of battery in electrolytic cell
gains weight in electrolytic cell
electrode where plating takes place in electroplating

14. Current and Voltage
the number of electrons that flow through the system per second is the current
unit = Ampere (A)
1 A of current = 1 Coulomb (C) of charge flowing by each second
1 A = x 1018 electrons/second
Electrode surface area dictates the number of electrons that can flow
the difference in potential energy between the reactants and products per Coulomb of charge is called the Voltage
unit = Volt (V)
1 V of force = 1 J of energy/Coulomb of charge
the voltage is what drives electrons through the external circuit
amount of force pushing the electrons through the wire is called the electromotive force, emf, (also units of V)
the cell potential under standard conditions is called the standard emf, E°cell
25°C, 1 atm for gases, 1 M concentration of solution
sum of the cell potentials for the half-reactions

15. Cell Potential
the difference in potential energy between the anode the cathode in a voltaic cell is called the cell potential
the cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode
the cell potential under standard conditions is called the standard emf, E°cell
25°C, 1 atm for gases, 1 M concentration of solution
sum of the cell potentials for the half-reactions

16. Cell Notation shorthand description of Voltaic cell
electrode | electrolyte || electrolyte | electrode
oxidation half-cell on left (anode), reduction half-cell on the right (cathode)
single | = phase barrier
if multiple electrolytes in same phase, a comma is used rather than |
often use an inert electrode
double line || = salt bridge

17. Fe(s) | Fe2+(aq) || MnO4-(aq), Mn2+(aq), H+(aq) | Pt(s)

18. Practice - Sketch and Label the Voltaic Cell Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s) Write the Half-Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions.
Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s)
anode
cathode

19. ox: Fe(s)  Fe2+(aq) + 2 e− Eox = +0.45 V
red: Pb2+(aq) + 2 e−  Pb(s) Ered = −0.13 V
tot: Pb2+(aq) + Fe(s)  Fe2+(aq) + Pb(s) Ecell = V

20. Measuring the Tendency to Reduce: Standard Reduction Potential
when two half-cells are connected, one side will oxidize and one side will reduce, but which will do what?
clearly the electrons will flow so that the half-reaction with the stronger tendency to reduce will reduce, making the other half-cell reaction oxidize
we cannot measure the absolute tendency of a half-reaction to reduce, we can only measure it relative to another half-reaction
To measure the tendency to reduce we measure the E (the voltage) in a voltaic cell where one of the electrodes is a standard hydrogen electrode (see right)
We assign a potential difference = 0.0 V to the following reaction
2H+(aq) +2e-  H2(g) Eored = 0.0V
E°cell = Eoox + Eored = Eoox V = Eoox
In so doing scientists can obtain standard reduction potentials as follows

22. The measured voltage can then be placed in a table like this one, where we write the voltage for the reduction reaction

23. Half-Cell Potentials ΔE°cell = E°oxidation + E°reduction
SHE reduction potential is defined to be exactly 0 V
half-reactions with a stronger tendency toward reduction than the SHE have a + value for E°red
half-reactions with a stronger tendency toward oxidation than the SHE have a - value for E°red
ΔE°cell = E°oxidation + E°reduction
E°oxidation = -E°reduction
when adding E° values for the half-cells, do not multiply the half-cell E° values (voltage is the energy per unit charge), even if you need to multiply the half-reactions to balance the equation
ΔGocell=-nFΔE°cell
Since ΔGocell < 0 to be spontaneous ΔE°cell > 0 for a redox reaction to be spontaneous if ΔE°cell < 0 the reaction will not be spontaneous

24. red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
Calculate E°cell for the reaction at 25°C Al(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l)
Separate the reaction into the oxidation and reduction half-reactions
ox: Al(s)  Al3+(aq) + 3 e−
red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
find the Eo for each half-reaction from the standard reduction table. If you need the oxidation potential of one Eoox=-Eored and sum to get Eocell
Eoox = −Eored = V
Eored = V
Eocell = (+1.66 V) + (+0.96 V) = V
reverse

25. red: Mg2+(aq) + 2 e−  Mg(s)
Predict if the following reaction is spontaneous under standard conditions Fe(s) + Mg2+(aq)  Fe2+(aq) + Mg(s)
Separate the reaction into the oxidation and reduction half-reactions
ox: Fe(s)  Fe2+(aq) + 2 e−
red: Mg2+(aq) + 2 e−  Mg(s)
look up the relative positions of the reduction half-reactions
red: Mg2+(aq) + 2 e−  Mg(s) Ered = V
ox: Fe(S)  Fe2+(aq) + 2 e− Eox =+0.45V
Ecell = Eox + Ered = 0.45V – 2.37V = -1.92V
Since Ecell < 0 the cell reaction is not spontaneous as written
reverse

26. Keeping track of electrons
Redox reactions involve the transfer of electrons from the donor to the acceptor
The donor loses electrons and is oxidized
The acceptor acquires electrons and is reduced
To know if a reaction is a redox reaction we need a way to keep track of how many valence electrons each element has
We define the oxidation state of an element to be +n (n integer) if it has n less electrons than it does as the free atom, and –p if it has p more electrons than it does in the free atom

27. Redox Reaction one or more elements change oxidation number
all single displacement (A+BX--> AX+B), and combustion,
some synthesis and decomposition
always have both oxidation and reduction
split reaction into oxidation half-reaction and a reduction half-reaction
aka electron transfer reactions
half-reactions include electrons
oxidizing agent is reactant molecule that causes oxidation
contains element reduced
reducing agent is reactant molecule that causes reduction
contains the element oxidized

28. Oxidation & Reduction oxidation is the process that occurs when
oxidation number of an element increases
element loses electrons
compound adds oxygen
compound loses hydrogen
half-reaction has electrons as products
reduction is the process that occurs when
oxidation number of an element decreases
element gains electrons
compound loses oxygen
compound gains hydrogen
half-reactions have electrons as reactants

29. Assigning Oxidation Numbers
The sum of the oxidation numbers Q of all the atoms in a compound/ion up to the charge on the compound/ion. This means ….
free elements have an oxidation state = 0
QNa = 0 and QCl = 0 in 2 Na(s) + Cl2(g)
monatomic ions have an oxidation state equal to their charge
QNa = +1 and QCl = -1 in NaCl
The sum of the oxidation numbers of all the atoms in a neutral compound is 0
the sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge on the ion
(a) Group I metals have an oxidation state of +1 in all their compounds
QNa = +1 in NaCl
(b) Group II metals have an oxidation state of +2 in all their compounds
QMg = +2 in MgCl2
F has an oxidation number QF = -1
H has an oxidation number QH = +1
O has an oxidation number QO = -2

30. Oxidation and Reduction
oxidation occurs when an atom’s oxidation state increases during a reaction
reduction occurs when an atom’s oxidation state decreases during a reaction
rule 4
rule 1
rule 5
rule 4
CH O2 → CO2 + 2 H2O –
oxidation
reduction

31. Oxidation–Reduction 2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
oxidation and reduction must occur simultaneously
if an atom loses electrons another atom must take them
the reactant that reduces an element in another reactant is called the reducing agent
the reducing agent contains the element that is oxidized
the reactant that oxidizes an element in another reactant is called the oxidizing agent
the oxidizing agent contains the element that is reduced
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent

32. Identify the Oxidizing and Reducing Agents in Each of the Following
3 H2S + 2 NO3– H+  3 S + 2 NO + 4 H2O
MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O

33. Identify the Oxidizing and Reducing Agents in Each of the Following
red ag
ox ag
3 H2S + 2 NO3– H+  3 S + 2 NO + 4 H2O
MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O
reduction
oxidation
ox ag
red ag
oxidation
reduction

34. Common Oxidizing Agents

35. Common Reducing Agents

36. Balancing Redox Reactions
assign oxidation numbers
determine element oxidized and element reduced
write ox. & red. half-reactions, including electrons
ox. electrons on right, red. electrons on left of arrow
balance half-reactions by mass
first balance elements other than H and O
add H2O where need O
add H+1 where need H
neutralize H+ with OH- in base
balance half-reactions by charge
balance charge by adjusting electrons
balance electrons between half-reactions
add half-reactions
check

37. Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution
Assign Oxidation States
Separate into half-reactions
ox: I-(aq)  I2(aq)
red: MnO4-(aq)  MnO2(s)
Assign Oxidation States
I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s)
Separate into half-reactions
ox:
red:

38. Ex 18.3 – Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution
Balance half-reactions by mass
in base, neutralize the H+ with OH-
ox: 2 I-(aq)  I2(aq)
red: 4 H+(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l)
4 H+(aq) + 4 OH-(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) + 4 OH-(aq)
4 H2O(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) + 4 OH-(aq)
MnO4-(aq) + 2 H2O(l)  MnO2(s) + 4 OH-(aq)
Balance half-reactions by mass
then H by adding H+
ox: 2 I-(aq)  I2(aq)
red: 4 H+(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l)
Balance half-reactions by mass
ox: 2 I-(aq)  I2(aq)
red: MnO4-(aq)  MnO2(s)
Balance half-reactions by mass
ox: I-(aq)  I2(aq)
red: MnO4-(aq)  MnO2(s)
Balance half-reactions by mass
then O by adding H2O
ox: 2 I-(aq)  I2(aq)
red: MnO4-(aq)  MnO2(s) + 2 H2O(l)

39. Ex 18.3 – Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution
Balance Half-reactions by charge
ox: 2 I-(aq)  I2(aq) + 2 e-
red: MnO4-(aq) + 2 H2O(l) + 3 e-  MnO2(s) + 4 OH-(aq)
Balance electrons between half-reactions
ox: { 2 I-(aq)  I2(aq) + 2 e- } x 3
red: {MnO4-(aq) + 2 H2O(l) + 3 e-  MnO2(s) + 4 OH-(aq) } x 2
ox: 6 I-(aq)  3 I2(aq) + 6 e-
red: 2 MnO4-(aq) + 4 H2O(l) + 6 e-  2 MnO2(s) + 8 OH-(aq)

40. Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution
Add the Half-reactions
ox: 6 I-(aq)  3 I2(aq) + 6 e-
red: 2 MnO4-(aq) + 4 H2O(l) + 6 e-  2 MnO2(s) + 8 OH-(aq)
tot: 6 I-(aq)+ 2 MnO4-(aq) + 4 H2O(l)  3 I2(aq)+ 2 MnO2(s) + 8 OH-(aq)
Check
Reactant
Count
Element
Product 6 I 2 Mn 12 O 8 H 8-
charge

41. Practice - Balance the Equation H2O2 + KI + H2SO4  K2SO4 + I2 + H2O

42. Practice - Balance the Equation H2O2 + KI + H2SO4  K2SO4 + I2 + H2O
ox: 2 I-1  I2 + 2e-1
red: H2O2 + 2e H+  2 H2O
tot 2 I-1 + H2O2 + 2 H+  I2 + 2 H2O
H2O2 + 2 KI + H2SO4  K2SO4 + I2 + 2 H2O

43. Practice - Balance the Equation ClO3-1 + Cl-1  Cl2 (in acid)

44. Practice - Balance the Equation ClO3-1 + Cl-1  Cl2 (in acid)
oxidation
reduction
ox: 2 Cl-1  Cl2 + 2 e-1 } x 5
red: 2 ClO e H+  Cl2 + 6 H2O} x 1
tot 10 Cl ClO H+  6 Cl2 + 6 H2O
1 ClO Cl H+1  3 Cl2 + 3 H2O

45. Standard Reduction Potential
a half-reaction with a strong tendency to occur has a large half-cell potential
when two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occur
we cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reaction
we select as a standard half-reaction, the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 V
standard hydrogen electrode, SHE

47. Half-Cell Potentials
SHE reduction potential is defined to be exactly 0 V
half-reactions with a stronger tendency toward reduction than the SHE have a + value for E°red
half-reactions with a stronger tendency toward oxidation than the SHE have a - value for E°red
E°cell = E°oxidation + E°reduction
E°oxidation = -E°reduction
when adding E° values for the half-cells, do not multiply the half-cell E° values, even if you need to multiply the half-reactions to balance the equation

50. red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
Calculate E°cell for the reaction at 25°C Al(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l)
Separate the reaction into the oxidation and reduction half-reactions
ox: Al(s)  Al3+(aq) + 3 e−
red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
find the Eo for each half-reaction and sum to get Eocell
Eoox = −Eored = v
Eored = v
Eocell = (+1.66 v) + (+0.96 v) = v

51. the reaction is spontaneous in the reverse direction
tot: Mg(s) + Fe2+(aq)  Mg2+(aq) + Fe(s)
ox: Mg(s)  Mg2+(aq) + 2 e−
red: Fe2+(aq) + 2 e−  Fe(s)
sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode

52. Predicting Whether a Metal Will Dissolve in an Acid
acids dissolve in metals if the reduction of the metal ion is easier than the reduction of H+(aq)
metals whose ion reduction reaction lies below H+ reduction on the table will dissolve in acid

53. Electrochemical Cell Summary
The cell consists of two half cells, one where oxidation occurs, and one where reduction occurs
A device for either harnessing the electrical power of a redox reaction (battery), or a device for using electricity to induce non-spontaneous redox reactions (electrolytic cell)
The anode and cathode are connect by a wire through which electrons can flow, and the electrolytes in each half cell are connected by a salt bridge through which cations and anions can flow
The electrode which loses electrons we call the anode, and the electrode that gains electrons we call the cathode
Each half cell consists of an electrode and an electrolyte e-
anode
Zn (s)--> Zn2+ (aq)+ 2e-
salt bridge
cathode
Cu2+(aq)+ 2e- --> Cu(s)

54. Electrochemical Cell Summary
The cell potential can be calculated knowing the standard reduction potentials. These can be used to find Eored for the reaction at the cathode, and Eoox (= - Eored). Then Eocell = Eoox+ Eored
This manifests as a potential difference Ecell, across the electrodes. Where -qEcell is the change in potential energy when an amount of negative charge (-q) passes from the anode to the cathode
The cell potential is related to the free energy of the reaction according to the relation Gcell = -nFEcell
The differing stability of reactants, (Zn(s), Cu2+(aq)), and products (Zn2+, and Cu(s)), creates a potential energy gradient through which the charges migrate (from high energy to low).
Zn(s) --> Zn2+(aq) + 2e- Eox= 0.76V
Zn2+(aq) + 2e- --> Zn(s) V
Cu2+(aq) + 2e- --> Cu(s) Ered=0.34V
Ecell = 0.76V+0.34V = 1.1V
Ecell=1.1 V e-
anode
Zn (s)--> Zn2+ (aq)+ 2e-
salt bridge
cathode
Cu2+(aq)+ 2e- --> Cu(s)

55. The Relationship between ΔGo and Eocell
Remember ΔGo is the maximum work the system can do
Eocell is the cell potential energy (standard emf) per unit of charge
Since the potential energy is the maximum amount of work that can be done on the surrounding we can write
The total charge q = nF, n = number moles of electrons in the balanced equation and F is Faraday’s constant. Then we can write
Δ 𝐺 𝑐𝑒𝑙𝑙 =Δ 𝐺 𝑐𝑒𝑙𝑙 𝑜 +𝑅𝑇𝑙𝑛𝑄
−𝑛𝐹Δ 𝐸 𝑐𝑒𝑙𝑙 =−𝑛𝐹Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 +𝑅𝑇𝑙𝑛𝑄
Δ 𝐸 𝑐𝑒𝑙𝑙 =Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 − 𝑅𝑇 𝑛𝐹 𝑙𝑛𝑄= Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 − 𝑉 𝑛 𝑙𝑜𝑔𝑄
Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 = 𝑉 𝑛 𝑙𝑜𝑔𝐾

56. E°cell, ΔG° and K for a spontaneous reaction
one proceeds in the forward direction with the chemicals in their standard states
ΔG° < 1 (negative)
E° > 1 (positive)
K > 1
ΔG° = −RTlnK = −nFE°cell
n is the number of electrons
F = Faraday’s Constant = 96,485 C/mol e−

57. I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq) ΔG°, (J)
Example Calculate ΔG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Given:
Find:
I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
ΔG°, (J)
Concept Plan:
Relationships:
E°ox, E°red
E°cell
ΔG°
Solve:
ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 V
red: I2(l) + 2 e− → 2 I−(aq) E° = V
tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 V
Answer:
since ΔG° is +, the reaction is not spontaneous in the forward direction under standard conditions

58. Nonstandard Conditions - the Nernst Equation
ΔG = ΔG° + RT ln Q
E = E° - (0.0592/n) log Q at 25°C
when Q = K, E = 0
use to calculate E when concentrations not 1 M

59. Example 18.7- Calculate K at 25°C for the reaction Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq)
Given:
Find:
Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq) K
Concept Plan:
Relationships:
E°ox, E°red
E°cell K
Solve:
ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 V
red: 2 H+(aq) + 2 e− → H2(aq) E° = V
tot: Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) E° = −0.34 V
Answer:
since K < 1, the position of equilibrium lies far to the left under standard conditions

60. E° at Nonstandard Conditions

61. Calculate Ecell at 25°C for the reaction 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + 3Cu2+(aq) + 4 H2O(l) Given [Cu2+] = M, [MnO4−] = 2.0 M, [H+] = 1.0 M
3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + 3Cu2+(aq) + 4 H2O(l)
[Cu2+] = M, [MnO4−] = 2.0 M, [H+] = 1.0 M
Ecell
Given:
Find:
Concept Plan:
Relationships:
E°ox, E°red
E°cell
Ecell
Solve:
ox: Cu(s) → Cu2+(aq) + 2 e− }x3 E° = −0.34 V
red: MnO4−(aq) + 4 H+(aq) + 3 e− → MnO2(s) + 2 H2O(l) }x2 E° = V
tot: 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)) E° = V
Check:
units are correct, Ecell > E°cell as expected because [MnO4−] > 1 M and [Cu2+] < 1 M

62. Concentration Cells
it is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different
the difference in energy is due to the entropic difference in the solutions
the more concentrated solution has lower entropy than the less concentrated
electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution till the concentrations are equal

63. Concentration Cell
when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow
when the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)
Cu(s)| Cu2+(aq) (0.010 M) || Cu2+(aq) (2.0 M)| Cu(s)

64. LeClanche’ Acidic Dry Cell
electrolyte in paste form
ZnCl2 + NH4Cl
or MgBr2
anode = Zn (or Mg)
Zn(s)  Zn2+(aq) + 2 e-
cathode = graphite rod
MnO2 is reduced
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e-
 2 NH4OH(aq) + 2 Mn(O)OH(s)
cell voltage = 1.5 V
expensive, nonrechargeable, heavy, easily corroded

65. Zn(s) + 2OH- Zn(OH)2(s) + 2 e-
Alkaline Dry Cell
same basic cell as acidic dry cell, except electrolyte is alkaline KOH paste
anode = Zn (or Mg)
Zn(s) + 2OH- Zn(OH)2(s) + 2 e-
cathode = graphite
MnO2 is reduced
2 MnO2(s) + 2 H2O(l) + 2 e-
2 Mn(O)OH(s) + 2 OH-(aq)
Overall reaction
Zn(s)+2MnO2(s)+2H2O(l) Zn(OH)2(s) + 2MnO(OH)(s)
cell voltage = 1.54 V
longer shelf life than acidic dry cells and rechargeable, little corrosion of zinc

66. Voltaic Cells as Batteries
Use chemical reactions to provide a source of energized electrons that can power electronics
Batteries in everything, phones, laptops, tablets, watches, cars
Actively seeking batteries that are light, powerful, and retain their effectiveness after many recharge cycles
Molten metal batteries developed for storing electricity in power stations (infinitely rechargable)
Huge area of scientific and commercial interest
Batteries are being made ever smaller for energy efficient processor applications
Nanotechnology, even biotechnology using viruses!
A benign bacteria virus serves as a template in water to grow manganese oxide nanowires that, when combined with palladium, increases the energy potential of a lithium battery

67. Voltaic Cells as Batteries Lead Storage Battery
Anode: Pb
Pb  Pb2+ + 2e-
Pb(s) + SO42-(aq)  PbSO4(s) + 2 e-
Cathode: Pb coated with PbO2
Pb4+ +2e- Pb2+
PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e- PbSO4(s) + 2 H2O(l)
Electrolyte: 30% H2SO4
cell voltage = 2.09 V
6 in series makes 12V
rechargeable, heavy

68. NiCad Battery electrolyte is concentrated KOH solution anode = Cd
Cd(s) + 2 OH-1(aq) ® Cd(OH)2(s) + 2 e-1 E0 = 0.81 V
cathode = Ni coated with NiO2
NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e-1 ® Ni(OH)2(s) + 2OH-1 E0 = 0.49 V
cell voltage = 1.30 V
rechargeable, long life, light – however recharging incorrectly can lead to battery breakdown

69. Ni-MH Battery Tro, Chemistry: A Molecular Approach
electrolyte is concentrated KOH solution
anode = metal alloy with dissolved hydrogen
oxidation of H from H0 to H+1
M∙H(s) + OH-1(aq) ® M(s) + H2O(l) + e-1 E° = 0.89 V
cathode = Ni coated with NiO2
NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e-1 ® Ni(OH)2(s) + 2OH-1 E0 = 0.49 V
cell voltage = 1.30 V
rechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad
Tro, Chemistry: A Molecular Approach

70. Voltaic Cells as Batteries Lithium Ion Battery
anode = graphite impregnated with Li ions
cathode = Li - transition metal oxide
Electrolyte is an organic solvent like dimethylcarbonate, with a Li salt like LiPF6
rechargeable, long life, very light, more environmentally friendly, greater energy density
High surface area electrodes
Tiny ions that move fast between electrodes

71. Voltaic Cells as Batteries
Why do we use Li-ion batteries in our computers and phones and not say Pb Acid Batteries?
We want to make batteries as portable (ie small and light) as possible

72. H2 Fuel Cell
like batteries in which reactants are constantly being added
so it never runs down!
Anode and Cathode both Pt coated metal (may change with cheaper technology)
Electrolyte is OH– solution
Anode Reaction:
2 H2 + 4 OH– → 4 H2O(l) + 4 e-
Cathode Reaction:
O2 + 4 H2O + 4 e- → 4 OH–
Toyota Mirai available in CA $499/month 312 miles per tank

73. Electrolysis
electrolysis is the process of using electricity to make non-spontaneous redox reactions happen
electrolysis is done in an electrolytic cell
electrolytic cells can be used to separate elements from their compounds
generate H2 from water for fuel cells
recover metals from their ores
Make halogen gasses from halide salts

75. Electrolytic Cell
uses electrical energy to overcome the energy barrier and cause a non-spontaneous reaction
must be DC source
the + terminal of the battery connects to the anode
the - terminal of the battery connects to the cathode
cations attracted to the cathode, anions to the anode
Oxidation at the anode, reduction at the cathode
some electrolysis reactions require more voltage than Etot, called the overvoltage

76. Applications of electrolysis: electroplating
In electroplating, the work piece is the cathode.
Cations are reduced at cathode and plate to the surface of the work piece.
The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution
Gold plated terminals in electronics (Au is the most conductive of all metals)

77. Electrochemical Cells: Voltaic vs Electrolytic
in all electrochemical cells, oxidation occurs at the anode, reduction occurs at the cathode
in voltaic cells,
anode is the source of electrons and has a (−) charge
cathode draws electrons and has a (+) charge
in electrolytic cells
electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery
electrons are forced toward the cathode, so it must have a source of electrons, the − terminal of the battery

78. Applications of Electrolysis: Making H2
With sufficient voltage across 2 Pt electrodes one can electrolyze H2O to get O2 and H2
This may become an important process if we move toward fuel cell driven cars
For example use wind power to electrolyze water to make H2 for cars

79. Electrolysis of Pure Compounds
must be in molten (liquid) state
electrodes normally graphite
cations are reduced at the cathode to metal element
anions oxidized at anode to nonmetal element

80. Electrolysis of NaCl(l)

81. Mixtures of Ions
when more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode
least negative or most positive E°red
when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode
least negative or most positive E°ox

82. Electrolysis of Aqueous Solutions
Complicated by more than one possible oxidation and reduction
possible cathode reactions
reduction of cation to metal
reduction of water to H2
2 H2O + 2 e-1  H2 + 2 OH-1 E° = stand. cond.
E° = pH 7
possible anode reactions
oxidation of anion to element
oxidation of H2O to O2
2 H2O  O2 + 4e-1 + 4H+1 E° = stand. cond.
E° = pH 7
oxidation of electrode
particularly Cu
graphite doesn’t oxidize
half-reactions that lead to least negative Etot will occur
unless overvoltage changes the conditions

83. Electrolysis of NaI(aq): Inert Electrodes
Write down ions/molecules/metals present and don’t forget water
Here Na+, I-, H2O are what is present (the electrodes here are inert)
Write down the reduction reactions involving these species
I2(aq)+2e-  2I-(aq) Ered = 0.54V
O2(g) + 4e-1 + 4H+1(aq) 2 H2O(l) Ered = V
2 H2O + 2 e-1  H2 + 2 OH-1 Ered = 0.41V
Na+(aq) + e-  Na(s) Ered = -2.71V
Rearrange the equations so the reactants are on the left (reverse the sign of E if you flip the equation – these will be the oxidation reactions
possible oxidations (anode –ve terminal)
2 I-1  I2 + 2 e-1 E° = −0.54 v
2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v
possible oxidations
2 I-1  I2 + 2 e-1 E° = −0.54 v
2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v
possible reductions (cathode +ve terminal)
Na+1 + 1e-1  Na0 E° = −2.71 v
2 H2O + 2 e-1  H2(g) + 2 OH-1E° = -0.41V
possible reductions
Na+1 + 1e-1  Na0 E° = −2.71 v
2 H2O + 2 e-1  H2 + 2 OH-1 E° = −0.41 v
brown liquid
overall reaction
2 I−(aq) + 2 H2O(l)  I2(aq) + H2(g) + 2 OH-1(aq)
bubbles
Ecell = -0.54V – 0.41V = -0.95V
The battery needs to at least be 1V to make this happen

84. Electrolysis of NaI(aq) with Inert Electrodes
possible oxidations
2 I-1  I2 + 2 e-1 E° = −0.54 v
2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v
possible oxidations
2 I-1  I2 + 2 e-1 E° = −0.54 v
2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v
possible reductions
Na+1 + 1e-1  Na0 E° = −2.71 v
2 H2O + 2 e-1  H2 + 2 OH-1 E° = −0.41 v
possible reductions
Na+1 + 1e-1  Na0 E° = −2.71 v
2 H2O + 2 e-1  H2 + 2 OH-1 E° = −0.41 v
overall reaction
2 I−(aq) + 2 H2O(l)  I2(aq) + H2(g) + 2 OH-1(aq)

85. Electrolysis in aqueous solutions: Overpotentials
There are (potentially) competing processes in the electrolysis of an aqueous solution consider now NaCl(aq) with C electrodes
Cathode
Anode
We choose those reactions that are thermodynamically favored
But…chlorine is evolved at the anode!

86. Overpotentials
Thermodynamics true at equilibrium but when the process has a current flowing kinetics plays a role
Overpotential represents the additional voltage that must be applied to drive the process
In the NaCl(aq) solution the overpotential for evolution of oxygen is greater than that for chlorine, and so chlorine is evolved preferentially
The limiting process in electrolysis is usually diffusion of the ions in the electrolyte
Overpotential will depend on the electrolyte and electrode
Driving the cell at the least current will give rise to the smallest overpotential

87. Faraday’s Law
the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current (A), and the length of time (s) the cell runs
charge = current x time
Q = I x t

88. Calculate the mass of Au that can be plated in 25 min using 5
Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction Au3+(aq) + 3 e− → Au(s)
3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min
mass Au, g
Given:
Find:
t(s), amp
charge (C)
mol e−
mol Au
g Au
Concept Plan:
Relationships:
Solve:
Check:
units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−

89. Corrosion
corrosion is the spontaneous oxidation of a metal by chemicals in the environment
since many materials we use are active metals (metals that like to make ions), corrosion can be a very big problem

90. Rusting rust is hydrated iron(III) oxide moisture must be present
water is a reactant
required for flow between cathode and anode
electrolytes (sea-water) promote rusting
enhances current flow
acids (acid rain) promote rusting
lower pH = lower E°red

91. O2(g) + 2H2O(l) + 4e-  4OH-(aq)
Rusting
Anode (inside the droplet):
Fe(s)  Fe2+(aq) + 2e-
Cathode (at the surface of the droplet:
O2(g) + 2H2O(l) + 4e-  4OH-(aq)
Within the droplet
Fe2+(aq) + 2OH-(aq)  Fe(OH)2(s)
Rust is then quickly produced by the oxidation of the precipitate at the edge of the droplet.
4Fe(OH)2(s) + O2(g)  2Fe2O3 •H2O(s) + 2H2O(l)

92. Preventing Corrosion Zn  Zn2+ + 2e- 0.76V Fe  Fe2+ + 2e- 0.45V
one way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment
paint
some metals, like Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corroding
another method to protect one metal is to attach it to a more reactive metal that is cheap
sacrificial electrode
galvanized steel (Zinc coating)
Zn  Zn2+ + 2e V
Fe  Fe2+ + 2e V

93. Sacrificial Anode Zn  Zn2+ + 2e- 0.76V Fe  Fe2+ + 2e- 0.45V
Galvanic anode on the hull of a ship

94. The Final Long Questions on the Following: Kinetics
Method of Initial Rates to find rate law
Arrhenius plot to find Activation energy Ea and the collision factor A
Acid-Base Titration of a Weak Acid
Calculate the initial pH of the acid given Ka
Calculate the pH in the buffer region using Henderson-Hasselbalch
Calculate the pH at the end point using Kb
Galvanic and Electrolytic Cell
An extra credit question on kinetics mechanisms (worth 15% of the final)
Multiple Choice Redo for Electrochemistry
A short (optional multiple choice on electrochemistry for those who want a second chance)
The Final